HOMEWORK

11

HOMEWORK

Page 704

Day 1

#29, 31, 35, 37, 39, 43, 44, 45

Day 2

#47, 49, 51, 53, 57, 61, 63

Day 3

#65, 67, 69, 71, 75, 77, 79,83, 97,89

Day 4

#93,95,97.99.101,111,113,117

22The Chemistry The Chemistry of Acids and of Acids and BasesBases

The Chemistry The Chemistry of Acids and of Acids and BasesBases

33

Acid and BasesAcid and Bases

44


55


66Acids

Have a sour taste. Vinegar is a solution of acetic acid. CitrusHave a sour taste. Vinegar is a solution of acetic acid. Citrusfruits contain citric acid.fruits contain citric acid.

React with certain metals to produce hydrogen gasReact with certain metals to produce hydrogen gas..

React with carbonates and bicarbonates to produce carbon React with carbonates and bicarbonates to produce carbon dioxide gasdioxide gas

Have a bitter taste.Have a bitter taste.

Feel slippery. Many soaps contain bases.Feel slippery. Many soaps contain bases.

Bases

77

Some Properties of Acids

Produce H+ (as H3O+) ions in water (the hydronium ion is a

hydrogen ion attached to a water molecule)

Taste sour

Corrode metals

Electrolytes

React with bases to form a salt and water

pH is less than 7

Turns blue litmus paper to red “Blue to Red A-CID”

88

Some Properties of Bases

Produce OHProduce OH-- ions in water ions in water

Taste bitter, chalkyTaste bitter, chalky

Are electrolytesAre electrolytes

Feel soapy, slipperyFeel soapy, slippery

React with acids to form salts and waterReact with acids to form salts and water

pH greater than 7pH greater than 7

Turns red litmus paper to blue “Turns red litmus paper to blue “BBasic asic BBlue”lue”

99

Some Common Bases

NaOHNaOH sodium hydroxidesodium hydroxide lyelye

KOHKOH potassium hydroxidepotassium hydroxide liquid soapliquid soap

Ba(OH)Ba(OH)22 barium hydroxidebarium hydroxide stabilizer for plasticsstabilizer for plastics

Mg(OH)Mg(OH)22 magnesium hydroxidemagnesium hydroxide “MOM” Milk of magnesia“MOM” Milk of magnesia

Al(OH)Al(OH)33 aluminum hydroxidealuminum hydroxide Maalox (antacid)Maalox (antacid)

1010

Acid/Base definitions

• Definition #1: Arrhenius (traditional)

Acids – produce H+ ions (or hydronium ions H3O+)

Bases – produce OH- ions

(problem: some bases don’t have hydroxide ions!)

1111Arrhenius acid is a substance that produces H+ (H3O+) in water

Arrhenius base is a substance that produces OH- in water

1212

Acid/Base Definitions

• Definition #2: Brønsted – Lowry

Acids – proton donor

Bases – proton acceptor

A “proton” is really just a hydrogen atom that has lost it’s electron!

1313

A Brønsted-Lowry acid is a proton donorA Brønsted-Lowry base is a proton acceptor

acidconjugate

basebase conjugate

acid

1414

ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIES

The Brønsted definition means NHThe Brønsted definition means NH33 is is aa BASEBASE in water — and water is in water — and water is itself anitself an ACIDACID

BaseAcidAcidBaseNH4

+ + OH-NH3 + H2OBaseAcidAcidBase

NH4+ + OH-NH3 + H2O

1515

Conjugate PairsConjugate Pairs

1616

Learning Check!

Label the acid, base, conjugate acid, and Label the acid, base, conjugate acid, and conjugate base in each reaction:conjugate base in each reaction:

HCl + OHHCl + OH-- Cl Cl-- + H + H22OO HCl + OHHCl + OH-- Cl Cl-- + H + H22OO

HH22O + HO + H22SOSO44 HSO HSO44-- + H + H33OO

++ HH22O + HO + H22SOSO44 HSO HSO44-- + H + H33OO

++

1717Acids & Base Acids & Base DefinitionsDefinitions

Lewis acid - a Lewis acid - a substance that substance that accepts an electron accepts an electron pairpair

Lewis base - a Lewis base - a substance that substance that donates an electron donates an electron pairpair

Definition #3 – Lewis Definition #3 – Lewis

1818

Formation ofFormation of hydronium ion hydronium ion is also an is also an excellent example.excellent example.

Lewis Acids & BasesLewis Acids & Bases

•Electron pair of the new O-H bond Electron pair of the new O-H bond originates on the Lewis base.originates on the Lewis base.

HH

H

BASE

••••••

O—HO—H

H+

ACID

1919

Lewis Acid/Base ReactionLewis Acid/Base Reaction

2020

Lewis Acid-Base Lewis Acid-Base Interactions in BiologyInteractions in Biology

• The heme group The heme group in hemoglobin in hemoglobin can interact with can interact with OO22 and CO. and CO.

• The Fe ion in The Fe ion in hemoglobin is a hemoglobin is a Lewis acidLewis acid

• OO22 and CO can act and CO can act as Lewis basesas Lewis bases

Heme group

2121The The pH scalepH scale is a way of is a way of expressing the strength expressing the strength of acids and bases. of acids and bases. Instead of using very Instead of using very small numbers, we just small numbers, we just use the NEGATIVE use the NEGATIVE power of 10 on the power of 10 on the Molarity of the HMolarity of the H++ (or (or OHOH--) ion.) ion.

Under 7 = acidUnder 7 = acid 7 = neutral 7 = neutral

Over 7 = baseOver 7 = base

2222

pH Scale

2323

pH of Common pH of Common SubstancesSubstances

2424

Strong Acids:Perchloric HClO4

Chloric, HClO3

Hydrobromic, HBrHydrochloric, HClHydroiodic, HINitric, HNO3

Sulfuric, H2SO4

Strong Acids:100% ionized (completely dissociated) in water.

HCl + H2O H3O+ + Cl-

2525What is a strong Base?

A base that is completely dissociated in water (highly soluble).

NaOH(s) Na+ + OH-

Strong Bases:

Group 1A metal hydroxides(LiOH, NaOH, KOH,RbOH, CsOH)

Heavy Group 2A metal hydroxides[Ca(OH)2, Sr(OH)2, andBa(OH)2]

2626

Weak Acids & Bases: “The Rest”

2727

Strong Acids:100% ionized (completely dissociated) in water.

HCl + H2O H3O+ + Cl-

Note the “one way arrow”.

Weak Acids:Only a small % (dissociated) in water.

HC2H3O2 + H2O H3O+ + C2H3O2

-

Note the “2-way” arrow.

Why are they different?

2828

Strong Acids:

HCl HCl HClHCl HCl

ADD WATER to MOLECULAR ACID

(H2O)

2929

Strong Acids:

(H2O)H3O+

H3O+

H3O+

H3O+

H3O+

Cl-

Cl-

Cl-

Cl-

Cl-

Note: No HCl molecules remain in solution, all have been ionized in water.

3030

HC2H3O2

HC2H3O2

HC2H3O2

HC

2H3O

2

HC2H3O2

H30

+ C2H3O2-

(H2O)

Weak Acid Ionization:

Note: At any given time only a small portion of the acid molecules are ionized and since reactions are running in BOTH directions the mixture composition stays the same.

This gives rise to an Equilbrium expression, Ka

H30+ C2H3O2

-HC2H3O2

3131

HNO3, HCl, H2SO4 and HClO4 are among the only known strong acids.

Strong and Weak Strong and Weak Acids/BasesAcids/Bases

The strength of an acid (or base) is determined by the amount of IONIZATION.

The strength of an acid (or base) is determined by the amount of IONIZATION.

3232


• Generally divide acids and bases into STRONG or Generally divide acids and bases into STRONG or WEAK ones.WEAK ones.

STRONG ACID:STRONG ACID: HNOHNO3 3 (aq) + H(aq) + H22O (l) --->O (l) --->

HH33OO+ + (aq) + NO(aq) + NO33- - (aq)(aq)

HNOHNO33 is about 100% dissociated in water. is about 100% dissociated in water.

3333

• Weak acidsWeak acids are much less than 100% ionized in are much less than 100% ionized in

water.water.

One of the best known is acetic acid = CHOne of the best known is acetic acid = CH33COCO22HH



3434

• Strong Base:Strong Base: 100% dissociated in 100% dissociated in water.water.

NaOH (aq) ---> NaNaOH (aq) ---> Na+ + (aq) + OH(aq) + OH- - (aq)(aq)


Other common strong Other common strong bases include KOH andbases include KOH and Ca(OH)Ca(OH)22..

CaO (lime) + HCaO (lime) + H22O -->O -->

Ca(OH)Ca(OH)22 (slaked lime) (slaked lime)CaOCaO

3535

More About WaterMore About Water

HH22O can function as both an ACID and a BASE.O can function as both an ACID and a BASE.

In pure water there can beIn pure water there can be AUTOIONIZATIONAUTOIONIZATION

Equilibrium constant for water = KEquilibrium constant for water = Kww

KKww = [H = [H33OO++] [OH] [OH--] =] = 1.00 x 101.00 x 10-14-14 at 25 at 25 ooCC

3636

More About WaterMore About Water

KKww = [H = [H33OO++] [OH] [OH--] = 1.00 x 10] = 1.00 x 10-14-14 at 25 at 25 ooCC

In a In a neutral neutral solution [Hsolution [H33OO++] = [OH] = [OH--]]

so Kso Kww = [H = [H33OO++]]22 = [OH = [OH--]]22

and so [Hand so [H33OO++] = [OH] = [OH--] = 1.00 x 10] = 1.00 x 10-7-7 M M

OH-

H3O+

OH-

H3O+

AutoionizationAutoionization

3737

Write equations for the following

• 1. sulfuric acid in water

• 2. ammonia in water

• 3. calcium hydroxide in water

• 4. hydroflouric acid in water

• 5. hydrogen sulfide in water

3838Calculating the pH

pH = - log [H+](Remember that the [ ] mean Molarity)

Example: If [H+] = 1 X 10-10

pH = - log 1 X 10-10

pH = - (- 10)

pH = 10

Example: If [H+] = 1.8 X 10-5

pH = - log 1.8 X 10-5

pH = - (- 4.74)

pH = 4.74

3939

Try These!Try These!

Find the pH of Find the pH of these:these:

1) A 0.15 M solution 1) A 0.15 M solution of Hydrochloric of Hydrochloric acidacid

2) A 3.00 X 102) A 3.00 X 10-7-7 M M solution of Nitric solution of Nitric acidacid

4040pH calculations – Solving for pH calculations – Solving for H+H+pH calculations – Solving for pH calculations – Solving for H+H+

If the pH of Coke is 3.12, [HIf the pH of Coke is 3.12, [H++] = ???] = ???

Because pH = - log [HBecause pH = - log [H++] then] then

- pH = log [H- pH = log [H++]]

Take antilog (10Take antilog (10xx) of both) of both sides and get sides and get

1010-pH -pH == [H[H++]][H[H++] = 10] = 10-3.12-3.12 = 7.6 x 10 = 7.6 x 10-4-4 M M *** to find antilog on your calculator, look for “Shift” or “2*** to find antilog on your calculator, look for “Shift” or “2nd nd

function” and then the log buttonfunction” and then the log button

4141pH calculations – Solving for pH calculations – Solving for H+H+

• A solution has a pH of 8.5. What is the A solution has a pH of 8.5. What is the Molarity of hydrogen ions in the Molarity of hydrogen ions in the solution?solution?

4242pOH

• Since acids and bases are Since acids and bases are opposites, pH and pOH are opposites, pH and pOH are opposites!opposites!

• pOH does not really exist, but it is pOH does not really exist, but it is useful for changing bases to pH.useful for changing bases to pH.

• pOH looks at the perspective of a pOH looks at the perspective of a basebase

pOH = - log [OHpOH = - log [OH--]]Since pH and pOH are on opposite Since pH and pOH are on opposite

ends,ends,pH + pOH = 14pH + pOH = 14

4343

pHpH [H+][H+] [OH-][OH-] pOHpOH

4444

[H[H33OO++], [OH], [OH--] and pH] and pHWhat is the pH of the What is the pH of the

0.0010 M NaOH solution? 0.0010 M NaOH solution?

[OH-] = 0.0010 (or 1.0 X 10[OH-] = 0.0010 (or 1.0 X 10-3-3 M) M)

pOH = - log 0.0010pOH = - log 0.0010

pOH = 3pOH = 3

pH = 14 – 3 = 11pH = 14 – 3 = 11

OR KOR Kww = [H = [H33OO++] [OH] [OH--]]

[H[H3OO++] = 1.0 x 10] = 1.0 x 10-11-11 M M

pH = - log (1.0 x 10pH = - log (1.0 x 10-11-11) = 11.00) = 11.00

4545The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H+ ion concentration of the rainwater?

The OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is the pH of the blood?

4646

[OH[OH--]]

[H[H++]] pOHpOH

pHpH

1010 -pOH

-pOH

1010 -pH-pH-Log[H

-Log[H++]]

-Log[OH

Log[OH

--]]

14 -

pOH

14 -

pOH

14 -

pH

14 -

pH

1.0

x 10

1.0

x 10-1

4-14

[OH[O

H-- ]]

1.0

x 10

1.0

x 10-1

4-14

[H[H

++ ]]

4747Calculating [H3O+], pH, [OH-], and pOH

A chemist dilutes concentrated hydrochloric acid to make two solutions: (a) 3.0 M and (b) 0.0024 M. Calculate the [H3O+], pH, [OH-], and pOH of the two solutions at 25°C.

4848Calculating [H3O+], pH, [OH-], and pOH

What is the [H3O+], [OH-], and pOH of a solution with pH = 3.67? Is this an acid, base, or neutral?

4949

• Weak base:Weak base: less than 100% ionized less than 100% ionized in waterin water

One of the best known weak bases is One of the best known weak bases is ammoniaammonia

NHNH3 3 (aq) + H(aq) + H22O (l) O (l) NH NH44+ + (aq) + OH(aq) + OH- - (aq)(aq)



5050

Weak BasesWeak Bases

5151

Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases

Consider acetic acid, HCConsider acetic acid, HC22HH33OO22 (HOAc) (HOAc)

HCHC22HH33OO22 + H + H22O O H H33OO++ + C + C22HH33OO22 --

AcidAcid Conj. base Conj. base

Ka [H3O+][OAc- ]

[HOAc] 1.8 x 10-5Ka

[H3O+][OAc- ][HOAc]

1.8 x 10-5

(K is designated K(K is designated Kaa for ACID) for ACID)

K gives the ratio of ions (split up) to molecules K gives the ratio of ions (split up) to molecules

(don’t split up) (don’t split up)

5252Ionization Constants for Ionization Constants for Acids/Bases Acids/Bases

AcidsAcids ConjugateConjugateBasesBases

Increase strength

Increase strength

5353

Equilibrium Constants Equilibrium Constants for Weak Acidsfor Weak Acids

Equilibrium Constants Equilibrium Constants for Weak Acidsfor Weak Acids

(Yes, all weak acids do this – DO NOT endeavor to (Yes, all weak acids do this – DO NOT endeavor to make this complicated!)make this complicated!)

Weak acid has KWeak acid has Kaa < 1 < 1

Leads to small [HLeads to small [H33OO++] and a pH of 2 - 7] and a pH of 2 - 7

Ka = XKa = X22 / or-x x = H / or-x x = H++ and conjugate base and conjugate base

5454

Calculate the H+ and pH for 0.250 M

HC2H3O3

5555Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid

Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid

You have 1.00 M HOAc. Calc. the You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hequilibrium concs. of HOAc, H33OO++, OAc, OAc--, ,

and the pH.and the pH.



You have 1.00 M HOAc. Calc. the You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hequilibrium concs. of HOAc, H33OO++, OAc, OAc--, ,

and the pH.and the pH.

Step 1.Step 1. Define equilibrium concs. in ICE Define equilibrium concs. in ICE

table.table.

[HOAc][HOAc] [H[H33OO++]] [OAc[OAc--]]

initialinitial

changechange

equilibequilib

1.001.00 00 001.001.00 00 00

-x-x +x+x +x+x-x-x +x+x +x+x

1.00-x1.00-x xx xx1.00-x1.00-x xx xx

5757Reaction of Weak Bases with Reaction of Weak Bases with WaterWater

The generic reaction for a base reacting with The generic reaction for a base reacting with water, producing its conjugate acid and water, producing its conjugate acid and hydroxide ion:hydroxide ion:

B + H2O BH+ + OH-

[ ][ ]

[ ]b

BH OHK

B

(Yes, all weak bases do this – DO NOT(Yes, all weak bases do this – DO NOTendeavor to make this complicated!)endeavor to make this complicated!)Kb = XKb = X22 /or-x X = OH /or-x X = OH- - and conjuate acid and conjuate acid

5858

Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases

Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases

Weak base has KWeak base has Kbb < 1 < 1

Leads to small [OHLeads to small [OH--] and a pH of 12 - 7] and a pH of 12 - 7

5959NOTE•Ka x Kb = Kw

•pH + pOH = 14

•H3O+ x OH- = 1 x 10-14

6060NOTE• On the AP test you might get the Ka

but need the Kb. If you are given an acid, you need Ka. If given a base, you need Kb. You must know if you are dealing with an acid or base.

6161

Relation Relation

of Kof Kaa, K, Kbb, ,

[H[H33OO++] ]

and pHand pH



Step 2.Step 2. Write KWrite Kaa expression expression

You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hof HOAc, H33OO++, OAc, OAc--, and the pH., and the pH.

Ka 1.8 x 10-5 = [H3O+][OAc- ]

[HOAc]

x2

1.00 - xKa 1.8 x 10-5 =

[H3O+][OAc- ][HOAc]

x2

1.00 - x

This is a quadratic. Solve using quadratic This is a quadratic. Solve using quadratic formula.formula.

or you can make an approximation if x is very or you can make an approximation if x is very small! (Rule of thumb: 10small! (Rule of thumb: 10-5-5 or smaller is ok) or smaller is ok)or you can make an approximation if x is very or you can make an approximation if x is very small! (Rule of thumb: 10small! (Rule of thumb: 10-5-5 or smaller is ok) or smaller is ok)



Step 3.Step 3. Solve KSolve Kaa expression expression


Ka 1.8 x 10-5 = [H3O+][OAc- ]

[HOAc]

x2

1.00 - xKa 1.8 x 10-5 =

[H3O+][OAc- ][HOAc]

x2

1.00 - x

First assume x is very small because First assume x is very small because KKaa is so small. is so small.

Ka 1.8 x 10-5 = x2

1.00Ka 1.8 x 10-5 =

x2

1.00

Now we can more easily solve this Now we can more easily solve this approximate expression.approximate expression.

6464

Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid

Step 3.Step 3. Solve KSolve Kaa approximateapproximate expressionexpression


Ka 1.8 x 10-5 = x2

1.00Ka 1.8 x 10-5 =

x2

1.00

x =x = [[HH33OO++] = [] = [OAcOAc--] = 4.2 x 10] = 4.2 x 10-3-3 M M

pH = - log [pH = - log [HH33OO++] = -log (4.2 x 10] = -log (4.2 x 10-3-3) =) = 2.372.37



Calculate the pH of a 0.0010 M solution of Calculate the pH of a 0.0010 M solution of formic acid, HCOformic acid, HCO22H.H.



Calculate the pH of a 0.0010 M solution of Calculate the pH of a 0.0010 M solution of formic acid, HCOformic acid, HCO22H.H.

HCOHCO22H + HH + H22O O HCO HCO22-- + H + H33OO++

KKaa = 1.8 x 10 = 1.8 x 10-4-4

Approximate solutionApproximate solution

[H[H33OO++] = 4.2 x 10] = 4.2 x 10-4-4 M, M, pH = 3.37pH = 3.37

Exact SolutionExact Solution

[H[H33OO++] = [HCO] = [HCO22--] = 3.4 x 10] = 3.4 x 10-4-4 M M

[HCO[HCO22H] = 0.0010 - 3.4 x 10H] = 0.0010 - 3.4 x 10-4-4 = 0.0007 M = 0.0007 M

pH = 3.47 pH = 3.47

6767Equilibria Involving A Weak Equilibria Involving A Weak BaseBase

You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.

NHNH33 + H + H22O O NH NH44++ + OH + OH--




KKbb = 1.8 x 10 = 1.8 x 10-5-5

Step 1.Step 1. Define equilibrium concs. in ICE tableDefine equilibrium concs. in ICE table

[NH[NH33]] [NH[NH44++]] [OH[OH--]]

initialinitial

changechange

equilibequilib

0.0100.010 00 000.0100.010 00 00


0.010 - x0.010 - x x x xx0.010 - x0.010 - x x x xx




KKbb = 1.8 x 10 = 1.8 x 10-5-5


[NH[NH33]] [NH[NH44++]] [OH[OH--]]

initialinitial

changechange

equilibequilib

0.0100.010 00 000.0100.010 00 00


0.010 - x0.010 - x x x xx0.010 - x0.010 - x x x xx




KKbb = 1.8 x 10 = 1.8 x 10-5-5


[NH[NH33]] [NH[NH44++]] [OH[OH--]]

initialinitial

changechange

equilibequilib

0.0100.010 00 000.0100.010 00 00


0.010 - x0.010 - x x x xx0.010 - x0.010 - x x x xx

7171Reaction of Weak Bases with Reaction of Weak Bases with WaterWater

The base reacts with water, producing its conjugate acid The base reacts with water, producing its conjugate acid and hydroxide ion:and hydroxide ion:

NHNH33 + H + H22O O NH NH44++ + OH + OH- - KKbb = 1.8 x 10 = 1.8 x 10-5-5

Kb = [NH4+][OH-

] [NH3]




KKbb = 1.8 x 10 = 1.8 x 10-5-5

Step 2.Step 2. Solve the equilibrium expressionSolve the equilibrium expression

Assume x is small, soAssume x is small, so x = [OHx = [OH--] = [NH] = [NH44

++] = 4.2 x 10] = 4.2 x 10-4-4 M M

and [NHand [NH33] = 0.010 - 4.2 x 10] = 0.010 - 4.2 x 10-4-4 ≈ 0.010 M ≈ 0.010 M

The approximation is validThe approximation is valid !!




KKbb = 1.8 x 10 = 1.8 x 10-5-5

Step 3.Step 3. Calculate pHCalculate pH

[OH[OH--] = 4.2 x 10] = 4.2 x 10-4-4 M M

so pOH = - log [OHso pOH = - log [OH--] = 3.37] = 3.37

Because pH + pOH = 14,Because pH + pOH = 14,

pH = 10.63pH = 10.63

7474

Types of Acid/Base Reactions: Types of Acid/Base Reactions: SummarySummary

7575pH testing

• There are several ways to test pHThere are several ways to test pH

–Blue litmus paper (red = acid)Blue litmus paper (red = acid)

–Red litmus paper (blue = basic)Red litmus paper (blue = basic)

–pH paper (multi-colored)pH paper (multi-colored)

–pH meter (7 is neutral, <7 acid, >7 pH meter (7 is neutral, <7 acid, >7 base)base)

–Universal indicator (multi-colored)Universal indicator (multi-colored)

– Indicators like phenolphthaleinIndicators like phenolphthalein

–Natural indicators like red cabbage, Natural indicators like red cabbage, radishesradishes

7676Paper testing

• Paper tests like litmus paper and pH Paper tests like litmus paper and pH paperpaper

– Put a stirring rod into the solution Put a stirring rod into the solution and stir.and stir.

– Take the stirring rod out, and Take the stirring rod out, and place a drop of the solution from place a drop of the solution from the end of the stirring rod onto a the end of the stirring rod onto a piece of the paperpiece of the paper

– Read and record the color change. Read and record the color change. Note what the color indicates. Note what the color indicates.

– You should only use a small You should only use a small portion of the paper. You can use portion of the paper. You can use one piece of paper for several one piece of paper for several tests.tests.

7777pH paperpH paper

7878Behavior of Salts (Acidic or Basic)

7979Behavior of Salts Acidic, basic, or neutral?NaFNaOClKBrNH4FLi2SLi2CO3

Fe(NO3 )3

8080

NOTE•Ka > Kb acidic

•Kb < Ka basic

•Ka = Kb neutral

8181Behavior of Salts WRITE EQUATIONS to show WHY THE FOLLOWING ARE ACIDIC, BASIC, OR

NEUTRAL.NaFNaOClKBrNH4NO2

NH4FLi2SZnCl2Li2CO3

8282

Amphoteric Salts

8383

% DISSOCIATION OF ACIDS & Bases

• % Diss of an acid = H+ / original conc x 100

• % Diss of a base = OH- / original conc x 100

8484Determine the pH, pOH, hydrogen ion concentration, and hydroxideion concentration for 0.125 M Na2 CO3.

8585Determine the pH for 0.015 M sodium fluoride.

8686Determine the percent ionization for 0.250 M KNO2.

8787

What about polyprotic acids?

8888What about polyprotic acids? For example sulfuric acid:

11. Determine the pH for 1.2 M H2 SO4

8989What about polyprotic acids? For example sulfuric acid:

11. Determine the pH for 0.012 M H2 SO4

9090

pH meter

• Tests the voltage of the Tests the voltage of the electrolyteelectrolyte

• Converts the voltage to Converts the voltage to pHpH

• Very cheap, accurateVery cheap, accurate

• Must be calibrated with Must be calibrated with a buffer solutiona buffer solution

9191pH indicators

• Indicators are dyes that can be added that will change color in the presence of an acid or base.

• Some indicators only work in a specific range of pH

• Once the drops are added, the sample is ruined

• Some dyes are natural, like radish skin or red cabbage

9292

ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations

ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations

HH22CC22OO44(aq) + 2 NaOH(aq) --->(aq) + 2 NaOH(aq) --->

acidacid basebase

NaNa22CC22OO44(aq) + 2 H(aq) + 2 H22O(liq)O(liq)

Carry out this reaction using aCarry out this reaction using a TITRATIONTITRATION..

Oxalic acid,Oxalic acid,

HH22CC22OO44

9393Setup for titrating an acid with a baseSetup for titrating an acid with a base

9494

TitrationTitrationTitrationTitration

1. Add solution from the buret.1. Add solution from the buret.2. Reagent (base) reacts with 2. Reagent (base) reacts with

compound (acid) in solution compound (acid) in solution in the flask.in the flask.

3.3. Indicator shows when exact Indicator shows when exact stoichiometric reaction has stoichiometric reaction has occurred. (Acid = Base)occurred. (Acid = Base)

This is called This is called NEUTRALIZATION.NEUTRALIZATION.

9595

35.62 mL of NaOH is 35.62 mL of NaOH is

neutralized with 25.2 mL of neutralized with 25.2 mL of

0.0998 M HCl by titration to 0.0998 M HCl by titration to

an equivalence point. What an equivalence point. What

is the concentration of the is the concentration of the

NaOH?NaOH?

LAB PROBLEM #1: Standardize a LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately solution of NaOH — i.e., accurately determine its concentration.determine its concentration.

LAB PROBLEM #1: Standardize a LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately solution of NaOH — i.e., accurately determine its concentration.determine its concentration.

9696

PROBLEM: You have 50.0 mL of PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M 3.0 M NaOH and you want 0.50 M NaOH. What do you do?NaOH. What do you do?

PROBLEM: You have 50.0 mL of PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M 3.0 M NaOH and you want 0.50 M NaOH. What do you do?NaOH. What do you do?

Add water to the 3.0 M solution to lower Add water to the 3.0 M solution to lower its concentration to 0.50 M its concentration to 0.50 M

Dilute the solution!Dilute the solution!

9797

PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you do?What do you do?


3.0 M NaOH 0.50 M NaOH

H2O

Concentrated Dilute

But how much water But how much water do we add?do we add?

9898

PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you doWhat do you do??

PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you doWhat do you do??

How much water is added?How much water is added?

The important point is that --->The important point is that --->

moles of NaOH in ORIGINAL solution = moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solutionmoles of NaOH in FINAL solution

9999

PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?and you want 0.50 M NaOH. What do you do?PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?and you want 0.50 M NaOH. What do you do?

Amount of NaOH in original solution = Amount of NaOH in original solution =

M • VM • V = =

(3.0 mol/L)(0.050 L) = 0.15 mol NaOH(3.0 mol/L)(0.050 L) = 0.15 mol NaOH

Amount of NaOH in final solution must also = Amount of NaOH in final solution must also = 0.15 mol NaOH0.15 mol NaOH

Volume of final solution =Volume of final solution =

(0.15 mol NaOH) / (0.50 M) = 0.30 L(0.15 mol NaOH) / (0.50 M) = 0.30 L

or or 300 mL300 mL

100100



Conclusion:Conclusion:

add 250 mL add 250 mL of waterof water to to 50.0 mL of 50.0 mL of 3.0 M NaOH 3.0 M NaOH to make 300 to make 300 mL of 0.50 M mL of 0.50 M NaOH.NaOH.

3.0 M NaOH 0.50 M NaOH

H2O

Concentrated Dilute

101101

A shortcutA shortcut

MM11 • V • V11 = M = M22 • V • V22

Preparing Solutions Preparing Solutions by Dilutionby Dilution

Preparing Solutions Preparing Solutions by Dilutionby Dilution

102102You try this dilution problem

• You have a stock bottle of hydrochloric acid, which is 12.1 M. You need 400 mL of 0.10 M HCl. How much of the acid and how much water will you need?

Date post:	30-Dec-2015
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