Chapter 9 Answer Key Practice Questions Only - Quia

CHEMISTRY 12 

Chapter 9 

Aqueous Solutions 

and Solubility Equilibria 

Solutions for Practice Problems 

Student Textbook page 424 

1. Problem 

Predict whether an aqueous solution of each salt is neutral, acidic, or basic. 

(a) NaCN 

(b) LiF 

(c) Mg(NO 3 ) 2 

(d) NH 4 I 

What Is Required? 

Predict whether each aqueous solution is acidic, basic, or neutral. 

What Is Given? 

The formula of each salt is given. 

Plan Your Strategy 

Determine whether the cation is from a strong or weak base, and whether the anion is 

from a strong or weak acid. Ions derived from weak bases or weak acids react with water 

and affect the pH of the solution. 

Act on Your Strategy 

(a) Sodium cyanide, NaCN, is the salt of a strong base (NaOH) and a weak acid 

(HCN). Only the cyanide ions react with water. The solution is basic. 

(b) Lithium fluoride, LiF, is the salt of a strong base (LiOH) and a weak acid (HF). 

Only the fluoride ions react with water. The solution is basic. 

(c) Magnesium nitrate, Mg(NO 3 ) 2 , is the salt of a strong base (Mg(OH) 2 ) and a 

strong acid (HNO 3(aq) ). Neither ion reacts with water, so the solution is neutral. 

(d) Ammonium iodide, NH 4 I, is the salt of a weak base (NH 3(aq) ) and a strong acid 

(HI (aq) ). Only the ammonium ions react with water. The solution is acidic. 

Check Your Solution 

Check that you have correctly identified whether the cation is from a strong or weak 

base, and whether the anion is from a strong or weak acid. 

2. Problem 

Is the solution of each salt acidic, basic, or neutral? For solutions that are not neutral, 

write equations that support your predictions. 

(a) NH 4 BrO 4 

(b) NaBrO 4 

(c) NaOBr 

(d) NH 4 Br 


Predict whether each aqueous solution is acidic, basic, or neutral. Write equations that 

represent the solutions that are acidic or basic. 

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 

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CHEMISTRY 12 


The formula of each salt is given. 


Determine whether the cation is from a strong or weak base, and whether the anion is 

from a strong or weak acid. Ions derived from weak bases or weak acids react with water 

and affect the pH of the solution. 


(a) Ammonium perbromate, NH 4 BrO 4 , is the salt of a weak base (NH 3(aq) ) and a 

strong acid (HBrO 4 ). Only the ammonium ions react with water. 

+ 

NH 4 (aq) + H 2 O (l) 

⇀↽ NH 3(aq) + H 3 O + (aq) 

The solution is acidic. 

(b) Sodium perbromate, NaBrO 4 , is the salt of a strong base (NaOH) and a strong acid 

(HBrO 4 ). Neither ion reacts with water, so the solution is neutral. 

(c) Sodium hypobromite, NaOBr, is the salt of a strong base (NaOH) and a weak acid 

(HOBr (aq) ). Only the hypobromite ions react with water. 

OBr − (aq) + H 2 O (l) 

⇀↽ HOBr (aq) + OH − (aq) 

The solution is basic. 

(d) Ammonium bromide, NH 4 Br, is the salt of a weak base (NH 3(aq) ) and a strong 

acid (HBr (aq) ). 

Only the ammonium ions react with water. 

+ 

NH 4 (aq) + H 2 O (l) 

⇀↽ NH 3(aq) + H 3 O + (aq) 

The solution is acidic. 


The equations that represent the reactions with water support the prediction that 

NH 4 BrO 4 and NH 4 Br dissolve to form an acidic solution and NaOBr dissolves to 

form a basic solution. Sodium hypobromite is the salt of a strong base-strong acid, 

so neither ion reacts with water and the solution is neutral. 

3. Problem 

K a for benzoic acid, C 6 H 5 COOH, is 6.3 × 10 −5 . K a for phenol, C 6 H 5 OH, is 

1.3 × 10 −10 . Which is the stronger base, C 6 H 5 COO − (aq) or C 6 H 5 O − (aq)? 

Explain your answer. 


You must determine which ion, C 6 H 5 COO − (aq) or C 6 H 5 O − (aq), is the stronger base. 


The K a of each conjugate acid is given. 


For a conjugate acid-base pair, K a K b = 1.0 × 10 −14 . Therefore, a small value of K a for 

an acid results in a large value of K b for the conjugate base. 


Because K a for phenol is smaller than K a for benzoic acid, K b for the phenolate 

ion must be larger than K b for the benzoate ion. Consequently, C 6 H 5 O − (aq), is 

the stronger base. 


Using the equation K a K b = 1.0 × 10 −14 , you can calculate the K b for each conjugate 

base. K b for C 6 H 5 COO − (aq) is 1.6 × 10 −10 . K b for C 6 H 5 O − (aq) is 7.7 × 10 −5 . 

This supports the reasoning that C 6 H 5 O − (aq) is the stronger base. 


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CHEMISTRY 12 

4. Problem 

Sodium hydrogen sulfite, NaHSO 3 , is a preservative that is used to prevent the 

discolouration of dried fruit. In aqueous solution, the hydrogen sulfite ion can act 

as either an acid or a base. Predict whether NaHSO 3 dissolves to form an acidic 

solution or a basic solution. (Refer to Appendix E for ionization data.) 


You must decide whether NaHSO 3 dissolves to form an acidic solution or a 

basic solution. 


K a for H 2 SO 3(aq) = 1.4 × 10 −2 

− 

K a for HSO 3 (aq) = 6.3 × 10 −8 


The hydrogen sulfite ion is amphoteric. Compare the equilibrium constants K a and K b 

− 

for HSO 3 (aq) acting as an acid and a base, to determine which reaction goes farthest 

to completion. 


The ion reactions with water are 

− 2− 

HSO 3 (aq) + H 2 O (l) 

⇀↽ SO 3 (aq) + H 3 O + (aq) K a = 6.3 × 10 −8 

− 

HSO 3 (aq) + H 2 O (l) 

⇀↽ H 2 SO 3(aq) + OH − (aq) K b = ? 

K b can be calculated using the value for K a of H 2 SO 3(aq) . 

K b = K w 

K a 

= 

1.0 × 10−14 

1.4 × 10 −2 

= 7.1 × 10 −13 

The equilibrium constant (K a ) for the hydrogen sulfite ion acting as an acid is greater 

than the equilibrium constant (K b ) for the ion acting as a base. Therefore, the solution 

is acidic. 


It is a common difficulty in this type of problem to identify correctly the base reaction 

and the corresponding value for K b . K b must be calculated using the K a value for 

− 

H 2 SO 3(aq) , because HSO 3 (aq) is the conjugate base of H 2 SO 3(aq) . 



5. Problem 

After titrating sodium hydroxide with hydrofluoric acid, a chemist determined that the 

reaction had formed an aqueous solution of 0.020 mol/L sodium fluoride. Determine 

the pH of the solution. 


You need to calculate the pH of the solution. 


[NaF] = 0.020 mol/L 

From Appendix E, K a for HF (aq) = 6.3 × 10 −4 


Step 1 Decide which ion reacts with water. Write the equation that represents the 

hydrolysis reaction. 


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CHEMISTRY 12 

Step 2 Determine the equilibrium constant for the ion involved in the 


Step 3 Divide the ion concentration by the appropriate equilibrium constant to 

determine whether the change in concentration of the ion can be ignored. 

Step 4 Set up an ICE table for the ion that is involved in the reaction with water. 

Let x represent the change in the concentration of the ion that reacts. 

Step 5 Write the equilibrium expression. Substitute the equilibrium concentrations 

into the expression, and solve for x. 

Step 6 Use the value of x to determine [H 3 O + ]. Then calculate the pH of 

the solution. 


Step 1 Sodium fluoride is the salt of a strong base (NaOH) and a weak acid (HF). 

Thus, only the fluoride ion reacts with water. 

F − (aq) + H 2 O (l) 

⇀↽ HF (aq) + OH − (aq) 

Step 2 K b for the fluoride ion can be calculated using K a for HF (aq) . 

K b = K w 

K a 

= 

1.0 × 10−14 

6.3 × 10 −4 

= 1.6 × 10 −11 

Step 3 NaF (aq) 

⇀↽ Na + (aq) + F − (aq) 

∴[F − ] = 0.020 mol/L 

Step 4 

F − 

K b 

= 

0.020 

1.6 × 10 −11 

= 1.2 × 10 9 

Because this value is much greater than 500, the change in concentration of 

F − (aq) can be ignored compared with its initial concentration. 

Concentration (mol/L) 

F − (aq) 

+ H 2 O (l) 

+ 

OH − (aq) 

Initial 

0.020 

0 

2HF (aq) 

161 

~0 

Change 

−x 

+x 

+x 

Equilibrium 

0.020 − x ≈ 0.020 

x 

x 

Step 5 K b = [HF][OH− ] 

[F − ] 

1.6 × 10 −11 = (x)(x) 

0.020 

x = √ 3.2 × 10 −13 

= 5.6 × 10 −7 mol/L 

Step 6 x = [OH − ] = 5.6 × 10 −7 mol/L 

pOH =−log[OH] − 

=−log(5.6 × 10 −7 ) 

= 6.25 

pH = 14.00 − pOH 

= 14.00 − 6.25 

= 7.75 


The pH of the solution is weakly basic. This is consistent with a dilute aqueous 

solution of the salt of a strong base and a weak acid. 

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR

CHEMISTRY 12 

6. Problem 

Part way through a titration, 2.0 × 10 1 mL of 0.10 mol/L sodium hydroxide has 

been added to 3.0 × 10 1 mL of 0.10 mol/L hydrochloric acid. What is the pH 

of the solution? 


You need to calculate the pH of the solution. 


20 mL of 0.10 mol/L NaOH (aq) has been added to 30 mL of 0.10 mol/L HCl (aq) . 


Step 1 Calculate the amount of each reagent using the following equation: 

Amount (mol) = concentration (mol/L) × volume (L) 

Step 2 Write the chemical equation for the reaction and determine the 

excess reagent. 

Step 3 Calculate the concentration of the excess reagent using the following equation: 

Concentration (mol/L) = 

amount in excess (mol) 

total volume (L) 

Step 4 Calculate the pH of the solution. 


Step 1 Amount NaOH (aq) = (0.10 mol/L) × (2.0 × 10 −2 L) 

= 2.0 × 10 −3 mol 

Amount HCl (aq) = (0.10 mol/L) × (3.0 × 10 −2 L) 

= 3.0 × 10 −3 mol 

Step 2 NaOH (aq) + HCl (aq) → NaCl (aq) + H 2 O (l) 

Because the acid and base react in a 1:1 ratio, the HCl (aq) is in excess. 

Step 3 Amount of excess HCl (aq) = (3.0 × 10 −3 mol) − (2.0 × 10 −3 mol) 

= 1.0 × 10 −3 mol 

Total volume = (2.0 × 10 −2 L) + (3.0 × 10 −2 L) = 5.0 × 10 −2 L 

[HCl] = 10 × 10−3 mol 

5.0 × 10 −2 L 

= 2.0 × 10 −2 mol/L 

Step 4 HCl (aq) is a strong acid. Therefore, [H 3 O + ] = 2.0 × 10 −2 mol/L. 

pH =−log[H 3 O + ] 

=−log(2.0 × 10 −2 ) 

= 1.70 


The solutions have the same molar concentration. Because the volume of hydrochloric 

acid is greater, the solution should be acidic and the pH less than 7. 

7. Problem 

0.025 mol/L benzoic acid, C 6 H 5 COOH, is titrated with 0.025 mol/L sodium 

hydroxide solution. Calculate the pH at equivalence. 


You need to calculate the pH at equivalence. 


You know that 0.025 mol/L C 6 H 5 COOH (aq) reacts with 0.025 mol/L NaOH (aq) . 

Tables of K a and K b values are in Appendix E. 


Step 1 Write the balanced chemical equation for the reaction. 

Step 2 Calculate the concentration of the salt formed, based on the amount 

(in mol) and the total volume of the solution. 


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CHEMISTRY 12 

Step 3 Decide which ion reacts with water. Write the equation that represents 

the reaction. 

Step 4 Determine the equilibrium constant for the ion that is involved in the 


Step 5 Divide the concentration of the ion identified in Step 3 by the appropriate 

ionization constant to determine whether the change in concentration of the 

ion can be ignored. 






the solution. 


Step 1 The following chemical equation represents the reaction. 

C 6 H 5 COOH (aq) + NaOH (aq) → NaC 6 H 5 COO (aq) + H 2 O (l) 

Step 2 The acid and base react in a 1:1 ratio, and the concentrations are the same. 

Therefore, the volume of each reagent must be the same and the total 

volume will be double the initial volume. 

Therefore, 

[NaC 6 H 5 COO] = 

0.025 mol/L 

2 

= 0.0125 mol/L 

Step 3 The salt forms Na + (aq) and C 6 H 5 COO − (aq) in solution. Na + (aq) is the cation 

of a strong base, so it does not react with water. C 6 H 5 COO − (aq) is the conjugate 

base of a weak acid, so it does react with water. The pH of the solution 

is therefore determined by the extent of the following reaction. 

C 6 H 5 COO − (aq) + H 2 O (l) → C 6 H 5 COOH (aq) + OH − (aq) 

Step 4 Therefore, 

K b for C 6 H 5 COO (aq) = K w 

K a 

= 

1.0 × 10−14 

6.3 × 10 −5 

Step 5 

Step 6 

= 1.6 × 10 −10 

[C 6 H 5 COO − ] 

= 

0.0125 

K b 1.6 × 10 −10 

= 7.8 × 10 7 

This is well above 500, so the change in [C 6 H 5 COO − ] can be ignored. 


C 6 H 5 COOH − (aq) 

+ H 2 O (l) 

+ 

OH − (aq) 

Initial 

0.0125 

0 

C 6 H 5 COOH (aq) 

163 

~0 

Change 

−x 

+x 

+x 

Equilibrium 

0.0125 − x ≈ 0.0125 

x 

x 

Step 7 K b = [C 6H 5 COOH][OH − ] 

[C 6 H 5 COO − ] 

1.6 × 10 −10 = (x)(x) 

0.0125 

x = √ 2.0 × 10 −12 

= 1.4 × 10 −6 mol/L 


CHEMISTRY 12 

Step 8 x = [OH − ] = 1.4 × 10 −6 mol/L 

pOH =−log[OH − ] 

=−log(1.4 × 10 −6 ) 

= 5.85 

pH = 14.00 − 5.85 

= 8.15 


The titration forms an aqueous solution of a salt derived from a strong base and a weak 

acid. The solution should be basic, which is supported by the calculation of the pH. 

8. Problem 

50.0 mL of 0.10 mol/L hydrobromic acid is titrated with 0.10 mol/L aqueous 

ammonia. Determine the pH at equivalence. 


You need to calculate the pH at equivalence. 


You know that 50 mL of 0.10 mol/L HBr (aq) reacts with 0.10 mol/L NH 3(aq) . 

Tables of K a and K b values are in Appendix E. 


Step 1 Write the balanced chemical equation for the reaction. 

Step 2 Find the volume of NH 3(aq) added to neutralize the hydrobromic acid based 

on the amount and concentration of HBr (aq) . 

Step 3 Calculate the concentration of the salt formed, based on the amount (in mol) 

and the total volume of the solution. 

Step 4 Decide which ion reacts with water. Write the equation that represents 

the reaction. 

Step 5 Determine the equilibrium constant for the ion that is involved in the 


Step 6 Divide the concentration of the ion identified in Step 4 by the appropriate 

ionization constant to determine whether the change in concentration of the 

ion can be ignored. 






the solution. 


Step 1 The following chemical equation represents the reaction. 

NH 3(aq) + HBr (aq) → NH 4 Br (aq) 

Step 2 Amount (in mol) of NH 3(aq) = 0.20 mol/L × 0.020 L 

= 4.0 × 10 −3 mol 

Step 3 NH 3(aq) and HBr (aq) react in a 1:1 ratio. Because [NH 3 ] = [HBr], the volume 

of aqueous ammonia added must be the same as the volume of hydrobromic 

acid. Therefore, the volume of NH 3(aq) is 50 mL, and the total volume of the 

solution is twice the volume of HBr (aq) used. Thus, the [NH 4 Br] must be half 

the initial [HBr]. 

Therefore, [NH 4 Br] = 0.050 mol/L 


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CHEMISTRY 12 

Step 4 The titration results in an aqueous solution of ammonium bromide, NH 4 Br (aq) . 

+ 

NH 4 (aq) is the conjugate acid of a weak base, so it reacts with water. Br − (aq) 

is the conjugate base of a strong acid, so it does not react with water. The solution 

will be acidic, and the pH is determined by the extent of the following 

reaction. 

+ 

NH 4 (aq) + H 2 O (l) → NH 3(aq) + H 3 O + (aq) 

Step 5 From Appendix E, K b for NH 3(aq) is 1.8 × 10 −5 . 

K a for the conjugate base, NH 4(aq) , can be calculated using the relationship 

K a K b = K w . 

K a = K w 

K b 

= 

1.0 × 10−14 

1.8 × 10 −5 

= 5.6 × 10 −10 

Step 6 

Step 7 

[NH + 4 ] 

= 

0.050 

K a 5.6 × 10 −10 

= 8.9 × 10 7 

This is well above 500, so the change in [NH + 4 

] can be ignored. 

+ 

Concentration (mol/L) NH 4 (aq) + H 2 O (l) 

+ 

H 3 O + (aq) 

Initial 

0.050 

0 

NH 3(aq) 

165 

~0 

Change 

−x 

+x 

+x 

Equilibrium 

0.050 − x ≈ 0.050 

x 

x 

Step 8 K a = [NH 3][H 3 O + ] 

[NH + 4 ] 

5.6 × 10 −10 = (x)(x) 

0.050 

x = √ 2.8 × 10 −11 

= 5.3 × 10 −6 mol/L 

Step 9 x = [H 3 O + ] = 5.3 × 10 −6 mol/L 

pH =−log[H 3 O + ] 

=−log(5.3 × 10 −6 ) 

= 5.28 


The titration forms an aqueous solution of a salt derived from a weak base and a strong 

acid. The solution should be acidic, which is supported by the calculation of the pH. 



9. Problem 

Write the balanced chemical equation that represents the dissociation of each compound 

in water. Then write the corresponding solubility product expression. 

(a) copper(I) chloride 

(b) barium fluoride 

(c) silver sulfate 

(d) calcium phosphate 


CHEMISTRY 12 

Solution 

First, write a balanced equation for the equilibrium between excess solid and dissolved 

ions in a saturated aqueous solution. Then use the balanced equation to write the 

expression for K sp . 

(a) CuCl (s) 

⇀↽ Cu + (aq) + Cl − (aq) 

K sp = [Cu + ][Cl − ] 

(b) BaF 2(s) 

⇀↽ Ba 2+ (aq) + 2F − (aq) 

K sp = [Ba 2+ ][F − ] 2 

(c) Ag 2 SO 4(s) 

⇀↽ 2Ag + (aq) + SO 4 

2− (aq) 

K sp = [Ag + ] 2 [SO 4 2− ] 

(d) Ca 3 (PO 4 ) 2(s) 

⇀↽ 3Ca 2+ (aq) + 2PO 4 

3− (aq) 

K sp = [Ca 2+ ] 3 [PO 4 3− ] 2 


The K sp expressions are based on balanced equations for saturated solutions of slightly 

soluble ionic compounds. The exponents in the K sp expressions match the corresponding 

coefficients in the chemical equations. The coefficient 1 is not written, following 

chemistry conventions. 

10. Problem 

Write a balanced dissolution equation and solubility product expression for silver 

carbonate, Ag 2 CO 3 . 

Solution 




Ag 2 CO 3(s) 

⇀↽ 2Ag + − 

(aq) + CO 3 (aq) 

K sp = [Ag + ] 2 [CO − 3 ] 


The K sp expression is based on the balanced equation between excess solid and dissolved 

ions. The exponents in the K sp expression match the corresponding coefficients in the 

chemical equation. 

11. Problem 

Write a balanced dissolution equation and solubility product expression for magnesium 

ammonium phosphate, MgNH 4 PO 4 . 

Solution 




MgNH 4 PO 4(s) 

⇀↽ Mg 2+ + 3− 

(aq) + NH 4 (aq) + PO 4 (aq) 

K sp = [Mg 2+ ][NH + 4 ][PO 3− 4 ] 


The K sp expression is based on the balanced equation between excess solid and dissolved 

ions. The exponents in the K sp expression match the corresponding coefficients in the 

chemical equation. 

12. Problem 

Iron (III) nitrate has very low solubility. 

(a) Write the solubility product expression for iron(III) nitrate. 

(b) Do you expect the value of K sp of iron(III) nitrate to be larger or smaller than the 

K sp for aluminum hydroxide, which has a slightly higher solubility? 


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CHEMISTRY 12 

Solution 

(a) Write a balanced equation for the equilibrium between excess solid iron(III) nitrate 

and dissolved ions in a saturated aqueous solution. Then use the balanced equation 

to write the expression for K sp . 

Fe(NO 3 ) 3(s) 

⇀↽ Fe 3+ − 

(aq) + 3NO 3 (aq) 

K sp = [Fe 3+ ][NO − 3 ] 3 

(b) Write a balanced equation for the equilibrium between excess solid aluminum hydroxide 

and dissolved ions in a saturated aqueous solution. Then use the balanced 

equation to write the expression for K sp . Finally, compare the two equilibrium 

expressions and decide which expression is larger. 

Al(OH) 3(s) 

⇀↽ Al 3+ (aq) + 3OH − (aq) 

K sp = [Al 3+ ][OH − ] 3 

The equilibrium expressions have the same form. In both cases, if x mol/L of salt 

dissolves, K sp is given by x 4 . Therefore, the most soluble salt will have the larger 

value of K sp . The value of K sp for iron(III) nitrate is smaller than K sp for aluminum 

hydroxide. 


Each K sp expression is based on the balanced equation between excess solid and dissolved 

ions. The exponents in the K sp expressions match the corresponding coefficients in the 

chemical equations. The equilibrium expressions are of the same form, so the least 

soluble salt has the lower value for K sp . 



13. Problem 

The maximum solubility of silver cyanide, AgCN, is 1.5 × 10 −8 mol/L at 25˚C. 

Calculate K sp for silver cyanide. 


You need to find the value of K sp for AgCN at 25˚C. 


You know the solubility of AgCN at 25˚C. 


Step 1 Write an equation for the dissolution of AgCN. 

Step 2 Use the equation to write the solubility product expression. 

Step 3 

Step 4 

Find the concentration (in mol/L) of each ion. 

Substitute the concentrations into the solubility product expression, 

and calculate K sp . 


Step 1 AgCN (s) 

⇀↽ Ag + (aq) + CN − (aq) 

Step 2 K sp = [Ag + ][CN − ] 

Step 3 [Ag + ] = [CN − ] = [AgCN] = 1.5 × 10 −8 mol/L 

Step 4 K sp = [Ag + ][CN − ] 

= (1.5 × 10 −8 )(1.5 × 10 −8 ) 

= 2.2 × 10 −16 

K sp for silver cyanide is 2.2 × 10 −16 at 25˚C. 


The value of K sp is less than the concentration of the salt, as expected. It has the correct 

number of significant digits. 


167

CHEMISTRY 12 

14. Problem 

A saturated solution of copper(II) phosphate, Cu 3 (PO 4 ) 2 , has a concentration of 

6.1 × 10 −7 gCu 3 (PO 4 ) 2 per 1.00 × 10 2 mL of solution at 25˚C. What is K sp for 

Cu 3 (PO 4 ) 2 at 25˚C ? 


You need to find the value of K sp for Cu 3 (PO 4 ) 2 at 25˚C. 


You know 6.1 × 10 −7 g of Cu 3 (PO 4 ) 2 is dissolved in 1.00 × 10 2 mL of solution 

at 25˚C. 


Step 1 Write an equation for the dissolution of Cu 3 (PO 4 ) 2 . 


Step 3 Find the concentration (in mol/L) of copper(II) phosphate. 

Next, find the concentration (in mol/L) of each ion. 

Step 4 Substitute the concentrations into the solubility product expression, 



Step 1 Cu 3 (PO 4 ) 2(s) 

⇀↽ 3Cu 2+ 3− 

(aq) + 2PO 4 (aq) 

Step 2 K sp = [Cu 2+ ] 3 [PO 3− 4 ] 2 

Step 3 The molar mass of Cu 3 (PO 4 ) 2 is 380.6 g/mol. 

Amount of Cu 3 (PO 4 ) 2 = 6.1 × 10−7 g 

380.6 g/mol 

= 1.6 × 10 −9 mol 

[Cu 3 (PO 4 ) 2 ] = 1.6 × 10−9 mol 

0.100 L 

= 1.6 × 10 −8 mol/L 

[Cu 2+ ] = 3 × [Cu 3 (PO 4 ) 2 ] 

= 3 × (1.6 × 10 −8 mol/L) 

= 4.8 × 10 −8 mol/L 

[PO 3− 4 ] = 2 × [Cu 3 (PO 4 ) 2 ] = 3.2 × 10 −8 mol/L 

Step 4 K sp = [Cu 2+ ] 3 [PO 3− 4 ] 2 

= (4.8 × 10 −8 ) 3 (3.2 × 10 −8 ) 2 

= 1.1 × 10 −37 

K sp for copper(II) phosphate is 1.1 × 10 −37 at 25˚C. 


Check that you wrote the balanced chemical equation and the corresponding K sp equation 

correctly. Pay attention to molar relationships and to the exponent of each term. 

Check the concentration calculations carefully. 

15. Problem 

A saturated solution of CaF 2 contains 1.2 × 10 20 formula units of calcium fluoride per 

litre of solution. Calculate K sp for CaF 2 . 


You need to find the value of K sp for CaF 2 . 


You know one litre of solution contains 1.2 × 10 20 formula units of CaF 2 . 


168

CHEMISTRY 12 


Step 1 Write an equation for the dissolution of CaF 2 . 


Step 3 Find the concentration (in mol/L) of calcium fluoride. Next, find the 

concentration (in mol/L) of each ion. 

Step 4 Substitute the concentrations into the solubility product expression, 



Step 1 CaF 2(s) 

⇀↽ Ca 2+ (aq) + 2F − (aq) 

Step 2 K sp = [Ca 2+ ][F − ] 2 

Step 3 The amount of CaF 2 is given by the following formula: 

Amount = 

number of formula units 

Avogadro’s constant 

= 

1.2 × 10 20 formula units 

6.0 × 10 23 formula units/mol 

= 2.0 × 10 −4 mol 

[CaF 2 ] = 2.0 × 10 −4 mol/L 

[Ca 2+ ] = [CaF 2 ] = 2.0 × 10 −4 mol/L 

[F − ] = 2 × [CaF 2 ] 

= 2 × (2.0 × 10 −4 mol/L) 

= 4.0 × 10 −4 mol/L 

Step 4 K sp = [Ca 2+ ][F − ] 2 

= (2.0 × 10 −4 )(4.0 × 10 −4 ) 2 

= 3.2 × 10 −11 

K sp for calcium fluoride is 3.2 × 10 −11 . 


Check that you wrote the balanced chemical equation and the corresponding K sp equation 

correctly. Pay attention to molar relationships and to the exponent of each term. 

Check the concentration calculations carefully. 

16. Problem 

The concentration of mercury(I) iodide, Hg 2 I 2 , in a saturated solution at 25˚C is 

1.5 × 10 −4 ppm. 

(a) Calculate K sp for Hg 2 I 2 . The solubility equilibrium is written as follows: 

Hg 2 I 2 

⇀↽ Hg 2+ 2 + 2I − 

(b) State any assumptions that you made when you converted ppm to mol/L. 


You need to find the value of K sp for Hg 2 I 2 at 25˚C. 


You know the concentration of Hg 2 I 2 in a saturated solution at 25˚C 

is 1.5 × 10 −4 ppm. 

You are given the dissolution equation. 


Step 1 Use the dissolution equation to write the solubility product expression. 

Step 2 Find the concentration (in mol/L) of mercury(I) iodide. Then, find the 

concentration (in mol/L) of each ion. 

Step 3 Substitute the concentrations into the solubility product expression, and 

calculate K sp . 


169

CHEMISTRY 12 


(a) Step 1 K sp = [Hg 2+ 2 ][I − ] 2 

Step 2 1 ppm is equivalent to 1 mg of solute dissolved in 1 kg of solution. 1 kg 

of water at 4˚C has a volume of 1 L. Therefore, 1.5 × 10 −4 mg of Hg 2 I 2 

is dissolved in 1.0 L of solution. 

1.5 × 10 −4 mg = 1.5 × 10 −7 g 

The molar mass of Hg 2 I 2 is 655 g/mol. 

Amount of Hg 2 I 2 = 1.5 × 10−7 g 

655 g/mol 

= 2.3 × 10 −10 mol 

[Hg 2 I 2 ] = 2.3 × 10 −10 mol/L 

[Hg 2+ 2 ] = [Hg 2 I 2 ] = 2.3 × 10 −10 mol/L 

[I − ] = 2 × [Hg 2 I 2 ] 

= 2 × (2.3 × 10 −10 mol/L) 

= 4.6 × 10 −10 mol/L 

Step 3 K sp = [Hg 2+ 2 ][I − ] 2 

= (2.3 × 10 −10 )(4.6 × 10 −10 ) 2 

= 4.9 × 10 −29 

K sp for mercury(I) iodide is 4.9 × 10 −29 at 25˚C. 

(b) 1 ppm is equivalent to 1 mg of solute dissolved in 1 kg of solution. This leads to 

two assumptions in this problem. Firstly, that the solution is so dilute it can be 

treated as 1 kg of water. Secondly, you assume that 1 kg of water has a volume 

of 1 L. This is only strictly true at 4˚C. 


Check that you wrote the K sp equation correctly. Pay attention to molar relationships 

and to the exponent of each term. Check the concentration calculations carefully. 



17. Problem 

K sp for silver chloride, AgCl, is 1.8 × 10 −10 at 25˚C. 

(a) Calculate the molar solubility of AgCl in a saturated solution at 25˚C. 

(b) How many formula units of AgCl are dissolved in 1.0 L of saturated silver 

chloride solution? 

(c) What is the percent (m/v) of AgCl in a saturated solution at 25˚C? 


You need to determine the solubility (in mol/L, in formula units, and expressed as a 

mass/volume percent) of AgCl at 25˚C. 


At 25˚C, K sp for AgCl is 1.8 × 10 −10 . 


(a) Step 1 Write the dissolution equilibrium equation. 

Step 2 Use the equilibrium equation to write an expression for K sp . 

Step 3 Set up an ICE table. Let x represent molar solubility. Use the stoichiometry 

of the equilibrium equation to write expressions for the equilibrium 

concentrations of the ions. 

Step 4 Substitute your expressions into the expression for K sp , and solve for x. 

(b) Convert the molar solubility to formula units per litre. 

(c) Convert the molar solubility to a percent (m/v). 


170

CHEMISTRY 12 


(a) Step 1 AgCl (s) 

⇀↽ Ag 2+ (aq) + Cl − (aq) 

Step 2 K sp = [Ag + ][Cl − ] 

Step 3 


AgCl (s) 

0 

Ag + (aq) 

+ 

Cl − (aq) 

Initial 

0 

Change 

+x 

+x 

Equilibrium 

x 

x 

Step 4 K sp = [Ag + ][Cl − ] = (x)(x) 

Therefore, 1.8 × 10 −10 = x 2 

x = √ 1.8 × 10 −10 

= 1.3 × 10 −5 mol/L 

The molar solubility of AgCl in water is 1.3 × 10 −5 mol/L. 

(b) 1.3 × 10 −5 mol = (1.3 × 10 −5 ) × (6.0 × 10 23 formula units/mol) 

= 7.8 × 10 18 formula units 

The solubility of AgCl in water is 7.8 × 10 18 formula units/L 

mass of solute (g) 

(c) Mass/volume percent = 

volume of solution (mL) × 100% 

Molar mass of AgCl = 143.3 g/mol 

Mass in 1 L of solution = (1.3 × 10 −5 mol/L) × (143.3 g/mol) 

= 1.9 × 10 −3 g/L 

Mass/volume percent = 1.9 × 10−3 g 

× 100% 

1000 mL 

The solubility of AgCl at 25˚C is 1.9 × 10 −4 % (m/v). 


Substitute the values of [Ag + ] and [Cl − ] into the K sp equation. You should get the given 

K sp . Check the number of significant digits given in the question, and the number of 

digits in the answers. 

18. Problem 

Iron(III) hydroxide, Fe(OH) 3 , is an extremely insoluble compound. K sp for Fe(OH) 3 

is 2.8 × 10 −39 at 25˚C. Calculate the molar solubility of Fe(OH) 3 at 25˚C. 


You need to determine the solubility (in mol/L) of Fe(OH) 3 at 25˚C. 


At 25˚C, K sp for Fe(OH) 3 is 2.8 × 10 −39 . 


Step 1 Write the dissolution equilibrium equation. 







171

CHEMISTRY 12 


Step 1 Fe(OH) 3(s) 

⇀↽ Fe 3+ (aq) + 3OH − (aq) 

Step 2 K sp = [Fe 3+ ][OH − ] 3 

Step 3 


Fe(OH) 3(s) 

0 

Fe 3+ (aq) 

+ 

3OH − (aq) 

Initial 

0 

Change 

+x 

+3x 

Equilibrium 

x 

3x 

Step 4 K sp = [Fe 3+ ][OH − ] 3 

= (x) × (3x) 3 

= 27x 4 

2.8 × 10 −39 = 27x 4 

√ 

x = 4 2.8 × 10 −39 

27 

= 1.0 × 10 −10 mol/L 

The molar solubility of Fe(OH) 3 in water is 1.0 × 10 −10 mol/L. 


Recall that x = [Fe 3+ ] eq and 3x = [OH − ] eq . Substitute these values into the K sp 

equation. You should get the given K sp . 

19. Problem 

K sp for zinc iodate, Zn(IO 3 ) 2 , is 3.9 × 10 −6 at 25˚C. Calculate the solubility (in mol 

and in g/L) of Zn(IO 3 ) 2 in a saturated solution. 


You need to determine the solubility (in mol/L and in g/L) of Zn(IO 3 ) 2 at 25˚C. 


At 25˚C, K sp for Zn(IO 3 ) 2 is 3.9 × 10 −6 . 








Step 5 Determine the solubility in g/L. 


Step 1 Zn(IO 3 ) 2(s) 

⇀↽ Zn 2+ − 

(aq) + 2IO 3 (aq) 

Step 2 K sp = [Zn 2+ ][IO − 3 ] 2 

Step 3 


Zn(IO 3 ) 2(s) 

0 

Zn 2+ (aq) 

+ 

2IO 3 

− 

(aq) 

Initial 

0 

Change 

+x 

+2x 

Equilibrium 

x 

2x 

Step 4 K sp = [Zn 2+ ][IO − 3 ] 2 

= (x) × (2x) 2 

= 4x 3 

3.9 × 10 −6 = 4x 3 


172

CHEMISTRY 12 

√ 

x = 3 3.9 × 10 −6 

4 

= 9.9 × 10 −3 mol/L 

The molar solubility of Zn(IO 3 ) 2 in water is 9.9 × 10 −3 mol/L. 

Step 5 Molar mass of Zn(IO 3 ) 2 = 415.2 g/mol 

9.9 × 10 −3 mol/L = (9.9 × 10 −3 mol/L) × (415.2 g/mol) 

= 4.1 g/L 

The solubility of Zn(IO 3 ) 2 in water is 4.1 g/L. 


Recall that x = [Zn 2+ ] eq and 2x = [IO − 3 ] eq . Substitute these values into the K sp 

equation. You should get the given K sp . The answers have the correct number of 

significant digits. 

20. Problem 

What is the maximum number of formula units of zinc sulfide, ZnS, that can dissolve 

in 1.0 L of solution at 25˚C? K sp for ZnS is 2.0 × 10 −22 . 


You need to determine the solubility (in formula units) of ZnS at 25˚C. 


At 25˚C, K sp for ZnS is 2.0 × 10 −22 . 





of the equilibrium equation to write expressions for the equilibrium concentrations 

of the ions. 


Step 5 Determine the number of formula units dissolved in 1.0 L. 


Step 1 ZnS (s) 

⇀↽ Zn 2+ (aq) + S 2− (aq) 

Step 2 K sp = [Zn 2+ ][S 2− ] 

Step 3 


ZnS (s) 

0 

Zn 2+ (aq) 

+ 

S 2− (aq) 

Initial 

0 

Change 

+x 

x 

Equilibrium 

x 

x 

Step 4 K sp = [Zn 2+ ][S 2− ] = x 2 

Therefore, 2.0 × 10 −22 = x 2 

x = √ 2.0 × 10 −22 

= 1.4 × 10 −11 mol/L 

The molar solubility of ZnS in water is 1.4 × 10 −11 mol/L. 

Step 5 1.4 × 10 −11 mol = (1.4 × 10 −11 mol) × (6.0 × 10 23 formula units/mol) 

= 8.4 × 10 12 formula units 

The solubility of ZnS in water is 8.4 × 10 12 formula units/L. 


Recall that x = [Zn 2+ ] eq = [S 2− ] eq . Substitute these values into the K sp equation. 

You should get the given K sp . 


173

CHEMISTRY 12 



21. Problem 

Determine the molar solubility of AgCl 

(a) in pure water 

(b) in 0.15 mol/L NaCl 


You need determine the solubility of AgCl (in mol/L) in water, and then in a solution 

of NaCl. 


From Appendix E, K sp for AgCl = 1.77 × 10 −10 . You know the concentration of the 

solution of NaCl. 








(b) Step 1 Set up an ICE table. The initial concentrations are based on the solution 

of NaCl. Let x represent the concentration of chloride (the common ion) 

that is contributed by AgCl. 

Step 2 Solve for x. 


(a) Step 1 AgCl (s) 

⇀↽ Ag + (aq) + Cl − (aq) 

Step 2 K sp = [Ag + ][Cl − ] = 1.77 × 10 −10 

Step 3 


AgCl (s) 

0 

Ag + (aq) 

+ 

Cl − (aq) 

Initial 

Change 

Equilibrium 

+x 

x 

0 

+x 

x 

Step 4 K sp = [Ag + ][Cl − ] 

1.77 × 10 −10 = x 2 

(b) Step 1 

x = √ 1.77 × 10 −10 

x =±1.33 × 10 −5 

The negative root has no physical meaning. The solubility of AgCl is 

1.33 × 10 −5 mol/L. 


AgCl (s) 

0 

Ag + (aq) 

+ 

Cl − (aq) 

Initial 

Change 

Equilibrium 

+x 

x 

0.15 

+x 

0.15 + x 

Step 2 Since K sp is very small, you can assume that x is much smaller than 0.15. 

To check the validity of this assumption, determine whether or not 0.15 is 

more than 500 times greater than K sp , as shown on the next page. 


174

CHEMISTRY 12 

0.15 

K sp 

= 

0.15 

1.77 × 10 −10 

= 8.5 × 10 8 > 500 

Therefore, in the (0.15 + x) term, x can be ignored. In other words, 

(0.15 + x) is approximately equal to 0.15. As a result, you can simplify 

the equation as follows: 

K sp = [Ag + ][Cl − ] 

1.77 × 10 −10 = (x)(0.15 + x) ≈ (x)(0.15) 

Therefore, 0.15x ≈ 1.77 × 10 −10 

x ≈ 1.18 × 10 −9 mol/L 

The solubility of AgCl in a solution of 0.15 mol/L NaCl is 

1.18 × 10 −9 mol/L. 


Your approximation in Step 6 was reasonable. The solubility x is much smaller than 

0.15. Le Châtelier’s principle predicts a smaller solubility in a solution containing a 

common ion, and this is confirmed by the calculations. 

22. Problem 

Determine the molar solubility of lead(II) iodide, PbI 2 , in 0.050 mol/L NaI. 


You need to determine the solubility of PbI 2 (in mol/L) in a solution of NaI. 


From Appendix E, K sp for PbI 2 = 9.8 × 10 −9 . You know the concentration of the 

solution of NaI. 




Step 3 Set up an ICE table. The initial concentrations are based on the solution of 

NaI. Let x represent the concentration of iodide (the common ion) that is 

contributed by NaI. 



Step 1 PbI 2(s) 

⇀↽ Pb 2+ (aq) + 2I − (aq) 

Step 2 K sp = [Pb 2+ ][I − ] 2 

Step 3 


Pb 2+ (aq) 

+ 

2I − (aq) 

Initial 

PbI 2(s) 

0 

0.050 

Change 

+x 

+2x 

Equilibrium 

x 

0.050 + 2x 

Step 4 Since K sp is very small, you can assume that 2x is much smaller than 0.050. 


more than 500 times greater than K sp , as shown. 

0.050 

K sp 

= 

0.050 

9.8 × 10 −9 

= 5.1 × 10 6 > 500 


175

CHEMISTRY 12 

Therefore, in the (0.050 + 2x) term, 2x can be ignored. In other words, 

(0.050 + 2x) is approximately equal to 0.050. As a result, you can simplify 


K sp = [Pb 2+ ][I − ] 2 

9.8 × 10 −9 = (x)(0.050 + 2x) 2 ≈ (x)(0.050) 2 

Therefore, (2.5 × 10 −3 )(x) ≈ 9.8 × 10 −9 

x ≈ 3.9 × 10 −6 mol/L 

The solubility of PbI 2 in a solution of 0.050 mol/L NaI is 3.9 × 10 −6 mol/L. 


Your approximation in Step 4 was reasonable. The solubility 2x is much smaller than 

0.050. You can substitute your calculated values of equilibrium concentrations into the 

expression you wrote for K sp . The calculated value should equal the given value for K sp , 

within the errors introduced by mathematical rounding. 

23. Problem 

Calculate the solubility of calcium sulfate, CaSO 4 , 


(b) in 0.25 mol/L Na 2 SO 4 


You need determine the solubility of CaSO 4 (in mol/L) in water, and then in a solution 

of Na 2 SO 4 . 


From Appendix E, K sp for CaSO 4 = 4.93 × 10 −5 . You know the concentration of the 

solution of Na 2 SO 4 . 









of Na 2 SO 4 . Let x represent the concentration of sulfate (the common ion) 

that is contributed by Na 2 SO 4 . 



(a) Step 1 CaSO 4(s) 

⇀↽ Ca 2+ 2− 

(aq) + SO 4 (aq) 

Step 2 K sp = [Ca 2+ ][SO 2− 4 ] 

= 4.93 × 10 −5 

Step 3 


CaSO 4(s) 

0 

Ca 2+ (aq) 

+ 

SO 4 

2− 

(aq) 

Initial 

Change 

Equilibrium 

Step 4 K sp = [Ca + ][SO 2− 4 ] = x 2 

4.93 × 10 −5 = x 2 

x = √ 4.93 × 10 −5 

x =±7.02 × 10 −3 

+x 

x 

0 

+x 

x 


176

CHEMISTRY 12 

The negative root has no physical meaning. The solubility of CaSO 4 is 

7.02 × 10 −3 mol/L. 

(b) Step 1 


CaSO 4(s) 

0 

Ca 2+ (aq) 

+ 

SO 4 

2− 

(aq) 

Initial 

0.25 

Change 

+x 

+x 

Equilibrium 

x 

0.25 + x 


To check the validity of this assumption, determine whether or not 0.25 

is more than 500 times greater than K sp , as shown below. 

0.25 

K sp 

= 

0.25 

4.93 × 10 −5 

= 5.1 × 10 3 > 500 

Therefore, in the (0.25 + x) term, x can be ignored. In other words, 

(0.25 + x) is approximately equal to 0.25. As a result, you can simplify 


K sp = [Ca 2+ ][SO 2− 4 ] 

4.93 × 10 −5 = (x)(0.25 + x) ≈ (x)(0.25) 

Therefore, 0.25x ≈ 4.93 × 10 −5 

x ≈ 1.97 × 10 −4 mol/L 

The solubility of AgCl in a solution of 0.15 mol/L NaCl is 

1.97 × 10 −4 mol/L. 


Your approximation in Step 6 was reasonable. The solubility x is a much smaller than 



24. Problem 

K sp for lead(II) chloride, PbCl 2 , is 1.6 × 10 −5 . Calculate the molar solubility of PbCl 2 . 


(b) in 0.10 mol/L CaCl 2 


You need to determine the solubility of PbCl 2 (in mol/L) in water, and then in a 

solution of CaCl 2 . 


You know that K sp for PbCl 2 is 1.6 × 10 −5 , and you know the concentration of the 

solution of CaCl 2 . 









of CaCl 2 . Let x represent the concentration of chloride (the common ion) 

that is contributed by CaCl 2 . 



177

CHEMISTRY 12 


(a) Step 1 PbCl 2(s) 

⇀↽ Pb 2+ (aq) + 2Cl − (aq) 

Step 2 K sp = [Pb 2+ ][Cl − ] 2 

= 1.6 × 10 −5 

Step 3 Concentration (mol/L) 

PbCl 2(s) 

0 

Pb 2+ (aq) 

+ 

2Cl − (aq) 

Initial 

Change 

Equilibrium 

+x 

x 

0 

+2x 

2x 

Step 4 K sp = [Pb 2+ ][Cl − ] 2 

(b) Step 1 

1.6 × 10 −5 = x(2x) 2 

1.6 × 10 −5 = 4x 3 

x = 3 √ 

4.0 × 10 

−6 

x = 1.6 × 10 −2 

The solubility of PbCl 2 is 1.6 × 10 −2 mol/L. 

Each formula unit of calcium chloride dissociates to form two chloride ions. 

Therefore, 0.10 mol/L CaCl 2 dissociates to give [Cl − ] = 0.20 mol/L. 


PbCl 2(s) 

0 

Pb 2+ (aq) 

+ 

2Cl − (aq) 

Initial 

0.20 

Change 

+x 

+2x 

Equilibrium 

x 

0.20 + 2x 



more than 500 times greater than K sp , as shown below. 

0.20 

K sp 

= 

0.20 

1.6 × 10 −5 

= 1.2 × 10 4 > 500 

Therefore, in the (0.20 + 2x) term, 2x can be ignored. In other words, 

(0.20 + 2x) is approximately equal to 0.20. As a result, you can simplify 


K sp = [Pb 2+ ][Cl − ] 2 

1.6 × 10 −5 = (x)(0.20 + 2x) 2 ≈ (x)(0.20) 2 

Therefore, (4.0 × 10 −2 )(x) ≈ 1.6 × 10 −5 

x ≈ 4.0 × 10 −4 mol/L 

The solubility of PbCl 2 in a solution of 0.10 mol/L CaCl 2 is 

4.0 × 10 −4 mol/L. 


Your approximation in Step 6 was reasonable. The solubility 2x is a much smaller than 




178

CHEMISTRY 12 


Student Textbook Page 441 

25. Problem 

A buffer solution is made by mixing 250 mL of 0.200 mol/L aqueous ammonia and 

400 mL of 0.150 mol/L ammonium chloride. Calculate the pH of the buffer solution. 


You need to determine the pH of the buffer solution. 


250 mL of 0.200 mol/L aqueous ammonia and 400 mL of 0.150 mol/L ammonium 

chloride are mixed. From Appendix E, K b for NH 3(aq) is 1.8 × 10 −5 . 


+ 

Step 1 Determine the initial concentrations of NH 3(aq) and NH 4 (aq) in the 

buffer solution. 

Step 2 Write the reaction for the dissociation of NH 3(aq) . Set up an ICE table, 

+ 

including the initial concentration of NH 4 (aq) in the buffer. 

Step 3 

Step 4 

Step 5 

Write the equation for K b , and substitute equilibrium terms into the equation. 

Solve the equation for x. Assume that x is small compared with the initial 

concentrations. Check the validity of this assumption when you find the 

value of x. 

Use the following equations: 


pH = 14.00 = pOH 


Step 1 When the solutions are mixed, the total volume is: 

(250 mL + 400 mL) = 650 mL 

[NH 3 ] = 

0.250 L × 0.200 mol/L 

0.650 L 

= 7.69 × 10 −2 mol/L 

[NH + 4 ] = 

0.400 L × 0.150 mol/L 

0.650 L 

= 9.23 × 10 −2 mol/L 

+ 

Step 2 Concentration (mol/L) NH + + OH − 3(aq) H 2 O (l) NH 4 (aq) 

(aq) 

Initial 

7.69 × 10 −2 

9.23 × 10 −2 ~0 

Change 

−x 

+x 

+x 

Equilibrium 

7.69 × 10 −2 − x 

9.23 × 10 −2 + x 

x 

Step 3 K b = [NH 4 + ][OH − ] 

[NH 3 ] 

= (9.23 × 10−2 + x)(x) 

(7.69 × 10 −2 − x) 

Step 4 Assume that (7.69 × 10 −2 − x) ≈ 7.69 × 10 −2 and 

(9.23 × 10 −2 + x) ≈ 9.23 × 10 −2 . 

K b = (9.23 × 10−2 )(x) 

(7.69 × 10 −2 = 1.8 × 10 

) 

−5 

Therefore, x = 1.5 × 10 −5 mol/L = [OH − ] 


179

CHEMISTRY 12 

Step 5 pOH =−log(1.5 × 10 −5 ) 

= 4.82 

pH = 14.00 − 4.82 

= 9.18 

The pH of the buffer solution is 9.18. 


The value of x (1.5 × 10 −5 ) is negligible compared with the initial concentration of 

each component. A buffer that is made using a weak base and its conjugate acid 

should have a pH that is greater than 7. 

26. Problem 

A buffer solution is made by mixing 200 mL of 0.200 mol/L aqueous ammonia and 

450 mL of 0.150 mol/L ammonium chloride. Calculate the pH of the buffer solution. 




200 mL of 0.200 mol/L aqueous ammonia and 450 mL of 0.150 mol/L ammonium 

chloride are mixed. From Appendix E, K b for NH 3(aq) is 1.8 × 10 −5 . 


+ 

Step 1 Determine the initial concentrations of NH 3(aq) and NH 4 (aq) in the 

buffer solution. 

Step 2 Write the reaction for the dissociation of NH 3(aq) . Set up an ICE table, 

+ 

including the initial concentration of NH 4 (aq) in the buffer. 

Step 3 

Step 4 

Step 5 

Write the equation for K b , and substitute equilibrium terms into the equation. 



value of x. 

Use the following equations: 


pH = 14.00 − pOH 


Step 1 When the solutions are mixed, the total volume is (250 + 400) = 650 mL. 

[NH 3 ] = 

0.200 L × 0.0200 mol/L 

0.650 L 

= 6.15 × 10 −2 mol/L 

[NH + 4 ] = 

0.450 L × 0.150 mol/L 

0.650 L 

= 1.04 × 10 −1 mol/L 

+ 

Step 2 Concentration (mol/L) NH + + OH − 3(aq) H 2 O (l) NH 4 (aq) 

(aq) 

Initial 

6.15 × 10 −2 

1.04 × 10 −1 ~0 

Change 

−x 

+x 

+x 

Equilibrium 

6.15 × 10 −2 − x 

1.04 × 10 −1 + x 

x 

Step 3 K b = [NH 4 + ][OH − ] 

[NH 3 ] 

= (1.04 × 10−1 + x)(x) 

(6.15 × 10 −2 − x) 


180

CHEMISTRY 12 

Step 4 Assume that (6.15 × 10 −2 − x) ≈ 6.15 × 10 −2 and 

(1.04 × 10 −1 + x) ≈ 1.04 × 10 −1 . 

K b = (1.04 × 10−1 + x)(x) 

(6.15 × 10 −2 = 1.8 × 10 

) 

−5 

Therefore, x = 1.1 × 10 −5 mol/L = [OH − ] 

Step 5 pOH =−log(1.1 × 10 −5 ) 

= 4.96 

pH = 14.00 − 4.96 

= 9.04 



The value of x (1.1 × 10 −5 ) is negligible compared with the initial concentration 

of each component. A buffer that is made using a weak base and its conjugate acid 

should have a pH that is greater than 7. 

27. Problem 

A buffer solution contains 0.200 mol/L nitrous acid, HNO 2(aq) , and 0.140 mol/L 

potassium nitrite, KNO 2(aq) . What is the pH of the buffer solution? 




0.200 mol/L HNO 2(aq) is mixed with 0.140 mol/L potassium nitrite, KNO 2(aq) . 

From Appendix E, K a for HNO 2 is 5.6 × 10 −4 . 


Step 1 Write the reaction for the dissociation of HNO 2 . Set up an ICE table, 

including the initial concentration of NO − 2 in the buffer. 

Step 2 Write the equation for K a , and substitute equilibrium terms into the equation. 

Step 3 Solve the equation for x. Assume that x is small compared with the initial 


value of x. 

Step 4 Use the equation pH =−log[H 3 O + ]. 


Step 1 HNO 2 + H 2 O (l) 

⇀↽ H 3 O + − 

(aq) + NO 2 (aq) 


HNO 2(aq) 

− 

+ H NO + H 3 O + 2 O (l) 2 (aq) 

(aq) 

0.140 

Initial 

0.200 

~0 

Change 

−x 

+x 

+x 

Equilibrium 

0.200 − x 

0.140 + x 

x 

Step 2 K a = [NO− 2 ][H 3O + ] 

[HNO 2 ] 

= 

(0.140 + x)(x) 

(0.200 − x) 

Step 3 Assume that (0.140 + x) ≈ 0.140 and (0.200 − x) ≈ 0.200. 

K a = (1.40)(x) = 5.6 × 10 

(0.200) 

−4 

∴x = 8.0 × 10 −4 mol/L 

Step 4 pH =−log(8.00 × 10 −4 ) 

= 3.10 



181

CHEMISTRY 12 



each component. A buffer that is made using a weak acid and its conjugate base 

should have a pH that is less than 7. 

28. Problem 

A buffer solution is prepared by dissolving 1.80 g of benzoic acid, C 6 H 5 COOH, and 

1.95 g of sodium benzoate, NaC 6 H 5 COO, in 800 mL of water. Calculate the pH of 

the buffer solution. 




1.8 g of C 6 H 5 COOH and 1.95 g of NaC 6 H 5 COO are dissolved in 800 mL of water. 

From Appendix E, K a for C 6 H 5 COOH is 6.3 × 10 −5 . 


Step 1 Determine the initial concentrations of C 6 H 5 COOH and C 6 H 5 COO − in 

the buffer solution. 

Step 2 Write the reaction for the dissociation of C 6 H 5 COOH. Set up an ICE table, 

including the initial concentration of C 6 H 5 COO − in the buffer. 

Step 3 

Step 4 

Step 5 

Write the equation for K a , and substitute equilibrium terms into the equation. 



value of x. 

Use the following equation: 

pH =−log[H 3 O + ] 


Step 1 The molar mass of C 6 H 5 COOH is 122.1 g/mol. 

1.80 g 

[C 6 H 5 COOH] = 

122.1 g/mol 

= 1.47 × 10 −2 mol 

The molar mass of NaC 6 H 5 COO is 144.1 g/mol. 

1.95 g 

[NaC 6 H 5 COO] = 

144.1 g/mol 

= 1.35 × 10 −2 mol 

Step 2 


Initial 

C 6 H 5 COOH (aq) 

1.47 × 10 −2 

+ H C + H 3 O + 6 H 5 COOH − 2 O (l) 

(aq) 

(aq) 

1.35 × 10 −2 

~0 

Change 

−x 

+x 

+x 

Equilibrium 

1.47 × 10 −2 − x 

1.35 × 10 −2 + x 

x 

Step 3 K a = [C 6H 5 COO − ][H 3 O 3 ] 

[C 6 H 5 COOH] 

= (1.35 × 10−2 + x)(x) 

(1.47 × 10 −2 − x) 

Step 4 Assume that (1.35 × 10 −2 + x) ≈ 1.35 × 10 −2 and 

(1.47 × 10 −2 − x) ≈ 1.47 × 10 −2 . 

K a = (1.35 × 10−2 + x)(x) 

(1.47 × 10 −2 = 6.3 × 10 

) 

−5 

∴x = 6.9 × 10 −5 mol/L 


182

CHEMISTRY 12 

Step 5 pH =−log(6.9 × 10 −5 ) 

= 4.16 




each component. A buffer that is made using a weak acid and its conjugate base 

should have a pH that is less than 7. 



29. Problem 

A solution contains 0.15 mol/L of NaCl and 0.0034 mol/L Pb(NO 3 ) 2 . Does a 

precipitate form? Include a balanced chemical equation for the formation of the 

possible precipitate. K sp for PbCl 2 is 1.7 × 10 −5 . 

Solution 

Write the ion product expression for lead(II) chloride. Then substitute the ion 

concentrations into the expression to determine Q sp . Compare Q sp with K sp , 

and predict whether or not a precipitate will form. 

PbCl 2(s) 

⇀↽ Pb 2+ (aq) + 2Cl − (aq) 

Ion product expression = [Pb 2+ ][Cl − ] 2 

Q sp = (0.0034) × (0.15) 2 

= 7.6 × 10 −5 

Because Q sp is greater than K sp , a precipitate will form. 


It seems reasonable that a precipitate formed, since K sp for lead(II) chloride is small 

compared with the concentration of chloride ions and lead ions. 

30. Problem 

One drop (0.050 mL) of 1.5 mol/L potassium chromate, K 2 CrO 4 , is added to 

250 mL of 0.10 mol/L AgNO 3 . Does a precipitate form? Include a balanced 

chemical equation for the formation of the possible precipitate. K sp for Ag 2 CrO 4 

is 2.6 × 10 −12 . 


Will silver chromate precipitate under the given conditions? 


You know the concentration and volume of the potassium chromate and 

silver nitrate solutions. For K 2 CrO 4 , c = 1.5 mol/L and V = 0.050 mL. 

For AgNO 3 , c = 0.10 mol/L and V = 250 mL. You also know K sp for 

silver chromate. 


Step 1 Determine the concentrations of silver ions and chromate ions in the 

reaction mixture. 

Step 2 Substitute the ion concentrations into the ion product expression for silver 

chromate to determine Q sp . 

Step 3 Compare Q sp with K sp , and predict whether or not a precipitate will form. 


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CHEMISTRY 12 


Step 1 Because the volume of the potassium chromate solution is much smaller 

than the volume of the silver nitrate solution, you can ignore it. In other 

words, assume that the final volume of the reaction mixture is 0.250 L. 

AgNO 3(aq) → Ag + − 

(aq) + NO 3 (aq) 

∴[Ag + ] = [AgNO 3 ] = 0.10 mol/L 

K 2 CrO 4(aq) → 2K + 2− 

(aq) + CrO 4 (aq) 

[CrO 4 2− ] = [K 2 CrO 4 ] = c × V initial 

V final 

= (1.5 mol/L)(0.050 mL × 1.0 × 10−3 L/mL) 

0.250 L 

= 3.0 × 10 −4 mol/L 

Step 2 The possible precipitate is silver chromate. 

2Ag + (aq) + CrO 2− 4 → Ag 2 CrO 4(s) 

Q sp = [Ag + ] 2 [CrO 2− 4 ] 

= (0.10) 2 × (3.0 × 10 −4 ) 

= 3.0 × 10 −6 

Step 3 Since Q sp > K sp , Ag 2 CrO 4 will precipitate until Q sp = 2.6 × 10 −12 . 


The units are correct in the calculation of the concentration of silver ions. It seems reasonable 

that a precipitate formed, since K sp for silver chromate is very small compared 

with the concentration of chromate ions and silver ions. 

31. Problem 

A chemist adds 0.010 g of CaCl 2 to 5.0 × 10 2 mL of 0.0015 mol/L sodium carbonate, 

Na 2 CO 3 . Does a precipitate of calcium carbonate form? Include a balanced chemical 

equation for the formation of the possible precipitate. 


Will calcium carbonate precipitate under the given conditions? 


You know the concentration and volume of the sodium carbonate solution. 

For Na 2 CO 3 , c = 0.0015 mol/L and V = 5.0 × 10 2 mL. For CaCl 2 , 0.010 g 

is added to the solution. From Appendix E, you also know that K sp for calcium 

carbonate is 3.36 × 10 −9 . 


Step 1 Determine the concentrations of calcium ions and carbonate ions in the 


Step 2 Substitute the ion concentrations into the ion product expression for calcium 

carbonate to determine Q sp . 



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CHEMISTRY 12 


Step 1 Na 2 CO 3(aq) → 2Na + 2− 

(aq) + CO 3 (aq) 

∴[CO 2− 3 ] = [Na 2 CO 3 ] = 0.0015 mol/L 

The molar mass of CaCl 2 is 111.0 g/mol. Therefore, 

0.010 g 

Amount CaCl 2 = 

111.0 g/mol 

= 9.0 × 10 −5 mol 

[CaCl 2 ] = 9.0 × 10−5 mol 

0.50 L 

= 1.8 × 10 −4 mol/L 

CaCl 2(aq) → Ca 2+ (aq) + 2Cl − (aq) 

Therefore, [Ca 2+ ] = [CaCl 2 ] = 1.8 × 10 −4 mol/L 

Step 2 The possible precipitate is calcium carbonate. 

Ca 2+ (aq) + CO 2− 3 → CaCO 3(s) 

Q sp = [Ca 2+ ][CO 2− 3 ] 

= (1.8 × 10 −4 ) × (0.0015) 

= 2.7 × 10 −7 

Step 3 Since Q sp > K sp , CaCO 3 will precipitate until Q sp = 3.36 × 10 −9 . 


The units are correct in the calculation of the concentration of silver ions. It seems 

reasonable that a precipitate formed, since K sp for calcium carbonate is small 

compared with the concentration of calcium ions and carbonate ions. 

32. Problem 

0.10 mg of magnesium chloride, MgCl 2 , is added to 2.5 × 10 2 mL of 0.0010 mol/L 

NaOH. Does a precipitate of magnesium hydroxide form? Include a balanced chemical 

equation for the formation of the possible precipitate. 


Will magnesium hydroxide precipitate under the given conditions? 


You know the concentration and volume of the sodium hydroxide solution. 

For NaOH, c = 0.0010 mol/L and V = 2.5 × 10 2 mL. For MgCl 2 , 0.10 mg 

is added to the solution. From Appendix E, you also know that K sp for magnesium 

hydroxide is 5.61 × 10 −12 . 


Step 1 Determine the concentrations of magnesium ions and hydroxide ions in the 


Step 2 Substitute the ion concentrations into the ion product expression for 

magnesium hydroxide to determine Q sp . 



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CHEMISTRY 12 


Step 1 NaOH (aq) → Na + (aq) + OH − (aq) 

Therefore, [OH − ] = [NaOH] = 0.0010 mol/L 

The molar mass of MgCl 2 is 95.2 g/mol. Therefore, 

Amount MgCl 2 = 0.10 mg × 10−3 g/mg 

95.2 g/mol 

= 1.1 × 10 −6 mol 

[MgCl 2 ] = 1.1 × 10−6 mol 

0.25 L 

= 4.2 × 10 −6 mol/L 

MgCl 2(aq) → Mg 2+ (aq) + 2Cl − (aq) 

Therefore, [Mg 2+ ] = [MgCl 2 ] = 4.2 × 10 −6 mol/L. 

Step 2 The possible precipitate is magnesium hydroxide. 

Mg 2+ (aq) + 2OH − (aq) → Mg(OH) 2(s) 

Q sp = [Mg 2+ ][OH − ] 2 

= (4.2 × 10 −6 ) × (0.0010) 2 

= 4.2 × 10 −12 

Step 3 Since Q sp < K sp , no precipitate forms. 


The units are correct in the calculation of the concentration of silver ions. It seems 

reasonable that no precipitate formed, since the concentration of magnesium ion 

is small. 



33. Problem 

1.0 × 10 2 mL of 1.0 × 10 −3 mol/L Pb(NO 3 ) 2 is added to 40 mL of 0.040 mol/L 

NaCl. Does a precipitate form? Include a balanced chemical equation for the 

formation of the possible precipitate. 


Decide whether a precipitate will form under the given circumstances. 

Write a balanced chemical equation for the formation of the possible precipitate. 


You know the initial concentration and volume of the lead(II) nitrate and sodium 

chloride solutions. For Pb(NO 3 ) 2 , c = 1.0 × 10 −3 mol/L and V = 1.0 × 10 2 mL. 

For NaCl, c = 0.040 mol/L and V = 40 mL. As well, you have the solubility 

guidelines in Table 9.3. 


Step 1 Decide whether a compound with low solubility forms when lead(II) nitrate 

and sodium chloride are mixed, using the solubility guidelines. 

Step 2 If an insoluble compound forms, write an equation that represents the 

reaction. Look up K sp for the compound in Appendix E. 

Step 3 Determine the concentrations of the ions that make up the compound. 

Step 4 Substitute the concentrations of the ions into the ion product expression for 

the compound to determine Q sp . 

Step 5 Compare Q sp with K sp , and predict whether or not a precipitate forms. 


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CHEMISTRY 12 


Step 1 When you mix the solutions, lead(II) chloride, PbCl 2 , and sodium nitrate, 

NaNO 3 , can also form. Sodium nitrate is soluble, but lead(II) chloride is 

not, according to guideline 2. 

Step 2 Pb(NO 3 ) 2(aq) + NaCl (aq) 

⇀↽ PbCl 2(s) + NaNO 3(aq) 

From Table 9.2, K sp for PbCl 2 is 1.7 × 10 −5 . 

Step 3 When calculating the volume of the reaction mixture, assume that it is equal 

to the sum of the volumes of the solutions. This is a reasonable assumption, 

because both solutions are dilute. 

[Pb 2+ ] = [Pb(NO 3 ) 2 ] = c × V initial 

V final 

= (1.0 × 10−3 mol/L)(100 mL) 

(100 mL + 40 mL) 

= 7.1 × 10 −4 mol/L 

[Cl − ] = [NaCl] = c × V initial 

V final 

= 

(0.040 mol/L)(40 mL) 

(100 mL + 40 mL) 

= 1.1 × 10 −2 mol 

Step 4 Q sp = [Pb 2+ ][Cl − ] 2 

= (7.1 × 10 −4 )(1.1 × 10 −2 ) 2 

= 8.6 × 10 −8 

Step 5 Since Q sp < K sp , no precipitate forms. 


The units in the calculation of the concentrations of the ions are correct. It seems reasonable 

that no precipitate forms, since the concentrations of ions are relatively small. 

34. Problem 

2.3 × 10 2 mL of 0.0015 mol/L AgNO 3 is added to 1.3 × 10 2 mL of 0.010 mol/L 

calcium acetate, Ca(CH 3 COO) 2 . Does a precipitate form? Include a balanced 

chemical equation for the formation of the possible precipitate. K sp for AgCH 3 COO 

is 2.0 × 10 −3 . 


Decide whether a precipitate will form under the given circumstances. Write a 

balanced chemical equation for the formation of the possible precipitate. 


You know the initial concentration and volume of the silver nitrate and calcium 

acetate solutions. For AgNO 3 , c = 0.0015 mol/L and V = 230 mL. For 

Ca(CH 3 COO) 2 , c = 0.010 mol/L and V = 130 mL. As well, you have the solubility 



Step 1 The K sp value for silver acetate indicates that it is not very soluble in water. 

Write an equation that represents the reaction to form silver acetate. 




Step 4 Compare K sp with Q sp , and predict whether or not a precipitate forms. 


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CHEMISTRY 12 


Step 1 2AgNO 3(aq) + Ca(CH 3 COO) 2(aq) 

⇀↽ 2AgCH 3 COO (s) + Ca(NO 3 ) 2(aq) 

Step 2 When calculating the volume of the reaction mixture, assume that it is equal 



[Ag + ] = [AgNO 3 ] = c × V initial 

V final 

= 

(0.0015 mol/L)(230 mL) 

(230 mL + 130 mL) 

= 9.6 × 10 −4 mol/L 

[CH 3 COO − ] = 2 × [Ca(CH 3 COO) 2 ] = 2 × c × V initial 

= 2 × 

(0.010 mol/L)(130 mL) 

(130 mL + 230 mL) 

= 7.2 × 10 −3 mol/L 

V final 

Step 3 Q sp = [Ag + ][CH 3 COO − ] 

= (9.6 × 10 −4 )(7.2 × 10 −3 ) 

= 6.9 × 10 −6 

Step 4 Since Q sp < K sp , no precipitate is formed. 


The units in the calculation of the concentrations of the ions are correct. It seems 

reasonable that no precipitate formed, since K sp for silver acetate is a relatively large 

value compared with the concentrations of the silver ions and acetate ions. 

35. Problem 

25 mL of 0.10 mol/L NaOH is added to 5.0 × 10 2 mL of 0.00010 mol/L cobalt(II) 

chloride, CoCl 2 . Does a precipitate form? Include a balanced chemical equation for 

the formation of the possible precipitate. 


You know the initial concentration and volume of the sodium hydroxide and 

cobalt(II) chloride solutions. For NaOH, c = 0.10 mol/L and V = 25 mL. For 

CoCl 2 , c = 0.00010 mol/L and V = 500 mL. As well, you have the solubility 



Step 1 Decide whether a compound with low solubility forms when sodium 

hydroxide and cobalt(II) chloride are mixed, using the solubility guidelines. 








Step 1 When you mix the sodium hydroxide and cobalt(II) chloride solutions, 

sodium chloride, NaCl, and cobalt(II) hydroxide, Co(OH) 2 , can also form. 

Sodium chloride is soluble, but cobalt(II) hydroxide is probably not, 

according to guideline 2. 

Step 2 NaOH (aq) + CoCl 2(aq) → NaCl (aq) + Co(OH) 2(s) 

From Appendix E, K sp for Co(OH) 2 is 5.92 × 10 −15 . 


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CHEMISTRY 12 

Step 3 

When calculating the volume of the reaction mixture, assume that it is equal 



[Co 2+ ] = [CoCl 2 ] = c × V initial 

V final 

= 

(0.00010 mol/L)(500 mL) 

(500 mL + 25 mL) 

= 9.5 × 10 −5 mol/L 

[OH − ] = [NaOH] = c × V initial 

V final 

= 

(0.10 mol/L)(25 mL) 

(25 mL + 500 mL) 

= 4.8 × 10 −3 mol/L 

Step 4 Q sp = [Co 2+ ][Cl − ] 2 

= (9.5 × 10 −5 )(4.8 × 10 −3 ) 2 

= 2.2 × 10 −9 

Step 5 Since Q sp > K sp , Co(OH) 2 precipitates until Q sp = 5.92 × 10 −15 . 



reasonable that a precipitate formed, since K sp for cobalt hydroxide is very small 

compared with the concentrations of the cobalt ions and hydroxide ions. 

36. Problem 

250 mL of 0.0011 mol/L Al 2 (SO 4 ) 3 is added to 50 mL of 0.022 mol/L BaCl 2 . 

Does a precipitate form? Include a balanced chemical equation for the formation 

of the possible precipitate. 


You know the initial concentration and volume of the aluminum sulfate and barium 

chloride solutions. For Al 2 (SO 4 ) 3 , c = 0.0011 mol/L and V = 250 mL. For BaCl 2 , 

c = 0.022 mol/L and V = 50 mL. As well, you have the solubility guidelines in 

Table 9.3. 


Step 1 Decide whether a compound with low solubility forms when calcium nitrate 

and sodium fluoride are mixed, using the solubility guidelines. 








Step 1 When you mix the aluminum sulfate and barium chloride solutions, 

aluminum chloride, AlCl 3 , and barium sulfate, BaSO 4 , can also form. 

Aluminum chloride is soluble, but barium sulfate is not, according to 

guideline 4. 

Step 2 Al 2 (SO 4 ) 3(aq) + 3BaCl 2(aq) → 2AlCl 3(aq) + 3BaSO 4(s) 

From Appendix E, K sp for BaSO 4 is 1.08 × 10 −10 . 


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CHEMISTRY 12 

Step 3 

When calculating the volume of the reaction mixture, assume that it is equal 



[Ba 2+ ] = [BaCl 2 ] = c × V initial 

V final 

= 

(0.022 mol/L)(50 mL) 

(50 mL + 250 mL) 

= 3.7 × 10 −3 mol/L 

[SO 4 2− ] = 3 × [Al 2 (SO 4 ) 3 ] = 3 × c × V initial 

V final 

= 3 × 

(0.0011 mol/L)(250 mL) 

(250 mL + 50 mL) 

= 2.8 × 10 −3 mol/L 

Step 4 Q sp = [Ba 2+ ][SO 2− 4 ] 

= (3.7 × 10 −3 )(2.8 × 10 −3 ) 

= 1.0 × 10 −5 

Step 5 Since Q sp > K sp , BaSO 4 precipitates until Q sp = 1.08 × 10 −10 . 



reasonable that a precipitate formed, since K sp for barium sulfate is very small 

compared with the concentrations of the barium ions and sulfate ions. 


190

Chapter 9 Answer Key Practice Questions Only - Quia

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