AQUEOUS IONIC REACTIONS AND

SOLUTION STOICHIOMETRY

GENERAL CHEMISTRY

LECTURE 5

Coverage

5.1 Solution Stoichiometry

5.1.1 Calculating Concentration of Solutions

5.3 Determining Different Types of Aqueous Ionic Equation

5.3.1 Precipitation Reactions

5.2.2 Converting Mole-Mass-Number Involving Solutions

5.3.3 Dilution of Solutions

5.2 Writing Equations for Aqueous Ionic Reactions

5.3.2 Acid-Base Reactions

5.3.3 Redox Reactions

Environmental and biochemical reactions takes place in SOLUTIONS

For dissolved substances CONCENTRATION to find the VOLUME that contains a given number of moles

Solutions = Solute + Solvent

The concentration of solutions is the amount of solute dissolved in a given amount of solution. (Intrinsic quantity)

Molarity = Moles of solute

Liters of solution

M = mol solute

L solution

Fundamentals of Solution Stoichiometry

Exercise No. 1 Calculating the Molarity of a Solution

1. Glycine (H2NCH2COOH) is the simplest amino acid. What is the molarity of an aqueous solution that contains 0.715 mol of glycine in 495 mL?

SOLUTION:

0.715 mol glycine

495 mL soln

1000 mL

1 L

= 1.44 M glycine x M =

2. Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution.

SOLUTION:

mol H2SO4

98.1 g H2SO4

1

1.75 L = 0.0728 M H2SO4 x 12.5 g H2SO4 x

M = mol solute

L soln

Exercise No. 2 Calculating the Molarity of a Solution

3. A solution is prepared by dissolving 25.0 mL ethanol, C2H5OH (d= 0.789 g/mL), in enough water to produce 250.0 mL solution. What is the molarity of ethanol in the solution?

Exercise No. 3 Calculating the Molarity of a Solution

Answer: 1.71 M C2H5OH

Calculating Mass of Solute in a Given Volume of Solution

4. A buffered solution maintains acidity as a reaction occurs. In living cells, phosphate ions play a key buffering role, so biochemists often study reactions in such solutions. How many grams of solute are in 1.75 L of 0.460 M sodium monohydrogen phosphate?

SOLUTION:

1.75 L x 0.460 mol Na2HPO4

1 L

141.96 g Na2HPO4

mol Na2HPO4

= 114 g Na2HPO4 x

Exercise No. 4

Converting a concentrated solution to a dilute solution.

Initial solution Final solution

Molesinitial = Molesfinal

MinitialVinitial = MfinalVfinal

Dilution

mL 100.0mL 0.10 0.12 2 MM

Preparing a Dilute Solution from a Concentrated Solution

5. If 10.0 mL of 12.0 M HCl is added to enough water to give 100. mL of solution, what is the concentration of the solution?

2211 VV MM

MM

M 20.1mL 100.0

mL 0.100.122

Exercise No. 5

mL 333or L 0.333

18.0

2.40 L 2.50V1

M

M

6. What volume of 18.0 M sulfuric acid is required to make 2.50 L of a 2.40 M sulfuric acid solution?

2211 V V MM

1

221

V V

M

M

Preparing a Dilute Solution from a Concentrated Solution Exercise No. 6

7. How much:

a) glucose, C6H12O6, in grams, must be dissolved in water to produced 75.0 mL of 0.350 M C6H12O6

b) Methanol, CH3OH (d = 0.792 g/mL), in milliliters, must be dissolved in water to produced 2.25L of 0.485 M CH3OH

c) Concentrated sulfuric acid (18 M), in milliliters, is needed to produced 2.30 M sulfuric acid solution.

Preparing Solutions Exercise No. 7

Preparation of Standard Aqueous Solution

Preparing a Diluted Solution

Aqueous Ionic Solutions

Sodium Chloride Aqueous Solution

Aqueous Ionic Solutions

HCl solution NaOH solution

Determining Moles of Ions in Aqueous Ionic Solutions

8. How many moles of each ion are in the following solutions?

(a) 5.00 mol of ammonium sulfate dissolved in water

(b) 78.5 g of cesium bromide dissolved in water

Exercise No. 8

(c) 7.42 x 1022 formula units of copper(II) nitrate dissolved in water

(d) 35 mL of 0.84 M zinc chloride

SOLUTION: (a) (NH4)2SO4(s) 2NH4+(aq) + SO4

2-(aq)

5.00 mol (NH4)2SO4 x 2 mol NH4

+

1 mol (NH4)2SO4

= 10.0 mol NH4+ and 5.0 mol SO4

2-

H2O

78.5 g CsBr x mol CsBr

212.8 g CsBr = 0.369 mol CsBr = 0.369 mol Cs+

and 0.369 mol Br-

(b) CsBr(s) Cs+(aq) + Br-(aq) H2O

Determining Moles of Ions in Aqueous Ionic Solutions

8. How many moles of each ion are in the following solutions?

(a) 5.00 mol of ammonium sulfate dissolved in water

(b) 78.5 g of cesium bromide dissolved in water

Exercise No. 8

7.42 x 1022 formula units x Cu(NO3)2

mol Cu(NO3)2

6.022 x 1023 formula units = 0.123 mol Cu(NO3)2

and 0.123 mol Cu2+

= 0.246 mol NO3-

(c) Cu(NO3)2(s) Cu2+(aq) + 2NO3

-(aq)

35 mL ZnCl2 x 1L

103 mL = 2.9 x 10-2 mol ZnCl2

(d) ZnCl2(aq) Zn2+(aq) + 2Cl-(aq)

0.84 mol ZnCl2 L

= 2.9 x 10-2 mol Zn2+ and 5.8 x 10-2 mol Cl-

x

(c) 7.42 x 1022 formula units of copper(II) nitrate dissolved in water

(d) 35 mL of 0.84 M zinc chloride

8. How many moles of each ion are in the following solutions?

SOLUTION:

Determining Moles of Ions in Aqueous Ionic Solutions Exercise No. 8

Types of Aqueous Ionic Reactions

1. Precipitation Reaction

2. Acid-Base Reaction

3. Oxidation-reduction reaction

Writing Equations for Aqueous Ionic Reactions

The molecular equation

shows all of the reactants and products as intact, undissociated compounds.

The total ionic equation

shows all of the soluble ionic substances dissociated into ions.

The net ionic equation

omits the spectator ions and shows the actual chemical change taking place.

An Aqueous Ionic Reaction and Its Equation.

The reaction of Pb(NO3)2 and NaI.

Double-displacement reaction (metathesis)

NaI(aq) + Pb(NO3)2(aq) PbI2(s) + NaNO3(aq)

2Na+(aq) + 2I-(aq) + Pb2+(aq) + 2NO3-(aq)

PbI2(s) + 2Na+(aq) + 2NO3

-(aq)

2NaI(aq) + Pb(NO3)2(aq) PbI2(s) + 2NaNO3(aq)

Precipitation Reaction

Pb2+(aq) + 2I-(aq) PbI2(s)

1. Most nitrate (NO3-) salts are soluble.

2. Most alkali metal (group 1A) salts and NH4+ are soluble.

3. Most Cl-, Br-, and I- salts are soluble (except Ag+, Pb2+, Hg2

2+).

4. Most sulfate salts are soluble (except BaSO4, PbSO4, Hg2SO4, CaSO4).

5. Most OH- are only slightly soluble (NaOH, KOH are soluble, Ba(OH)2, Ca(OH)2 are marginally soluble).

6. Most S2-, CO32-, CrO4

2-, PO43- salts are only slightly soluble,

except for those containing the cations in Rule 2.

Simple Rules for Solubility

Predicting Whether a Precipitation Reaction Occurs; Writing Ionic Equations

9. Predict whether a reaction occurs when each of the following pairs of solutions are mixed. If a reaction does occur, write balanced molecular, total ionic, and net ionic equations, and identify the spectator ions.

(a) Potassium fluoride(aq) + strontium nitrate(aq)

(b) Ammonium perchlorate(aq) + sodium bromide(aq)

SOLUTION:

(a) KF(aq) + Sr(NO3)2 (aq) 2KNO3(aq) + SrF2 (s)

2K+(aq) + 2F-(aq) + Sr2+(aq) + 2NO3-(aq) 2K+(aq) + 2NO3

-(aq) + SrF2 (s)

2F-(aq) + Sr2+(aq) SrF2 (s)

(b) NH4ClO4(aq) + NaBr (aq) NH4Br (aq) + NaClO4(aq)

All reactants and products are soluble so no reaction

occurs.

Exercise No. 9

Calculating Amounts of Reactants and Products for a Reaction in Solution

10. A 25.00-mL pipetful of 0.250 M K2CrO4 (aq) is added to an excess of AgNO3 (aq). What mass of Ag2CrO4 will precipitate from the solution?

Exercise No. 10

Answer: 2.07 g Ag2CrO4

Calculating Amounts of Reactants and Products for a Reaction in Solution

Exercise No. 11

11. Solve for the following:

A. How many grams of sodium sulfide are required to react completely with 27.8 mL of 0.163 M silver nitrate?

B. How many grams of the solid product are obtained from the reaction in part A.

12. A 10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change).

a) What precipitate will form?

b) What mass of precipitate will form?

Solving Limiting-Reactant Problems for Reactions in Solution Exercise No. 12

Acid-Base/Neutralization Reaction

Writing Ionic Equations for Acid-Base Reactions

13. Write balanced molecular, total ionic, and net ionic equations for each of the following acid-base reactions and identify the spectator ions.

(a) Strontium hydroxide(aq) + perchloric acid(aq)

(b) Barium hydroxide(aq) + sulfuric acid(aq)

SOLUTION:

(a) Sr(OH)2(aq) + 2HClO4(aq) Sr(ClO4)2(aq) + 2H2O(l)

Sr2+(aq) + 2OH-(aq) + 2H+(aq) + 2ClO4-(aq) Sr2+(aq) + 2ClO4

-(aq) + 2H2O(l)

2OH-(aq)+ 2H+(aq) 2H2O(l)

(b) Ba(OH)2(aq) + H2SO4(aq) BaSO4(s) + 2H2O(l)

Ba2+(aq) + 2OH-(aq) + 2H+(aq) + SO42-(aq) BaSO4(s) + 2H2O(l)

Exercise No. 13

Calculating Amounts of Reactants and Products for a Reaction in Solution

14. Specialized cells in the stomach release HCl to aid digestion. If they release too much, the excess can be neutralized with antacids. A common antacid contains magnesium hydroxide, Mg(OH)2, which reacts with the acid to form water and magnesium chloride solution. As a government chemist testing commercial antacids, you use 0.10 M HCl to simulate the acid concentration in the stomach. How many liters of stomach acid react with a tablet containing 0.10 g of magnesium hydroxide?

SOLUTION: Mg(OH)2(s) + HCl(aq) MgCl2(aq) + H2O(l)

0.10 g Mg(OH)2 x mol Mg(OH)2

58.33 g Mg(OH)2

2 mol HCl

1 mol Mg(OH)2 x

1L

0.10 mol HCl

= 3.4 x 10-2 L

2 2

x

Exercise No. 14

An acid-base titration.

Start of titration

Excess of acid

Point of

neutralization

Slight excess of

base

Finding the Concentration of Acid from an Acid-Base Titration

15. You perform an acid-base titration to standardize an HCl solution by placing 50.00 mL of HCl in a flask with a few drops of indicator solution. You put 0.1524 M NaOH into the buret, and the initial reading is 0.55 mL. At the end point, the buret reading is 33.87 mL. What is the concentration of the HCl solution?

SOLUTION: NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)

(33.87 mL - 0.55 mL) x 1L

103 mL = 0.03332 L NaOH

0.03332 L x 0.1524 M = 5.078 x 10-3 mol NaOH

From the stoichiometric ratio in the balanced equation,

5.078 x 10-3 mol NaOH = 5.078 x 10-3 mol HCl

5.078 x 10-3 mol HCl

0.05000 L

= 0.1016 M HCl

Exercise No. 15

Finding the Concentration of Acid from an Acid-Base Titration

15. You perform an acid-base titration to standardize an HCl solution by placing 50.00 mL of HCl in a flask with a few drops of indicator solution. You put 0.1524 M NaOH into the buret, and the initial reading is 0.55 mL. At the end point, the buret reading is 33.87 mL. What is the concentration of the HCl solution?

Alternative Solution:

NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)

Exercise No. 15

mole NaOH = mole HCl

MVNaOH = MVHCl

MHCl = MVNaOH/ VHCl

MHCl = (0.1524 M) (33.87-0.55 mL) / 50.00 mL

MHCl=101.6 M HCl

Ate equivalence point:

16. What is the molarity of a barium hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH)2 solution?

2

2

2 Ba(OH) 0593.0Ba(OH) mL 3.38

Ba(OH) mmol 27.2M

OH 2 + BaCl HCl 2 + Ba(OH) 222

HCl mmol 4.54 = HCl) HCl)(0.103 mL (44.1 M

22 Ba(OH) mmol 2.27

HCl mmol 2

Ba(OH) mmol 1HCl mmol 54.4

Finding the Concentration of Acid from an Acid-Base Titration Exercise No. 16

16. What is the molarity of a barium hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH)2 solution?

OH 2 + BaCl HCl 2 + Ba(OH) 222

Finding the Concentration of Acid from an Acid-Base Titration Exercise No. 16

Alternative Solution:

2 mole Ba(OH)2 = mole HCl

2MVBa(OH)2 = MVHCl

MBa(OH)2 = MVHCl/ 2VBa(OH)2

MBa(OH)2 = (0.103 M) (44.1 mL) / (38.3 mL x 2)

MBa(OH)2 =0.0593 M Ba(OH)2

Ate equivalence point:

Solving Limiting-Reactant Problems for Reactions in Solution

Mercury and its compounds have many uses, from fillings for teeth (as an alloy with silver, copper, and tin) to the industrial production of chlorine. Because of their toxicity, however, soluble mercury compounds, such as mercury(II) nitrate, must be removed from industrial wastewater. One removal method reacts the wastewater with sodium sulfide solution to produce solid mercury(II) sulfide and sodium nitrate solution. In a laboratory simulation, 0.050 L of 0.010 M mercury(II) nitrate reacts with 0.020 L of 0.10 M sodium sulfide. How many grams of mercury(II) sulfide form?

Hg(NO3)2(aq) + Na2S(aq) HgS(s) + NaNO3(aq) 2

0.050 L Hg(NO3)2 1 mol HgS

1 mol Hg(NO3)2

0.10 mol

L x x

232.7 g HgS

1 mol HgS x = 0.12 g HgS

0.020 L Na2S 1 mol HgS

1 mol Na2S x x

0.10 mol

L

232.7 g HgS

1 mol HgS x = 0.47 g HgS

Additional Exercises