Calculations in Chemistry - CH. 13-14

7/26/2019 Calculations in Chemistry - CH. 13-14

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2011 ChemReview.net v. k6 Page ii

Calculations In Chemistry

* * * * *

Module 13 Ionic Equations and Precipitates

Module 14 Acid-Base Neutralization

* * * * *Module 13 Ionic Equations and Precipitates .........................................................326

Lesson 13A: Predicting Solubility for Ionic Compounds ...................................................326Lesson 13B: Total and Net Ionic Equations..........................................................................330Lesson 13C: Predicting Precipitation.....................................................................................334Lesson 13D: Precipitate and Gravimetric Calculations.......................................................341

Module 14 Acid-Base Neutralization.......................................................................349Lesson 14A: Ions in Acid-Base Neutralization.....................................................................349Lesson 14B: Balancing Hydroxide Neutralization..............................................................353Lesson 14C: Neutralization and Titration Calculations......................................................361Lesson 14D: Neutralization Calculations for Ratio Units...................................................365Lesson 14E: Neutralization Calculations in Parts ...............................................................369Lesson 14F: Carbonate Neutralization .................................................................................377

For additional modules, visit www.ChemReview.Net
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2011 ChemReview.net v. k6 Page 326


Pretests: On Module 13 lessons, if you think you know the topic, try the last problem in thelesson. If you can do that problem, you may skip the lesson.

Lesson 13A: Predicting Solubility for Ionic Compounds

Prerequisites: If you have any problems translating between the names, solid formulas,and separated formulas for ionic compounds during Module 13, review your ion name andformula flashcards from Lessons 7B and 7C.

* * * * *

Solubility Terminology

All ioniccompounds dissolve to someextent in water. Some dissolve only slightly, somedissolve to a substantial extent, and some ionic compounds have borderline solubility at

room temperature. In addition, the solubility of ionic compounds is temperaturedependent: some dissolve to a greater extent in warmer solvents and some dissolve less.

Those messy realities aside, mostion combinations can generally be classified as solubleorinsolublein water. A commonly accepted definition is

if 0.10 moles or more of a solid dissolve per liter of solution at room temperature,the solid is termed soluble;

if the solid dissolves less than 0.10 moles per liter, it is considered either slightlysolubleor insoluble.

Most introductory chemistry courses ask that you memorize a set of solubility rules thatwill allow you to predict the solubility of many ion combinations. Although there are somepatterns to solubility based on the Periodic Table, there are limited simple rules. Thesolubility scheme on the following page will cover most ions customarily assigned.

However,

if your course requires that you memorize a particular chart, memorize itinstead ofthis one.

If you are allowed to consulta particular solubility chart on quizzes and tests, use itin place of this one.

If the following chart conflicts in its predictions from a chart you are assigned inyour class, use the one that you are assigned.

A Solubility Scheme

The scheme below is hierarchical: higher rules takeprecedenceover those beneath them. Topredict the solubility of ion combinations, apply the rules in order from the top.

If only one ion of the two in a compound is in this table, you may presume that thecompound will follow the rule for the ion that is shown. (This will not always be accurate,but is a best guess. With solubility, there are exceptions.)


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Commit to long-term memory:

Positive Ions NegativeIons

1. (alkali metals)+, NH4+ NO3

, CH3COO, ClO3; ClO4

(nitrate, acetate, chlorate, perchlorate)

Soluble

2. Pb2+, Hg22+, Ag+ CO3

2, PO43, S2, CrO4

2 Insoluble

3. Cl, Br, I (except Cu+) Soluble

4. Ba2+(except OH) OH(except Ba2+, Sr2+, Ca2+) Insoluble

5. SO42(except Sr2+, Ca2+) Soluble

Exceptions include: Column 2 sulfides and aluminum sulfide decompose in water.CH3COOAg, Ag2SO4and Hg2SO4are moderately soluble.

Memorization Tips1. Focus especially on the rules near the top. In this scheme, the higher the rule, the more

often it will be used.

2. Notice the patterns in the last column and in the empty boxes.

3. Most people best recall what they hear repeatedly (especially if it rhymes). You maywant to recite the ion names in a series, as well as write them.

Using the Scheme to Make Predictions

When using any solubility scheme, if you are unsure of which ionsare in a compound, you

should write out the separatedformula that showsthe ion charges. Atoms that have two

possible ion charges, such as Cu+and Cu2+, can have differing solubilities.

Lets try a few predictions.

Q. For the questions below, use the solubility scheme of your choice. Write youranswer and your reason.

1. Is Ba(NO3)2soluble?

2. Is PbCl2soluble or insoluble?

3. Is Hg(NO3)2 soluble or insoluble?

* * * * * (See How To Use These Lessons, Point 1,on page 1).

Answers

1. Yes. Ba(NO3)2Ba2++ 2 NO3

, Ba2+ion is insoluble by Rule 4, but all nitrates are

soluble by Rule 1, and the higher rule has precedence.

2. PbCl2Pb2++ 2 Cl. Pb2+ion by Rule 2 is insoluble. Chloride ion in Rule 3 is

soluble, but rule 2 takes precedence. PbCl2is insoluble.


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3. Hg(NO3)2Hg2++ 2 NO3

. The above solubility scheme makes no specific

predictions for Hg2+, the mercuric[mercury (II)] ion, but based on Rule 1 that allnitrates are soluble, predict soluble. If only one ion in a pair is in the table, base yourprediction on the rule for that ion.

Practice

1. If you are required to memorize solubility rules in your course, do so beforeyou do theproblems below, then write your solubility scheme from memory in your notebook.

2. Label each ion combination as soluble or insoluble, and state a reason for yourprediction. Check your answers after each one or two parts.

a. K+ + Br

b. Sr2+ + Cl

c. Ca2+ + CO32

d. Ag+ + CrO42

3. Write the separated ions for these combinations, then label the ion combination assoluble or insoluble, and state a reason for your prediction. (If you are using a differentscheme, your reason may differ, but your answer will probablybe the same.)

Do every other part, and check your answers frequently. Do more if you need morepractice. If you have trouble writing the separated ions, redo Lessons 7B and 7C.

a. Lead (II) Bromide

Pb2+

+ 2 Br

; Insoluble-- Rule 2 for Pb2+

(example)

b. Barium Carbonate

c. Sodium Hydroxide

d. SrBr2

e. Silver nitrate

f. Ammonium hydroxide

g. Fe3(PO4)2

h. Pb(CH3COO)2

i. NiCl2


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j. BaSO4

k. RbBr

l. Fe2S3

4. Which ions in combination with Pb2+are soluble compounds?

5. Are all nitrates soluble? Are all phosphates insoluble?

ANSWERS

These answers are based on the solubility scheme above. You will have different rules if you use a differentscheme, but you should have the same answer in mostcases, and a similar reason.

2. a. K+ + Br Soluble. Rule 1, first column metal ions are always soluble

b. Sr2+ + Cl Soluble. Rule 3, chloride

c. Ca2+ + CO32 Insoluble. Rule 2, carbonates (if only one ion is in table, use it)

d. Ag+ + CrO42 Insoluble. Rule 2, chromate ion and/or silver ion.

3. b. Barium Carbonate Ba2+ + CO32 Insolubleby Rule 2 for carbonates

c. Sodium Hydroxide Na+ + OH Solubleby Rule 1 for alkali-metal ions

d. SrBr2 Sr2+ + 2 Br Solubleby Rule 3 for bromides

e. Silver nitrate Ag+ + NO3

Solubleby Rule 1 for nitrates

f. Ammonium hydroxide NH4+ + OH Solubleby Rule 1 for ammonium ion

g. Fe3(PO4)2 3 Fe2+ + 2 PO4

3 Insolubleby Rule 2 for phosphate ions

h. Pb(CH3COO)2 Pb2+ + 2 CH3COO

Solubleby Rule 1 for acetates

i. NiCl2Ni2+ + 2 Cl Solubleby Rule 3 for chlorides

j. BaSO4 Ba2+ + SO4

2 Insolubleby Rule 4 for barium ions

k. RbBr Rb+ + Br Solubleby Rule 1 for alkali-metal ions

l. Fe2S3 2 Fe3+ + 3 S2 Insolubleby Rule 2 for sulfide ions

4. Which ions in combination with Pb2+would make soluble compounds? Nitrate, acetate, chlorate, andperchlorate.

5. Are all nitrates soluble? YES. That rule is used frequently.

Are all phosphates insoluble? NO. Alkali-metal and ammonium-ion phosphates are soluble. Rule 1 hasprecedence.

* * * * *


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Lesson 13B: Total and Net Ionic Equations

Prerequisites: Lessons 7C, 10B, and 13A.

* * * * *

Mixing IonsWhen solutions of different ions are mixed, chemical reactions can occur.

One type of reaction is precipitation. When solutions of different soluble ionic compoundsare mixed, the ions can trade partners: new combinations of positive and negative ionsare possible. If a new combination is possible that is insoluble in water, it will precipitate:it will form a cloud of solid particles in the solution.

For example, when a 0.1 M solution of sodium chloride (table salt) is mixed with a 0.1M solution of silver nitrate, a bright white cloud immediately forms. Solid particlesfrom the cloud slowly settle to the bottom of the solution. Over time, in light, thesurface of the solid slowly turns grey.

The precipitate formed is silver chloride (AgCl), a solid that reacts with light.

The total ionic equationfor this reaction is

Na+(aq)+ Cl

(aq) + Ag+

(aq)+ NO3

(aq)AgCl(s) + Na+

(aq) + NO3

(aq)

The (aq) means that the stateof a particle is aqueous, which means dissolved in water. Insolution reactions in these lessons, if no state is shown after an ion or particle formula, youshould assume the state is (aq). The (s) abbreviates the solid state, which is the state of anyprecipitate.

Note that the sodium and nitrate ions did not change in the reaction. Ions that are presentduring a reaction but do not change are termed spectator ions. In totalionic equations, the

spectator ions are includedon each side, and they will be the sameon each side. Includingthe spectators helps us to see all of the reactant and product compounds.

However, as in math equations, terms that are the same on each side of an equation can becancelled. Canceling the spectators gives the netionic equation, which shows only theparticles that changetheir state in the reaction.

For the reaction above, cancel the spectators and write the netionic equation.

* * * * *

Ag+(aq)+ Cl

(aq) AgCl(s)

Practice A: Put a by and do Part a on each of these. Do the part bs during your nextpractice session.

1. For these total ionic equations, circle the precipitate, cross out the spectators, and writethe net ionic equation.

a. Pb2+(aq) + 2 NO3(aq) + Cu

2+(aq) + 2 Cl

(aq) Cu2+

(aq) + 2 NO3

(aq)+ PbCl2(s)


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b. 6 Na+(aq)+ 2 PO43

(aq)+ 3 Mg2+

(aq)+ 3 SO42

(aq)Mg3(PO4)2(s)+ 6 Na+

(aq)+ 3 SO42

(aq)

2. Balance these equations for precipitation reactions. (For rules, see Lesson 10B).

a. Fe(NO3)3(aq) + NaOH(aq) Fe(OH)3(s)+ NaNO3(aq)

b. Ca(NO3)2(aq) + K3PO4(aq)

KNO3(aq) + Ca3(PO4)2(s)3. Molecular formulas with an aqueous state, such as in Problem 2 above, are often

written to represent ionic solids dissolved in a solution. In reality, however, if an ionicsolid dissolves in water, its ions will separate in the solution. Totaland netionicequations show separated-ion formulas -- unless the ions react and precipitate.

Re-write the balanced equations for problems 2a and 2b using separated-ion formulas.This will be the totalionic equation for each reaction.

4. Write the netionic equations for the reactions in 2a and 2b.

Balancing Total Ionic EquationsFor precipitation reactions, you will often be asked to balance a total ionic equation. Tobalance properly, you must first balance each of the reactant and product formulas forcharge, then balance again to account for ratiosofreaction.

In the following equation, the brackets show the original ion combinations on the left of thearrow and the new possible combinations on the right.

[ Ca2++ NO3] + [ K++ PO4

3] Ca3(PO4)2(s) + [ K+ + NO3

]

To balance, first add coefficients insidethe brackets so that the charges are balanced for eachsubstance formula, and then check your answer below. (For reactions conducted in

aqueous solutions, if no state is shown, assume (aq).)* * * * *

[ 1Ca2++ 2NO3] + [ 3K++ 1PO4

3] Ca3(PO4)2(s) + [ 1K+ + 1 NO3

]

Now add coefficients in front of the brackets so that allof the atoms balance.

* * * * *

To form the precipitate, 3 calcium atoms and 2 phosphate ions are needed, so:

3[1Ca2+ + 2NO3] + 2 [3K++ 1 PO4

3]1 Ca3(PO4)2(s)+ 6[1K+ + 1NO3

]

Write the total ionic equationby removing the brackets. Check that the total ionic equation is

balanced for atoms. In totalionic equations, check also that the charges on each side totalto zero as well.

* * * * *

3 Ca2++ 6 NO3+ 6 K+ + 2 PO4

3 1 Ca3(PO4)2(s) + 6 K

++ 6NO3

Check: 3 Ca, 6 N, 6K, 2 P, 26 O on each side. Zero net charge on each side. Balanced.


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Practice B: Put a by and do Problem 1. Do Problem 2 during your next practicesession. Re-write these equations as balanced totalionic equations.

1. [ K+ + SO42] + [ Sr2+ + NO3

] SrSO4(s) + [ K+ + NO3

]

2. [ Fe2+ + Br ] + [ Na+ + PO43] [ Na+ + Br ] + Fe3(PO4)2(s)

Balancing Equations That Omit Some Spectator Ions

Until this point, most of the equations we have encountered have contained either

formulas for neutral compounds, or

separated ions, with a total charge of zeroon each side of the equation.

In reactions that involve ions, equations may be written that omitall of the spectator ions, asin net ionic equations, or omit someof the spectator ions.

For example, the reaction of silver nitrate and magnesium chloride solutions can berepresented with ionic solid and molecular formulas as

2 AgNO3(aq) + MgCl2(aq) 2 AgCl(s) + Mg(NO3)2(aq) (1)

Separated-ion formulas better represent soluble substances dissolved in water. Thetotal ionic equation for the above reaction is

2 Ag+ + 2 NO3 + Mg2+ + 2 Cl 2 AgCl(s) + Mg

2+ + 2 NO3

(2)

However, it is also common to write a mixture of the above two forms andleave outsomeof the spectators. The above reaction may be written as

2 AgNO3 + 2 Cl

2 AgCl(s) + 2 NO3

(3)In this equation, the spectator magnesium ions have been left out. The reaction can bewritten in this format to emphasize that silver nitrate solution will form a precipitatewhen mixed with anysolution that contains chloride ions.

Note that thispartialionic equation (3), unlike the total ionic equation (2), does not have azero charge on each side. However, it does have a balancedcharge: the total charge on eachside is negative two. Thats an important rule:

Equations are considered balanced if they have the same number and kind of atoms andthe same netchargeon each side.

In balancing equations, the equation can be considered balanced even if some of the ionsthat must be present to balance charge have been left out. In thepartialionic equations thatare frequently written, the atom counts and total charge must simply be the sameon eachside.


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Leaving Out the Spectators

Leaving out the spectators, when writing net or partial ionic equations, results in anequation that is quicker to write than the total ionic equation. Without the spectator ions,an equation focuses on the most important particles: those that change in the reaction.

In a similar fashion, in a laboratory a container may be labeled Ag+ or OH

. However,in all stable matter containing charged particles, the charges must balanced. When the labelon a container shows a singleion, spectator ions mustalso be present in the container so thatthe net charge of the container substances is zero. Leaving out the spectators oncontainer labels and in equations is a simplification: a shortcut that is employed to focusattention on the ions that may react in type of reaction being studied.

Practice C: Balance these partial ionic equations. Check answers as you go. Do everyother letter, and the rest in your next practice session.

1. Co(NO3)2 + OH Co(OH)2(s) + NO3

2. AgNO3

+ CrO42 Ag2CrO4(s) + NO3

3. CO32 + HCl H2CO3 + Cl

4. HCO3 + OH H2O + CO3

2

5. Al3++ NO + H2O Al + NO3

+ H+

ANSWERS

Practice A: Assume (aq) if no state is shown.

1a. Pb2+(aq) + 2 NO3(aq) + Cu

2+(aq) + 2 Cl

(aq) Cu2+

(aq) + 2 NO3

(aq) + PbCl2(s)

Net ionic equation: Pb2+(aq) + 2 Cl

(aq) PbCl2(s)

1b.. 6 Na++ 2 PO43+ 3 Mg2+ + 3 SO4

2 Mg3(PO4)2(s) + 6 Na+ + 3 SO4

2

Net ionic equation: 3 Mg2+(aq) + 2 PO43

(aq) Mg3(PO4)2(s)

2a. 1 Fe(NO3)3(aq) + 3 NaOH(aq) 1 Fe(OH)3(s)+ 3 NaNO3(aq)

2b. 3 Ca(NO3)2 + 2 K3PO4 6 KNO3 + 1 Ca3(PO4)2(s)

3a. 1 Fe3++ 3 NO3+ 3 Na+ + 3 OH1 Fe(OH)3(s)+ 3 Na

+ + 3 NO3

3b. 3 Ca2++ 6 NO3+ 6 K+ + 2 PO4

3 6 K

+ + 6NO3+ 1 Ca3(PO4)2(s)

4a. Net ionic equation: 1 Fe3+(aq) + 3 OH

(aq)1 Fe(OH)3(s)

4b. Net ionic equation: 3 Ca2+(aq) + 2 PO43

(aq) 1 Ca3(PO4)2(s)


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b. When the two solutions above are mixed, two newcombinations of positive andnegative ions are possible. Each positive ion can attract either its original negativepartnerorthe new negative ion with which it has been mixed. Fill in the blanksbelow to show the newpossible ion combinations, then add coefficients to balanceeach new combination for charge.

1 Ca2+ + ____ _______

________ + ____ NO3

In solution reactions, if no state is shown, assume (aq).

* * * * *

b. 1 Ca2+ + 1 CO3

2

1 Na+ + 1 NO3

* * * * *c. After the arrows, write solidformulas for these new combinations.

d. Based on ion solubility rules, label each of the solids as soluble or insoluble.

e. Apply the rule: When two solutions of soluble ions are mixed,

if a new combination is possible that is insoluble, it willprecipitate.

If the new combination is soluble, its ions remain separated.

Will there be a precipitate in this case? If so, what is its solid formula?

Write answers for the above three steps, and then check your answers below.

* * * * *

c. 1Ca2++ 1 CO32CaCO3 Insoluble by Rule 2. Will precipitate.

1 Na++ 1 NO3

NaNO3 Soluble by Rule 1. Ions stay separated.

e. The precipitate is solid CaCO3.

* * * * *f. To write the totalionic equation, write the 4 original reactant ions. Then, for

products, write the formula for the solid precipitate, plus the separated ions for thenew combination that did not precipitate.

Fill in the blanks to give the totalionic equation for the reaction above.

1 Ca2++___ ________+___ ______ + 1CO32

__________(s) +___ ______ + ___ ______

* * * * *

f. Ca2++ 2 NO3+ 2 Na++ CO3

2 CaCO3(s) +2 Na

++ 2 NO3

g. In your total ionic equation, cross out the spectator ions that are the same on eachside: these ions did not react. Rewrite the equation without the spectators. This isthe netionic equation.

* * * * *


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g. Netionic equation: Ca2+(aq) + CO32

(aq) CaCO3(s)

To write netionic equations, omit spectator ions. Write a balanced equation showingonly the particles that changed their state in the reaction.

* * * * *

Predicting Precipitation The Chart MethodThe chart method below can help in predicting whether a precipitate will form. Copy thefollowing chart into your notebook.

Solid Dissolved Ions Mixed Possible PossibleFormulas before mixing new ion combinations Precipitates

1.

2.

Lets use an example to see how the chart is used.

Q2. When solutions of Pb(NO3)2and KI are mixed, a yellow precipitate forms. Using

the steps on the previous page, fill in the chart above and write the formula for the

precipitate.

If you need a hint, read a part of the answer below and then again.

* * * * *

Answer

Ionic compounds in solution are often represented using solid formulas, but when ionicsolids dissolve, the ions separate. In these two solutions are these ions:

Solid Dissolved Ions

Formulas before mixing

1. Pb(NO3)2 Pb2++ 2 NO3

2. KI K+ + I

After they are mixed, the ions can trade partners. Two newcombinations of positive andnegative ions are possible.

Solid Dissolved Ions Mixed PossibleFormulas before mixing new ion combinations

Pb(NO3)2 Pb2++ 2 NO3

Pb2+ + I

KI K+ + I K+ + NO3

These new combinations mayor may notform solid precipitates. Add the possible solidformulas.

* * * * *


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Solid Dissolved Ions Mixed Possible PossibleFormulas before mixing new ion combinations Precipitates


Pb2+ + 2 I PbI2

KI K+ + I K+ + NO3 KNO3

Note that the ion coefficients are not the same before mixing and after. You must adjust thecoefficientsso that each new ion combination is balanced for atoms and charge.

Will a precipitate form? If a new combination is possible that is insoluble, it willprecipitate.Label each of the possible precipitates in the chart as soluble or insoluble.

* * * * *

PbCl2is insoluble by rule 2 (Pb2+); KNO3is solubleby rule 1. The result is

Solid Dissolved Ions Mixed Possible PossibleFormulas beforemixing new combinations Precipitates


Pb2+ + 2 I

PbI2 (Insoluble, forms ppt.)

KI K+ + I K+ + NO3 KNO3 (Soluble, ions stay dissolved)

* * * * *

Q3. For the above reaction, write the total and then the net ionic equation.

* * * * *

Answer

The chart shows the ionsbefore mixing (column 2) and the products after mixing (column 3or4). To write the total ionic equation, copy all of the column two ionson a line on yourpaper, followed by a reaction arrow. To the right of the arrow, write the new products that

form, shown in column 3 or column 4. Adjust the coefficients so that both each ioncombination and the overall equation are balanced for atoms and charge.

* * * * *

To make the precipitate, we need 2I. That means we must have 2K+to start).

Pb2++ 2 NO3 + 2K++ 2 I PbI2(s) + 2 K

+ + 2 NO3

That is the balanced totalionic equation. Now write the netionic equation.

* * * * *

Pb2+(aq)+ 2 I

(aq) PbI2(s) Net ionic equations show the ions that change.

Practice A: Do one now and one next time.

1. When AgNO3solution is mixed with Na2CrO4solution, a blood red precipitate

appears. Fill in a chart like the one above for this reaction, and then write the total and

net ionic equations.


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2. When potassium hydroxide and cobalt nitrate solutions are mixed, an intense blueprecipitate forms. Complete the chart, then write the total and net ionic equations.

The 3-Line Method

The chart method can be used for as long as needed to gain confidence that you arearriving at correct answers. At that point, you can try a quicker 3 line method (draft,total ionic, and net ionic equations) that handles more of the steps in your head. Lets learnthe method with the following example.

Q4. Write the total and net ionic equation identifying the white precipitate that formswhen solutions of barium chloride and potassium sulfate are mixed.

Write then check your answers to the steps below.

1. On one line, write the separated-ion formulas for the ions being mixed.

* * * * *

{ Ba2+ + 2 Cl } + { 2 K+ + SO42 }

2. After the arrow, write the newpossible ion combinations.

* * * * *

{ Ba2++ 2 Cl} + { 2 K++ SO42} { Ba2++ SO42} + { K+ + Cl}

3. Use solubility rules to label each newcombination as solubleor insoluble.

4. If a new combination of ions is possible that is insoluble, it will precipitate. Write thesolid formula for any precipitate that would form.

* * * * *{ Ba2++ 2 Cl} + { 2 K++ SO4

2} { Ba2++ SO42} + { K+ + Cl}

INsoluble Soluble

BaSO4(s)

5. Thats the rough draft. Copy the draft result without the brackets, adjusting thecoefficients as needed, to write the totalionic equation.

* * * * *

Ba2++ 2 Cl+ 2 K++ SO42 BaSO4(s)+ 2 K

++ 2 Cl

6. Write the netionic equation.* * * * *

Ba2+(aq) + SO42

(aq) BaSO4(s)

* * * * *


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Flashcards: Add these to your collection. Run each until perfect for 3 days.

But remember, you must also memorize a) a detailed solubility scheme and b) a method topredict whether mixed ions will precipitate, and what the precipitate formula will be.

One-way cards (with notch) Back Side -- Answers

Chemical equations must be balanced for Atoms and chargeNet ionic equations leave out Spectator ions

When will mixing two soluble ionic compoundsproduce a precipitate?

If a new possible combination is insoluble

When will mixing two soluble ionic compoundsnot result in a precipitate?

If both new possible combinations are soluble

Two-way cards (withoutnotch):

Ion combinations with these positiveions willalways be soluble

(alkali metals)+, NH4+

Ion combinations with these negative ions willalways be soluble nitrate, acetate, chlorate, perchlorate

Practice B

1. For these, write total and net ionic equations. Try using the 3-line method.

a. Ferric nitrate and rubidium hydroxide

b. MgSO4(aq)and BaCl2(aq)

2. Combining solutions of magnesium chloride and sodium sulfate,

a. what will be the names of the two new combinations that are possible?

b. Which of the new combinations are soluble in aqueous solutions?

c. Which of the new combinations will precipitate?

3. When these solutions are mixed: NiBr2(aq)+ K3PO4(aq)

a. What are the solid formulas for the two new combinations that are possible?

b. Will a precipitate form in the mixture? If so, what is its name and formula?

c. Write the total and the net ionic equations for the reaction.


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ANSWERS

( Assume the state is (aq) if not shown.)

Practice A

Problem 1

Solid Dissolved Ions Mixed Possible Possible

Formula before mixing new combinations Precipitates

1. AgNO3 Ag++ NO3

2 Ag+ + CrO42 Ag2CrO4 (Insoluble, ppt.)

2. Na2CrO4 2 Na++ CrO4

2 Na+ + NO3 NaNO3 (Soluble, does notppt.)

Totalionic equation:

2 Ag++ 2 NO3 + 2Na++ CrO4

2 Ag2CrO4(s) + 2 Na+ + 2 NO3

Netionic equation: 2 Ag+(aq) + CrO42

(aq) Ag2CrO4(s)

Problem 2

Solid Dissolved Ions Mixed Possible PossibleFormula before mixing new combinations Precipitates

Potassium HydroxideK++ OH K+ + NO3 KNO3(Soluble, does notppt.)

Cobalt Nitrate Co2++ 2 NO3 Co2+ + 2 OH Co(OH)2(Insoluble, ppt.)

Total ionic equation: 2 K+ + 2 OH + Co2++ 2 NO3Co(OH)2(s) + 2 K

+ + 2 NO3

Netionic equation: Co2+(aq) + 2 OH

(aq) Co(OH)2(s)

Practice B1a. { Fe3+ + 3 NO3

} + { Rb+ + OH} { Fe3++ 3 OH} + { Rb+ + NO3}

Insoluble Soluble

Fe(OH)3(s)

(Thats the rough draft. To write total and net ionic equations, remove brackets and adjust coefficients sothat all atoms and charges balance.)

Totalionic: Fe3+ + 3 NO3 + 3 Rb+ + 3 OH

Fe(OH)3(s)+ 3 Rb

+ + 3 NO3

Netionic equation: Fe3+(aq) + 3 OH

(aq) Fe(OH)3(s)

1b. { Mg2+

+ SO42

} + { Ba2+

+ 2 Cl

} { Mg2+

+ 2 Cl

} + { Ba2+

+ SO42

}

Soluble Insoluble

BaSO4(s)

Total ionic: Mg2+ + SO42

+ Ba2+

+ 2 Cl

Mg2+ + 2 Cl + BaSO4(s)

NET ionic equation: Ba2+(aq) + SO4

(aq) BaSO4(s)


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2. a. Magnesium sulfate and sodium chloride. b. Both c. Neither

3. { Ni2+ + 2 Br} + { 3 K+ + PO43} { 3 Ni+2 + 2 PO4

3} + { K+ + Br}

Insoluble Soluble

Ni3(PO4)2(s)

a. KBr and Ni3(PO4)2 b. Yes. Ni3(PO4)2(nickel phosphate)

c. Total: 3 Ni2+ + 6 Br + 6 K+ + 2 PO43

1 Ni3(PO4)2(s) + 6 K+ + 6 Br

Net ionic equation: 3 Ni2+(aq)+ 2 PO43

(aq) Ni3(PO4)2(s)

(Note how the coefficients are balanced above. When writing each combination at each stage, adjustcoefficients to balance the combination for atoms and charge. Then, for the total ionic equation, adjust thecoefficients again to balance each ion combination plus all of the atoms and charges on each side.

If the coefficients do not balance easily, it may indicate an error and you should double-check your work.)

* * * * *

Lesson 13D: Precipitation and Gravimetric Calculations

Timing: If there is a gap between the time that you are asked to predict which ioncombinations precipitate and when you are assigned calculationsinvolving precipitates,delay this lesson until precipitation calculations are assigned.

* * * * *

Gravimetric Calculations

Stoichiometry answers how muchof the reactants are used up and products are formed inchemical reactions. Once a count (usually in moles)of any one reaction component isdetermined, a count of other components can be solved using simple-whole-number ratios:the coefficients of the balanced equation.

In some precipitation calculations, volume and concentration are used to solve for moles.In others, precipitates are dried and weighed to determine moles; a process termedgravimetricbecause the moles of precipitate are determined by its weight, which is theattraction of its mass by gravity.

Precipitation/Gravimetric Stoichiometry

In most respects, precipitation/gravimetric calculations follow the same rules for solutionstoichiometry that you used in Lessons 12C and 12D, but in several ways they often differ.

1. In most previous equations to balance, you were supplied with the formulas of thereactants and products.

In precipitation calculations, you will often be asked to both predict all of the productformulas and identify which products precipitate.


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2. Precipitation calculations can involve measurements for both a solid formula andtheions formed when the solid dissolves in water.

3. For most reaction calculations, you can count the particles of one or two substancesdirectly from the supplied DATA. In many gravimetric calculations, you must firstidentify the formula of a precipitate before you can count any particles.

To handle these differences for precipitation reactions, we will add these special steps toour stoichiometry process.

Inprecipitationcalculations, do WDBB steps 3 and 4 as follows.

3. Balancethe equation.

a. Write and balance aprecipitationequation in threeparts:

Reactants in solidformulasReactants as separated ionsProducts

b. As you write theproductformulas, decide whether a new combination will besoluble and separated, or insoluble and precipitate. Use one of the methods

for predicting product formulas and precipitates in Lesson 13C.

In the products, write solidformulas for insoluble precipitates and separated-ionsfor combinations that are soluble.

c. Once the precipitate formula is identified, in the WANTED and/or DATA,replaceppt. with the formula for the precipitate. If this results ingramsof aformulain WANTED or DATA, add the molar mass of theppt. to the DATA.

4: Write the mol-mol bridge.

In the mol-mol bridge, include all chemical formulas that are written after unitsinthe WANTED and DATA. This may result in threeor moreterms with

coefficients, moles, and chemical formulas that are equal, instead of two.

Cover the answer below, then apply the steps to the following problem. If you get stuck,readpartof the answer below, then try again.

Q. If 14.0 mL of Pb(NO3)2solution is reacted with excess MgCl2solution, and the

resulting precipitate when dried weighs 0.314 grams, what was the original [Pb2+]?

* * * * *

Since there are two reactants, and one is in excess, the other must be limiting. If you knowthe limiting reactant, solve by conversion stoichiometry.

1. WANT: ? mol Pb2+

= (a ratio unit is WANTED)L Pb2+soln.

Concentration is a ratio that is constant in a well-mixed solution. The ratio for anysample will be the same as the ratio for the whole solution. We can easily measure theliters in a sample, but we cannot count the moles of ions directly.

However, if we precipitate the ions, weigh the precipitate, and identify the precipitate

formula, we can find the moles of Pb2+ ions in the original sample. We can then divide


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to find the mol/L ratio for that sample, and that is the WANTED unit. Since weighing

the dried precipitate is one step in this process, this is a gravimetriccalculation.

2. DATA: 14.0 mL Pb(NO3)2soln. = 0.314gppt. (Equivalent: 2SUA-R)

In the calculation of a WANTED ratiobased on a reaction, two DATA amounts will be

equivalentin the reaction (Lesson 12C).The MgCl2is in excess. Amounts of reactants in excess can be ignored in the DATA,

since they do not determine the amounts of reactants used up or products formed

(Lesson 10F).

3. Balance. Since this is a precipitation reaction, apply the special steps above.

a. Write the balanced equation for precipitation in 3 parts: Reactants as solidformulas, reactants in separated-ions, and products as either precipitates or asseparated, soluble spectator ions, depending on their state in the products.

SolidReactants: 1 Pb(NO3)2 + 1 MgCl2

SeparatedReactants: 1 Pb2+ + 2 NO3 + 1 Mg2++2Cl

Products: 1 PbCl2(s) + 1 Mg2+ + 2 NO3

Adjust coefficients so that all three parts balance for atoms and charge.

* * * * *

b. Decide the precipitate formula.

Upon mixing the two solutions, two new ion combinations can form.

One new combination is lead ion with chloride ion. By the solubility rules in

Lesson 13A, the combination of Pb2+ and Cl is insoluble. A possible new

combination that is insoluble willprecipitate, so PbCl2will precipitate. The other new possible combination is Mg2+and NO3

ions. All

combinations that include nitrate ion are soluble. Write these ions as

separated on the products side.

The only precipitate is therefore PbCl2.

c. In the DATA, replaceppt. with its solidformula:

DATA: 14.0 mL Pb(NO3)2soln. = 0.314g ppt. PbCl2(s) (Equivalent)

Now that the DATA includes grams of a formula, what should be added to theDATA that will likely be needed to solve?

* * * * *

278.1gPbCl2= 1 mol PbCl2 (grams prompt)

4. Bridgeconversion. In precipitation reactions, write mole ratios between allof thechemical formulas which are written after unitsin the WANTED or DATA.


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In this problem, the WANTED and DATA involve measurements of threechemical

formulas: Pb2+, Pb(NO3)2, and PbCl2. Write the bridge ratios for those threeformulas

using the coefficients in the balanced equations above.

* * * * *

Bridge: 1 mol Pb2+

(aq) = 1 mol Pb(NO3)2(aq) = 1 mol PbCl2(s)Any twoof those three terms can be used if needed as a bridge conversion.

5. SOLVE. For reaction calculations in which the answer unit is a ratio, our strategy is tofind a value for the top and bottom WANTED units separately, then divide (see Lesson12D). The steps are

a. Label the DATA equality that has two amounts that are equivalent.

b. Solve for the WANTED unit that is notmoles first, using as agiventhe side ofthe equivalency that measures the WANTED formula or sample.

c. Solve for molesWANTED using the otherhalf of the equivalency as thegiven.

c. Divide the two WANTED amounts.* * * * *

The WANTED unit that is notmoles is L Pb2+. We pick one of the terms in the

equivalency to convert to L Pb2+. Which one?

In stoichiometry, we pick as a given the side of the equivalency that includes the

WANTED substance formula, but neither of the two equivalent amounts has the

formula Pb2+. However, the labelsPb(NO3)2and Pb2+are equivalent. One label

includes the spectator ions, and the other does not, but the Pb(NO 3)2solution and Pb2+

solution are the same solution. The mL volumeof Pb(NO3)2is one side of the

equivalency, so use that term as yourgiven.

? L Pb2+soln. = 14.0 mL Pb(NO3)2soln.= 14.0 x 103L Pb(NO3)2

= 14.0 x 103LPb2+soln. A good general rule is:

If you get stuck, add more complete labelsto the WANTED and DATA.

In this case, by labeling all of the volume units as soln., it is a indication that they arerelated. Now solve for the other WANTED unit and then check below.

* * * * *

Starting from the other half of the equivalency as given,

? mol Pb2+= 0.314 g PbCl2 1 mol PbCl2 1 molPb2+ = 1.129x 103mol Pb2+

278.1 g PbCl2 1 molPbCl2

Finish the problem, and then check below.

* * * * *


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The sample of Pb2+solution was found to have 1.129x 103mol Pb2+in

14.0 x 103LPb2+soln. Knowing those amounts, find the WANTED ratio.

* * * * *

? molPb2+ = 1.129x 103mol Pb2+ = 0.0806molPb2+

LPb2+ soln. 14.0 x 103L Pb2+soln. LPb2+ soln.

* * * * *

Flashcards: Add these to your collection. Run each until perfect for 3 days, then movethem to stack 2 (see Lesson 6E).

One-way cards (with notch) Back Side -- Answers

In gravimetric or precipitation calculations,

what special rule applies to the balance step?

Write solidreactantsseparatedreactantsproducts

In gravimetric or precipitation calculations,

when you identify the ppt. formula

If gramsppt is in WANTED or DATA,

add the molar mass to the DATA

Practice: Memorize the special rules for precipitate stoichiometry, then apply them tothese problems. If you get stuck, read a part of the answer and try again.

1. A 25.0 mL sample of a K2SO4solution is reacted with an excess amount of BaCl2

solution. Assuming that all of the sulfate ions precipitate, if the mass of the dried

precipitate is 1.167 grams, what was the original [K2SO4]?

2. 50.0 mL of a solution containing Pb2+requires 12.0 mL of a 0.200 M NaCl solution to

precipitate all of the lead ions. What was the original [Pb2+]?

ANSWERS

1. If there are two reactants and one is in excess, the other is limiting and controls how much product can beformed. Solve using conversion stoichiometry, starting with WDBB.

1. WANTED: ? mol K2SO4L K2SO4soln.

2. DATA: 25.0 mL K2SO4 soln. = 1.167 g ppt. (Equivalent 2SUA-R)

(The problem is about a chemical reaction. The WANTED and DATA involve 2 substances, with theppt. being one. Thats the stoichiometry prompt. All stoichiometry begins with the same 4 steps.)

3. Balance.a. (In precipitation, balance in 3 parts)

Solid Reactants: 1K2SO4(aq) + 1 BaCl2(aq)

Separated Reactants: 2 K+(aq)+ 1 SO42

(aq) + 1 Ba2+(aq)

+ 2 Cl(aq)

Products: 1BaSO4(s)+ 2 K+

(aq)+ 2 Cl

(aq)


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b. (Identify the precipitate. Since KCl is soluble, the ppt. must be BaSO4.

c. Adjust the DATA.)

DATA: 25.0 mL K2SO4 soln. = 1.167 g ppt. BaSO4(s) (equivalent 2SUA-R)

(When a ratio unit is WANTED, all of the DATA will be equalities. Here, the mL K2SO4soln. is what

determined the grams of ppt. that formed. What additional key information can now be added to the DATA?* * * * *

233.4 gBaSO4= 1 mol BaSO4 (grams of a formula= g prompt)

4. Bridge. 1mole K2SO4=1mole BaSO4

(K2SO4and BaSO4are the only formulas that are written after unitsin the WANTED or DATA.

* * * * *

5. SOLVE. (Since the answer unit is a ratio, find amounts for the two WANTED units separately.

a. Label the equivalency in the DATA.

b. First find the WANTED unit that is not moles, choosing as a giventhe side of the equivalency thatmeasures the formula WANTED.)

? L K2SO4 = 25.0 mL K2SO4 = 25.0 x 103LK2SO4 (Substitute what prefix means)

c. (Find moles WANTED using the other half of the equivalency as a given.)

* * * * *

? mol K2SO4= 1.167 g BaSO41 mol BaSO4 1 mol K2SO4 = 5.000 x 103 mol K2SO4

233.4 g BaSO4 1 mol BaSO4

d. (Find the final WANTED unit.)

* * * * *

? molK2SO4 = 5.000 x 10

3molK2SO4 = 0.200 mol K2SO4LK2SO4 25.0 x 10

3L K2SO4soln LK2SO4

2. Since you can count the moles for only one substance (NaCl), that substance will be your givenwhen you

solve for moles Pb2+. Knowing the given, solve using conversion stoichiometry, starting with WDBB.

1. WANT: ? mol Pb2+ =

L Pb2+soln.

2. DATA: 50.0 mLPb2+soln = 12.0 mL NaCl soln. (equivalent 2SUA-R)

0.200 mol NaCl = 1 L NaCl soln. (M prompt a ratio)

(If a ratio unit is WANTED, all of the DATA will be equalities. The mL Pb2+

. is what determined the mLNaCl that reacted. They are proportional and equivalent.)


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3. Balance.

a. (Write the balanced equation for precipitate formation in three parts: solid, separated, products.)

1Pb2+ + 2NaCl 1Pb2+ + 2 Na+ + 2 Cl 1PbCl2(s) + 2 Na+

(The anion that was initially combined with lead ion is not known, but it is not needed as long as

the above equations are balanced for atoms and charge.

b. Identify: The precipitate must be PbCl2.

c. Adjust: Since the ppt. amount is not measured in the DATA, there is no DATA to adjust.)

4. Bridge. (The WANTED and DATA include measurements (units and formulas) for only 2 substance

formulas: Pb2+and NaCl.)

Bridge: 1 mol Pb2+ = 2 mol NaCl (If needed, adjust your work and solve from here.)

* * * * *

5. SOLVE. (Since the answer unit is a ratio, find amounts for the two WANTED units separately.

a. Label the equivalency in the DATA.b. First find the WANTED unit not moles, choosing as a giventhe side of the equivalency that

measures the formula WANTED.)

? L Pb2+soln. = 50.0 mLPb2+ = 50.0x 103L Pb2+ (Substitute what prefix means)

c. (Using other half of equivalency as given, find moles WANTED using stoichiometry steps.)

? molPb2+= 12.0 mL NaCl 103L 0.200 mol NaCl 1mol Pb2+ = 1.200x 103

1 mL 1 L NaCl 2 mol NaCl mol Pb2+

d. (Find the final WANTED unit.

Above, in the sample of the WANTED solution, you foundmol Pb2+and L Pb2+soln.)

? molPb2+ = 1.200 x 103mole Pb2+ = 0.0240 M Pb2+

LPb2+ 5.000x102L Pb2+

* * * * *


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Summary: Solubility and Precipitation

1. Most ionic solids can be characterized as soluble or insoluble in water. The rules forsolubility must be memorized, and exceptions occur. Some frequently used rules are

a. compounds containing alkali-metal atoms, and NH4+, NO3

, CH3COO, ClO3

,

and ClO4 ions are nearly always solublein water;

b. exceptin the above cases, compounds containing Pb2+, Hg22+, Ag+, CO3

2,

PO43, S2, or CrO4

2 ions will generally be insoluble in water.

2. When two solutions of soluble ions are mixed,

if a new combination is possible that is insoluble, it willprecipitate;

if a new combination is soluble, its ions will remain separated.

3. Totalionic equations show all of the ions and precipitates present when two solublesolutions are mixed. Netionic equations include the ions and solids that change in areaction, and omit spectator ions that do not change.

4. In precipitation calculations:

a. Write a balanced equation for precipitate formation in 3 parts.

Reactants as solid formulasReactants as separated ionsProducts

b. Use solubility rules to determine which ions precipitate.

c. Use stoichiometry steps to solve.

5. In general, to understand the reactions of ionic compounds,

Re-write the reactants in their separated-ions format;

Look for reactions among the new possible ion combinations.

# # # #


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Timing: This module covers acid-base neutralization. Other acid-base topics, includingpH, weak acids and bases and buffers are covered in Modules 29 - 32.

There are two types of neutralization calculations that may be assigned in your course. If all of the neutralization calculations that you are assigned supplychemical

formulasfor theproductsin the reactions, you do notneed to complete Module 14.This type of calculation was covered in Lesson 12C and 12D.

If your course assigns in which you mustpredict theproductformulas for acid-baseneutralization, complete Module 14.

* * * * *

Lesson 14A: Ions In Acid-Base Neutralization

Prerequisites: Lessons 7B, 7C, and the acid naming lesson (7D).

Pretest: If you can do each of the 3 following problems, you may skip Lessons 14A and14B. Check answers at the end of this lesson.

Assuming complete neutralization of all protons and basic ions, write product formulasand balance these.

a. KHC2O4 + KOH

b. H3Unk + Mg(OH)2

c. HCO3 + OH

* * * * *

Terminology

Many substances can be classified as acids or bases (and some can act as both). A variety ofdefinitions exist for acids and bases, with each definition helpful in certain types ofreactions and calculations.

Because acids and bases can react with many substances, they are often termed corrosive:they may damage a surface, metal, or fabric. When an acid or base is neutralized, at leastsome of its reactive ions are used up, and its corrosive power is reduced.

In an acid-base neutralizationreaction, an acid and a base are reactants. When mixedtogether, they react, and both are used up to some extent. In the process, both the acid and

the base are said to be neutralized.For the limited purpose of studying neutralization, we will define

an acidas a substance that creates H+ions when dissolved in water, and

a baseas a substance that can react with (use up) H+ions.


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Electrolytes

Because acids and bases form ions when they dissolve in water, they are termedelectrolytes: their solutions can conduct electricity. Strong acids and strong bases aretermed strong electrolytes because they ionize ~100%.

The Structure of H+: A ProtonNeutral hydrogen atoms contain one proton and one electron. (About 0.01% of naturallyoccurring H atoms also contain one or two neutrons, but the neutrons have no impact onthe types of chemical reactions in which hydrogen atoms participate.)

An H+ion is a hydrogen atom without an electron, so most H+ions in terms of structure

are single protons. The terms H+ionandprotonare often used interchangeably to describe

the active particle in an acid.

Memorize: the ion formed by acids = H+ = proton

In aqueous solutions, the proton released by an acid is nearly always is found attached to a

water molecule, forming a hydroniumion(H3O+). This reaction can be represented as

1 H+ + 1 H2O 1 H3O+ (goes ~ 100%)

The symbols H+ and H3O+in most cases are considered to be equivalent. For now, to

simplify work, we will use H+ as the symbol for the ion contributed to solutions by acids.

Identifying Acids and Bases

From a chemical formula, how can you tell whether a substance is an acid or a base? It isnot always easy to tell. A general rule is:

If Identify A Compound Is As An Acid or A Base

Write the balanced equation for the substance formula separating intofamiliarions.

Compounds that ionize to form H+ions are acids.

Compounds that contain OHor CO32ions can act as bases.

Compounds that contain HCO3 ions can act as both.

Practice A: For the reaction in which the compounds below ionize in water, writebalanced equations showing the ions formed, then label each initial reactant as either anacidor a base. Answers are at the end of the lesson. For help, review Lesson 7C and 7D.

a. LiOH H2O

b. Na2CO3

c. Perbromic acid d. Ca(OH)2


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e. HCN f. KHCO3

g. Nitrous acid h. Chromic acid

Identifying Acidic HydrogensHydrogen atoms in compounds can be divided into two types.

Acidichydrogens are generally defined as those that react with hydroxide ions, and

non-acidichydrogens are those that do not.

Compounds often contain both acidic and non-acidic hydrogens.

For example, in CH3COOH (acetic acid), the H atom at the end of the formula reacts

with NaOH, but the other three H atoms do not. The H at the endis the acidic

hydrogen. The other H atoms are notacidic hydrogens.

In an acid formula, how can you predict which hydrogens will be acidic and which will

not? In most cases, the rules are:

If one or more H atoms is written at thefrontof a formula, while other H atoms arenot, the H atoms at thefrontare acidic, and the others are not.

Example: Acetic acid is often written as HC2H3O2. Only the H in front is

acidic.

The H at the endof a COOH(also written CO2H) group is acidic.

Examples: In C6H5COOH, only the H at the end is acidic.

In C3H7CO2H, only the H at the end is acidic.

If H is the secondatom in the formula, written after a metal atom and before otheratoms, it is acidic.

Examples: In KHC8H4O4, thefirstH (and only the first) is acidic.

In NaH2PO4, the two Hs after the metal atom are acidic.

If a substance with only one H reacts with hydroxide ion, the H is acidic.

Practice B

1. Draw an arrow toward and countthe acidic hydrogens in these compounds.

a. NaH2PO4 b. C12H25COOH c. H2C4H4O6

d. H3AsO4 e. KHC8H4O4 f. NaHSO4


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2. Fill in the blanks to show the number of protons formed when these compounds ionizein water.

a. C3H7CO2H ___ H+

b. HC2H3O2 ___ H+

c. NaH2(C3H5O(COO)3) ___ H+

ANSWERS

Pretest: Coefficients of 1 may be omitted as understood.

a. 1 KHC2O4 + 1 KOH 1 H2O + 1 K2C2O4

b. 2 H3Unk + 3 Mg(OH)2 6 H2O + 1 Mg3(Unk)2

c. HCO3

+ OH

H2O + CO32

Practice A

a. LiOH Li + + OH Base b. Na2CO3 2 Na+ + CO3

2 Base

c. HBrO4 H+ + BrO4

Acid d. Ca(OH)2 Ca2+ + 2OH

Base

e. HCN H+ + CN Acid f. KHCO3 K+ + HCO3

Acid orBase

g. HNO2 H+ + NO2

Acid h. H2CrO4 2 H+ + CrO4

2 Acid

Practice B

1. a. NaH2PO4 Two b. C12H25COOH One c. H2C4H4O6 Two

d. H3AsO4 Three e. KHC8H4O4 One f. NaHSO4 One

2 a. C3H7CO2H 1 H+ b. HC2H3O2 1 H

+

c. NaH2(C3H5O(COO)3) 2 H+

* * * * *


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Lesson 14B: Balancing Hydroxide Neutralization

Prerequisites: Lesson 14A.

* * * * *

Hydroxide Neutralization

In an aqueous solution, for the reaction of an acid and a hydroxide, the neutralizationreaction forms liquid water. Water can be written as HOH or HOH or H2O.

H+ + OH H-OH(l) (1)

The reaction of an acid and a hydroxide is driven to completionby the formation of water: alow potential energy molecule. Whenever a physical or chemical system can go to lowerpotential energy, there is a strong tendency to do so.

The reaction of acids with compounds containing hydroxide ions can be represented by the

general equation

An acid + a basecontaining OH H2O + a salt (2)

Historically in chemistry, the term saltin reaction (2) was a general term for any product,in addition to water, formed when an acid and base react. In modern usage, salt is oftenused as a synonym for ionic compound.

A typical acid-hydroxide neutralization is the reaction of hydrochloric acid with sodiumhydroxide to form sodium chloride (also known as table salt).

HCl(aq) + NaOH(aq) H-OH(l) + NaCl(aq) (3)

The state (aq) means aqueous(dissolved in water). Most (but not all) acid-base reactions are

carried out in water. In the acid-base solution reactions in these lessons, you may assumethat the water is liquid and the formulas for other compounds and ions are aqueous(aq)unless otherwise noted. Writing water in the form HOH form helps to emphasize thereaction that occurs between acids and hydroxide ions.

Equation (3) above is one way that this reaction is represented. However, both HCl and

NaOH, when dissolved in water, separate completely into ions. The table salt in the

solution after the reaction also exists as separate ions of Na+ and Cl. Re-writing the

equation to show the separated ions that actually exist in the solution, the reaction is

H+ + Cl + Na+ + OH HOH(l) + Na

+ + Cl (4)

Note in reaction (4) that the sodium and chloride ions arespectators: they do not changeduring the reaction.

Predicting All Products of Hydroxide Neutralization

If formulas are supplied for all the reactants and products, neutralization equations can bebalanced by trial and error using the methods in Lesson 10B. However, in acid-baseneutralization problems, often the product formulas are notsupplied. In these cases, youcan oftenpredictthe products and balance the reaction equation by


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Re-writing the acid and base as separated ions, and

Predicting one of the neutralization products.

In hydroxide neutralization, a key rule is

If an acid reacts with OH, one of the products is HOH.

Knowing that water is one product, you can usually determine the other productformula(s). Complete the steps below in your notebook.

Q. Write a balanced equation for the complete neutralization of H3PO4(phosphoric

acid) by Ca(OH)2(calcium hydroxide).

Steps

1. Write the acid and base reactants in their solid(molecular) formulas.

2. On the line below, re-write each reactant inside parentheses ( ) in its separated-ionformat.

3. After the reactants, add

H

OH+ ______ + ______.* * * * *

Solid: H3PO4 + Ca(OH)2

Separated: ( 3 H++ PO43) + ( Ca2++ 2 OH ) H-OH+ _____ + _____

4. Add lowest-whole-number coefficients in front of the parentheses ( ) to balance the H+,

OH, and H-OH.

* * * * *

Separated: 2 ( 3 H+ + PO43) + 3( Ca2++2 OH ) 6H-OH+ ____ + ____

^ ^ ^

[ ( 6H+

)

( 6OH

)

( 6H-OH) ]The total H+ions must equal the total OH ions must equal the total H2O.

5. Add in the products side blanks the formulas and coefficient totals for the ions on the

left that are not H+and OH.

6. Move the left side coefficients that are infrontof the ( ), plus the coefficient of the water,up to the top line.

7. Finish adding coefficients to the top equation. To write a molecular formula for theremaining product, use the rules for writing ionic solidformulas in Lesson 7C.

* * * * *

Solid: 2 H3PO

4 + 3 Ca(OH)

2 6 H

2O + 1Ca

3(PO

4)2(s)

Separated: 2 ( 3 H++ PO43) + 3( Ca2++ 2 OH) 6 H2O + 2PO4

3 + 3Ca2+

In most neutralization reactions, the ions that are not H+ and OHwill simply remain

as aqueous spectator ions, dissolved in the final solution. In this problem, however,

calcium and phosphate ions are an insoluble combination. If a combination can form

that is insoluble, it will precipitate. In this reaction, we have both a neutralization and a

precipitation.


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8. Check that the final equation is balanced.

Practice A. Balance by inspection or the methods above. Assume that the acids andbases are completely neutralized. Do every other part, and more if you need more practice.

1. Write the product formulas in the solid (molecular) format and balance the equation.

a. HNO3 + KOH

b. KOH + H2SO4

c. H2SO4 + Al(OH)3

2. Write reactant and product formulas, then balance the equation. Write final formulas inthe ionic-solid (molecular) format.

a. Barium hydroxide plus sulfuric acid

b. Hydrochloric acid plus magnesium hydroxide

Balancing Neutralization with Spectators Omitted

In all stable substances and mixtures, the total of all of the positive and negative chargesmust add up to zero. However, as a shortcut, spectator ions are often omitted whenrepresenting substances or writing reaction equations. In such cases, the writtenchargesmay notadd up to zero.

For example, in the lab, solutions of acids and bases may be labeled as simply H+ or

OH to emphasize the reactive particles. In those solutions, however, there must

also be other ions that balance the overall charge.

Similarly, in writing equations for neutralization reactions,partialionicequations may bewritten that leave out spectators. In these equations, the total charge may not be zero oneach side, but as long as it is the same on both sizes, the equation is considered to bebalanced.

For a written reaction equation to be considered balanced, what is required is that thetotal chargeis the sameon each side (but not necessarily zero) and that the number andkind of atoms is the same on each side.

To balance a neutralization equation in which some spectators have been left out, use the

same steps as above: separate the formulas into familiar ions, one reactant including H+

and the other including a basic ion, then balance the atoms and charge on each side.

Try this example.

Q. Write the products and balance: HSO4 + OH

* * * * *


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Inside parentheses, break the reactants into familiar ions. When the base is ahydroxide, form wateron the right.

Add coefficients that balance the particles that form water.

Balance the particles that dont react.

Make sure that the atoms balance and the total charge is the same in all threeparts.

1 HSO4 + 1 OH 1( 1 H++ 1 SO4

2 ) + 1OH 1 H-OH + 1 SO42

Atoms balance, and in all three parts, the totalcharge is 2. Balanced.

Practice B. Check answers as you go. Do every other letter, and more for more practice.

1. Write product formulas and balance the equation. Assume all acids and bases aretotally neutralized. Products may be molecules or ions.

a. H+ + Al(OH)3

b. HSO4 + KOH

c. KHSO4 + OH

d. CH3COOH + OH

Balancing Only the Reactants

To solve most neutralization stoichiometry, we will need only the coefficients for thereactants: the acid and the base. Thispartialbalancing can be completed if know theformula for oneproduct, and in the case of acid-hydroxide neutralization, we do. One

product is always H-OH.For this problem, write H-OHas a product, balance the leftside, and then check below.

Q. What is the ratio of reaction for H2SO4 + Al(OH)3

* * * * *

Either balance by inspection (total H+= total OH) or by using these steps.

1. Below the two reactants, write the number of acidic and basic ionsin the reactants.

2. On the right, add one product of the reaction.

H2SO4 + Al(OH)3 HOH +

( 2 H+) + (3 OH

) HOH +

3. Add the coefficients to balance the atoms and charge.

H2SO4 + Al(OH)3 HOH +

3 ( 2 H+ ) + 2 (3 OH) 6HOH +


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4. Transfer the coefficients to the original reaction equation.

3 H2SO4 + 2 Al(OH)3 6 HOH +

3 ( 2 H+ ) + 2 (3 OH) 6 HOH +

The reactantratios will be all that is needed to solve most neutralization stoichiometry.

Practice C. Assume all reactants are completely neutralized. Balance by inspection orusing the steps above. Do every other problem, and the rest in your next practice session.

1. Balance the coefficients of the two reactants.

a. CsOH + H2SO4

b. Ca(OH)2 + nitric acid

c. HSO4 + OH

2. Add ratios of reaction for these reactants.a. Acetic acid + NaOH

b. Calcium hydroxide + HC2H3O2

c. NaHC2O4 + Al(OH)3

d. Potassium hydroxide + KHC8H4O4

Balancing with an Unknown Formula

In some neutralization calculations, the formula for an acid or the base is not supplied, but

the numberof protons in the acid, or basic ions in the base, is provided. In these cases, theacid formula can be represented as

H3R, where the R stands for an unknown,group, often containing carbon, or

H3Unk , where the Unk stands for unknown,

and H3represents three acidic hydrogens. Using these formulas, you can usually balance

the acid-base reactant ratio, and that is all that you need in solving most neutralization

stoichiometry. Try this example.

Q. A solid acid has an unknown formula but is known to contain three acidic hydrogens.What will be the ratio for the reaction of this acid with NaOH?

* * * * *Answer

The two reactants can be written as: H3Unk + NaOH

Complete the reactant balancing.

* * * * *


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H3Unk + NaOH 1( 3 H+ ) + 3 ( OH ) 3 HOH + .

The acid-baseratio must be 1 H3Unk + 3 NaOH

A formula could be written for the other products, but it would be speculation. The acidanion could remain intact, or it could decay in some fashion. However, for most

neutralization stoichiometry, formulas for the other products are not required: the acid-base ratio on the left side of the equation will be all that is needed to solve.

Summaryfor neutralization rules so far.

1. Acid-base neutralizationis an ionicreaction. To understand ionic reactions, writethe separated-ion formulas.

2. Ions: Acids contain H+. The reacting particle in acids = H+ = a proton.

Bases include compounds with hydroxide (OH), carbonate (CO32), and

hydrogen carbonate (HCO3) ions.

3. Products: For acids + OH, one product is water: H-OH. H++ OHH-OH

4. Balancing: Topredictthe products and balancethe equations,

Write the separated-ion formulas in ( ).

Write one product.

Finish by balancing atoms and charge.

5. To balancewhen a formula is unknown,

If a substance has 2 acidic hydrogens, write its formula as H2Unk or H2R.

If a base has 3 hydroxides, write Unk(OH)3.

Practice D: Learn the rules above, then do these problems. Assume that all reactantsare completely neutralized.

1. Supply the ratios of reaction for the two reactants.

a. H2Unk + Sr(OH)2

b. H2SO4 + Unk(OH)3

c. HCl + UnkOH

d. H2Unk + Unk(OH)3

2. Add coefficients to balance the two reactants.

a. An unknown acid with two acidic hydrogens is totally neutralized by potassiumhydroxide.

b. An unknown base with three hydroxide ions is totally neutralized by nitric acid.


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ANSWERS

Practice A: Coefficients of one may be omitted. Any coefficient ratioswhich are the same as these are notincorrect, but lowestwhole-number ratios are preferred.

1 a. 1 HNO3 + 1 KOH 1 H2O + 1 KNO3

b. 2 KOH + 1 H2SO4 2H2O + 1 K2SO4

c. 3 H2SO4 + 2 Al(OH)3 1Al2(SO4)3 + 6H2O

2. a. 1Ba(OH)2 + 1 H2SO4 2H2O + 1 BaSO4

b. 2 HCl + Mg(OH)2 2H2O + 1 MgCl2

Practice B

1. a. 3H+ + 1Al(OH)3 3 H-OH + Al3+

b. 1 HSO4 + 1 KOH 1H-OH + 1K+ + 1 SO4

2

c. 1 KHSO4+ 1OH

1H-OH + 1K+ + 1 SO42

d. 1 CH3COOH + 1OH

1H-OH + 1CH3COO

Practice C: To balance the reactants, write one product of the reaction.

1. a. 2 CsOH + 1 H2SO4 2 H-OH +

b. 1 Ca(OH)2 + 2 HNO3 2 H-OH +

c. 1 HSO4 + 1 OH 1 H-OH +

2. a. 1 CH3COOH + 1 NaOH 1 H2O +

b. 1 Ca(OH)2 + 2 HC2H3O2 2 H2O +

c. 3 NaHC2O4 + 1 Al(OH)3 3 H2O +

d. 1 KOH + 1 KHC8H4O4 1 H2O +

Practice D

1. a. 1 H2Unk + 1 Sr(OH)2 2 H-OH +

b. 3H2SO4 + 2 Unk(OH)3 6 H2O +

c. 1 HCl + 1 UnkOH H-OH +

d. 3H2Unk + 2 Unk(OH)3

2. a. 1 H2Unk + 2 KOH 2 H2O +

b. 3HNO3 + 1 Unk(OH)3 3 H2O +

* * * * *


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Lesson 14C: Neutralization and Titration Calculations

Solution Stoichiometry and Acid-Base Reactions

In any reaction, at the point where the molesof the reactants have been mixed in the exactratio that matches the coefficients of the balanced equation, the reactants arestoichiometrically equivalent. Calculations involving these amounts can be solved using thestoichiometry steps in Lessons 9E and 12C.

For a reaction in which an acid and base have been mixed in stoichiometrically equivalentamounts, both the acid and base are said to be exactly neutralized.

Neutralization calculations can be divided into two types: those that supply

all of the reactant and product formulas, or

only the reactant formulas.

If the formulas for all of the reactants and products are supplied, any equation can bebalanced by trial and error, and calculations involving reaction amounts can then be solvedby stoichiometry. In problems that supplied both the reactant and product formulas, yousolved several neutralization calculations in Lessons 12C and 12D.

We are now prepared to solve calculations in which the acid-hydroxide formulas aresupplied but the product formulas are not. Using the balancing strategies in Lesson 14B,the coefficients of the acidand basecan be determined even when the formula of the non-water product may not be certain. In most neutralization calculations, the ratio of reactionof the reactantsis all that is needed to solve by conversion stoichiometry.

Apply the acid-hydroxide balancing and stoichiometry steps to this example.

Q. How many milliliters of a 0.200 M sodium hydroxide solution are needed to

neutralize all of the acidic protons in 2.34 grams of arsenic acid (H 3AsO4) ?If you get stuck, read aportionof the answer until you are unstuck, then try again.

* * * * *

Reaction calculations for two substances that are at an equivalence point (as when an acidand base are both exactly neutralized) are solved by conversion stoichiometry. Start withWDBB.

* * * * *

Answer: Your paper should look like this, minus the (comments).

WANT: ? mL NaOH soln.

DATA: 0.200 mol NaOH soln. = 1 L NaOH soln.2.34gH3AsO4 (single unitgiven)

141.9gH3AsO4= 1 mol H3AsO4 (g prompt)

(Include formulas after all units to distinguish the two substances being measured.)

* * * * *


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Balance: 1H3AsO4 + 3NaOH 3HOH + Na3AsO4

(In hydroxide neutralization calculations, the exact non-water product formula is not

needed. The bridge ratio is based on balancing to form H2O. The product formula

Na3AsO4 is a best guess for an unfamiliar ion; additional reactions might occur.)

Bridge: 1mol H3AsO4 = 3mol NaOH

SOLVE: (Want a single unit?).

* * * * *

?mL NaOH = 2.34g H3AsO4 1 mol H3AsO4 3 mol NaOH 1 L NaOH 1mL =

141.9 g H3AsO4 1 mol H3AsO40.200 mol NaOH 103L

= 247 mL NaOH soln.

Lets review the logic of why we do these steps. We want to know the mL of base that willneutralize a given amount of an acid.

From the grams of the acid and its g/mol, we can find the moles of acid in thesample.

From the balanced equation, we can find the moles of base that neutralize the acid.

From the moles of base needed and its mol/L, we can find the liters and mL ofbase that are needed to exactly neutralize the acid.

The goal of learning is to be able to solve problems. The steps of stoichiometry allow us toprecisely solve problems that are frequently encountered in medicine, engineering, andscience.

Practice A

1. If all of the acidic hydrogens in a 2.00 M H3PO4solution are neutralized by 1.50 liters of

0.500 M KOH, how many mL of H3PO4solution were neutralized?

Titration Terminology

Titrationis an experimental technique that can supply the data needed for stoichiometrycalculations. In titration, calibrated buretsare used to precisely measure the amounts ofsolution added as a chemical reaction takes place.

Indicatorsare dyes used in titration that change color at the instant the molesof tworeacting particles are equal or have reacted in a simple-whole-number ratio. When thisequivalence point is reached, a change in indicator color signals the endpointof thetitration.

Anacid-base titrationis simply a neutralization in which the amounts of acid and base arecarefully measured. In neutralization, if one or both of the reactants is a strongacid or base,


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the reaction goes to completion: it will proceed until the limiting reactant is completelyused up.

When titrating a weakacid or base, the opposite solution must be strong, and a carefulselection of the indicator dye will be required to show a sharp endpoint. However, for allacids and bases, if one or both is strong, stoichiometry calculations can be done using the

same steps.

Acid-Hydroxide Titration Calculations

In the titration of an acid and a hydroxide, because OH is a strong base, the reaction goesto completion. The acid and base ions react as soon as they are mixed, and as additionalsolution is added, the reaction continues to occur until the equivalence point is reached. Atthat point, the indicator changes color.

For an acid with one acidic hydrogen, at the endpoint the moles of H +supplied by the acid

solution equal the moles of OHsupplied by the base solution.

At an acid-hydroxide endpoint,moles H+from acid= moles OHfrom base

For acids that contain more than one acidic proton, neutralization produces a series ofequivalence points. For many of these polyproticacids, by carefully selecting anappropriate indicator, it may be possible to titrate to an equivalence points for differentnumbers of acidic hydrogens in the acid.

At the endpoint of the titration, you should assume that the moles of acid and base havebeen mixed in the exact ratio shown by the coefficients of the balanced equation for theneutralization, and both are exactly used up (at the uncertainty indicated by the significantfigures in the data). At this point, the acid and base are stoichiometrically equivalent: bothreactants are limiting. Calculations using titrated amounts can then be solved byconversion stoichiometry.

For calculations involving areactionto an equivalencepointor titrationto an endpoint, useconversion stoichiometry. Start with WDBB.

Practice B

1. A 25.0 mL sample of 0.145 M HCl is titrated by a 0.200 M OHsolution. How many

milliliters of the base must be added to reach the point where the indicator changes

color?

2. Oxalic acid (H2C2O4, 90.03 gmol1

) is a solid at room temperature. Each oxalic acidparticle contains two acidic hydrogens. A sample of oxalic acid is titrated by sodium

hydroxide, and an indicator is chosen that changes color at the point when both of the

acidic hydrogens are neutralized. How many moles of NaOH are needed to titrate

0.100 g of oxalic acid crystals?

* * * * *


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ANSWERS

Practice A

1. WANT: ? mL H3PO4 (you want a single unit)

DATA: 2.00 mol H3PO

4 = 1 L H

3PO

4(M prompt)

1.50 L KOH soln. (the single unit given)

0.500 mol KOH = 1 L KOH (M prompt)

Since all volumes in this problem are for aqueous solutions, you may label one volume as soln. butomit the other solution labels after volumes as understood.)

Balance: 1 H3PO4 + 3 KOH 3H2O + 1 K3PO4

Bridge: 1 mol H3PO4 = 3 mol KOH

SOLVE:

* * * * *

? mL H3PO4 = 1.50 L KOH 0.500 mol KOH 1 mol H3PO4 1 L H3PO4 1mL =

1 L KOH 3 mol KOH 2.00 mol H3PO4 103L

= 125 mL H3PO4

Practice B

1. WANT : ? mL OHsolution

DATA: 25.0 mL HCl (Single-unit given)

0.145 mol HCl = 1 L HCl (M prompt)

0.200 mol OH= 1 L OH (M prompt)

Balance: 1 HCl + 1 OH 1 H2O + 1 Cl

Bridge: 1 mol HCl = 1 mol OH

SOLVE:

* * * * *

? mL OH= 25.0 mL HCl 103L 0.145 molHCl 1 mol OH 1 L OH 1mL =

1 mL 1 L HCl 1 mol HCl 0.200 mol OH 103L

= 18.1 mL OH solution

2. WANT: ? mol NaOHDATA: 0.0500 mol NaOH = 1 L NaOH

0.100 gH2C2O4 (Single-unit given)

90.03 gH2C2O4 = 1 molH2H2C2O4

Balance: 1 H2C2O4 + 2 NaOH 2 H2O + Na2C2O4 (2 acidic hydrogens)

Bridge: 1 mol H2C2O4 = 2 mol NaOH


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SOLVE:

? mol NaOH = 0.100 g H2C2O4 1 molH2C2O4 2 mol NaOH = 2.22x 103mol NaOH

90.03 g H2C2O4 1 mol H2C2O4

* * * * *

Lesson 14D: Solving Neutralization For Ratio Units

In Lesson 14C, we used neutralization and titration data to solve for a WANTED singleunit. The data gathered in neutralization can also be used to solve for ratio units. Our goalis to answer questions such as

Given a solution of an acid or base, how can we determine its concentration?

Given a sample of an unknown acid or base, how can we get close to proving itsidentity by calculating its molar mass?

We can answer these questions by performing neutralization titration in the laboratory,

then applying the steps of conversion stoichiometry to our laboratory data.

For a reaction that goes to completion, if a ratiounit is WANTED, lets review the rulesfrom Lesson 12D.

Solving Reaction Calculations For A Ratio

The Process

When a characteristic ratio for a solution or substance is WANTED, a sample isreacted, amounts of each substance are measured at an equivalence point, and theWANTED ratio is calculated. The ratio for this sample will be true for any sample ofthe solution or substance.

The Calculation Steps:

If a ratio-unitis WANTED, all of the DATA will be in equalities, and one of theequalities will be an equivalency. Write WDBB (steps 1-4) and then

Step 5. Label the equivalency in the DATA.

Step 6. Solve separately for the top and bottom WANTED amounts.

a. Solve for the WANTED unit that is notmoles first. Use as agiventhe sideof the equivalency that is an amount of the WANTED formula.

b. Solve for molesWANTED using single-unit stoichiometry with the otherhalf of the equivalency as yourgiven.

Step 7. Divide the two WANTED amounts to find the WANTED ratio.

In the following calculations, the product formulas are not supplied, but by applyng therules for acid-hydroxide balancing, you will be able to write the key bridge conversion thatallows us to solve stoichiometry. Using the steps above, try this example.


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Q. An unknown acid has three acidic hydrogens (call it H3R). For titration an

indicator is chosen that changes color at the point where all three acidic

hydrogens are neutralized. If 19.29 mL of 0.120 M KOH is needed to titrate 0.1484

grams of the acid to an endpoint, what is the molar mass of the acid?

If you get stuck, read aportionof the answer until you are unstuck, then try again.

* * * * *

Answer

1. WANT: ? g H3R (write the unitWANTED)mol H3R

The unit of molar mass is g/mol. For data about two substances, label each unitwiththe substance it is measuring. You WANT grams of acidpermole of acid.

2. DATA: 0.1484 g H3R = 19.29 mL KOH (2A-R: equivalentat endpt.)

0.120 mol KOH = 1 L KOH

3. Balance. 1 H3R + 3 KOH 3H2O + K3R

4. Bridge. 1mol H3R = 3 mol KOH

If needed, adjust your work and finish from here.

* * * * *

5. Label the equivalence (see above.)

6. a. In this WANTED ratio, the amount notmoles is g acid. The grams of acid in thesample are supplied in the equivalency. That was easy.

(not moles:) ? g H3R in sample = 0.1484 g H3R

* * * * *6. b. Start: ? molH3R= 19.29 mL KOH (other half of equivalency)

To find the moles of acid in the sample, we use single-unit stoichiometry. First findthe molesof basethat reacted with the acid. Then use the balanced equation to findthe moles of acid that reacted with the base.

? mol WANTED = unit base >> mol base >> mol WANTED (stop)

? mol H3R= 19.29 mL KOH 103L 0.120 mol KOH 1 molH3R =

1 mL 1 L KOH 3 mol KOH

= 7.716x 104

mol H3R = the moles of acidin the sample, based on themoles of KOH that reacted with the acid.

* * * * *

7. You want the ratio of the grams of acid and moles of acid in the sample that reacted.Apply the fundamental rule: let the units tell you what to do.


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? gH3R = gH3Rin sample = 0.1484 gH3R = 192 gH3R

mol H3R mol H3R in sample 7.716x 104mol H3R mol H3R

The molar mass of a substance is constant: a ratio that is characteristic. The concentrationof a well-mixed solution is another ratio that is constant. In cases where a ratio WANTED

is constant, if amounts of the two WANTED units in a samplecan be found, that ratio willbe true for all samples.

Stoichiometry and CSI

Consider what we were able to determine in the above calculation.

Many molecules that are important in chemical and biological systems, including legal andillegal drugs, are acids or bases. In the above problem, we did not know the chemicalformula for the acid. However, by finding the ratio of thegramsof acid per moleof the acid(its molar mass), we can get closeto identifying the acid.

The acid can be weighed to find thegramsin a sample, but nearly always we do not have away to count the number of particles (the moles) in a sample directly. However, for acids

and bases, we can determine the moles in a sample by titrating with an acid or basesolution of a known concentration and applying stoichiometry.

In titration, when the indicator changes color, the molesof two reacting particles are eitherequal or in a whole-number ratio.

Knowing the grams and moles in a sample, we can find the grams to moles ratio: the molarmass. From the molar mass of a substance and its melting point or other physicalconstants, tables in chemistry reference media will identify many unknown substanceswith near certainty.

The ability to identify an unknown acid or base is a skill that is marketable in forensiccriminology, medicine, and other rewarding careers.

Practice. Do Problems 1 and 2 today. Save Problem 3 for your next study session.

1. If 21.82 mL of 0.110 M HCl is required to titrate a 25.00 mL sample of KOH solution toan endpoint, what is the [base]?

2. In a titration of an organic acid that has two acidic hydrogens (call it H2R), an

indicator is selected that changes color after oneof the two acidic hydrogens is

neutralized. An endpoint is reached at 31.22 mL of 0.160 M NaOH. If the mass of the

acid sample was 1.380 grams, what is the molar mass of the acid?

3. Review Lessons 14A-D and prepare flashcards that cover fundamentals.


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ANSWERS

1. 1. WANT: ? mol KOH (KOH is the base)L KOH

2. DATA: 25.00 mL of KOH = 21.82 mL HCl (Equivalent at endpoint)

0.110 mol HCl = 1 L HCl soln. (M prompt)

3. Balance: 1 HCl + 1 KOH 1 H2O + 1 KCl

4. Bridge: 1 mol HCl = 1 mol KOH

In stoichiometry, if a ratiounit is WANTED, solve for the topand bottomWANTED units separately usingthe following steps.

5. Label the equivalency in the DATA equalities.

6. Solve separately for the two WANTED units.

6a. Solve for the WANTED unit that is not moles, using as a giventhe side of the equivalency with theWANTED formula.

* * * * *

(not moles:) ?L KOH = 25.00 mL KOH = 25.00 x 103L KOH

6b. Solve for moles of base in the sample based on the moles of acidthat reacted: start with the otherhalfof the equivalencyas your given.

* * * * *

? mol KOH = 21.82 mL HCl 103L 0.110 mol HCl 1 mol KOH = 2.40 x 103molKOH

1 mL 1 L HCl soln. 1 mol HCl

7. Find the WANTED ratio. Let the WANTED units tell you how to arrange the amounts found in Step 6.

* * * * *

? molKOH = 2.400 x 103 molKOH = 0.0960 molKOH

L KOH 25.00 x 103 L KOH L KOH

2. 1. WANT: ? g H2R (a ratio unit)

mol H2R

* * * * *

2. DATA: 1.380 g H2R = 31.22 mL NaOH (2A+R: equivalent when reaction stops)

0.160 mol NaOH = 1 L NaOH (M prompt)

Strategy: Titration to an endpoint means stoichiometrically equivalent amounts. To solve, use

conversion stoichiometry.

3. Balance: 1H2R + 1 NaOH 1 H2O + 1 NaHR (Only the first H is neutralized)

4. Bridge: 1 mol H2R = 1 mol NaOH (Use the mole to mole coefficients of the reaction )

If needed, adjust your work and finish.

* * * * *


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In stoichiometry, if a ratiounit is WANTED, find the topand bottomWANTED units separately .

5. Label the one equivalency in the DATA equalities.

6a. Solve for the WANTED unit that is not moles, using as a giventhe side of the equivalency with theWANTED formula.

* * * * *

(WANTED non-moles:) ? g H2R =1.380 g H2R (the grams of acid in the sample)

6b. (Solve for moles WANTED. Start with the otherhalf of the equivalencyas your given.)

* * * * *

? mol H2R = 31.22 mL NaOH 103L 0.160 mol NaOH 1 molH2R = 4.995 x 103mol H2R

1 mL 1 L NaOH 1 mol NaOH

7. Find the WANTED ratio. Let the WANTED units tell you how to arrange the amounts found in Step 6.

* * * * *

? g H2R = 1.380 g H2R = 276 g H2Rmol H2R 4.995 x 10

3mol H2R mol H2R

3. Your flashcards might include the following.One-way cards (with notch) Back Side -- Answers

Names and formulas for 3 strong acidsHydrochloric: HCl, sulfuric: H2SO4,

nitric: HNO3

Ion symbol for a proton H+

At the endpoint of acid-hydroxide neutralization moles H+reacted = moles OHreacted

Formula for an unknown base with 3 hydroxides Unk(OH)3

Formula for an unknown with 2 acidic protons H2Unk

Solve calculations to an endpoint with Conversion stoichiometry

Two-way cards (withoutnotch):

An acid A substance that produces H+in water

A base A s

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Calculations in Chemistry - CH. 13-14

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