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The Advanced Placement

Examination in Chemistry

Part II - Free Response Questions & Answers

1970 to 2005

Solubility & Ksp

Solve the following problem

AgBr (s) → Ag

(aq) + Br

(aq) K sp

= 3.3× 10

Ag

(aq) + 2 NH 3

(aq) → Ag(NH 3

2

(aq) K = 1.7× 10

(a) How many grams of silver bromide, AgBr, can be

dissolved in 50 milliliters of water?

(b) How many grams of silver bromide can be

dissolved in 50 milliliters of 10 molar ammonia

solution?

Answer:

(a) [Ag

][Br

] = K

sp

= 3.3× 10

= X

2

X = 5.7× 10

M = [Ag

] = mol/L AgBr that dissolve

1. 050 L ∞

2. 7 ∞ 1 0

− 7

mol A gBr

1 L

∞

187. 8 g A gBr

1 mol A gBr

=

= 5.4× 10

g AgBr

(b) AgBr (s) → Ag

(aq) + Br

(aq) K sp

= 3.3× 10

Ag

(aq) + 2 NH 3

(aq) → Ag(NH 3

2

(aq) K = 1.7× 10

AgBr + 2 NH 3 → Ag(NH 3 ) 2

• Br

K = K

sp

× K = 5.6× 10

[Ag(NH 3

2

] = [Br

] = X M; [NH

3

] = (10 - 2X) M

K =

[ Ag ( NH

3

)

][ Br

−

]

[ NH

3

]

2

=

X _ X

( 10 − 2 X )

2

= 5. 6 ∞ 1 0

− 6

X = 2.4× 10

M = [Br

] = mol/L dissolved AgBr

(2.4× 10

mol/L)(187.8 g/mol)(0.050 L) = 0.22 g

AgBr

1972 D

(a) How many moles of Ba(IO 3

2

is contained in 1.

liter of a saturated solution of this salt at 25°. Ksp

of Ba(IO 3

2

= 6.5× 10

(b) When 0.100 liter of 0.060 molar Ba(NO 3

2

and

0.150 liter of 0.12 molar KIO 3

are mixed at 25°C,

how many milligrams of barium ion remains in

each milliliter of the solution? Assume that the

volumes are additive and that all activity

coefficients are unity.

Answer:

(a) Ba(IO 3

2

↔ Ba

2+

+ 2 IO

3

K

sp

= [Ba

2+

][IO

3

]

2

= 6.5× 10

[Ba

2+

] = X; [IO 3

] = 2X; (X)(2X)

2

= 6.5× 10

X = 5.5× 10

M = mol/L of dissolved Ba(IO 3

2

(b) initial mol Ba

2+

= (0.060 mol/L)(0.100L) = 0.

mol

initial mol IO 3

= (0.150L)(0.120 mol/L) = 0.

mol

after reaction, essentially all Ba

2+

reacts while IO 3

= {0.0180 - (2)(0.0060)} mol = 0.0060 mol/0.

L = 0.024M [IO 3

]

[ Ba

2

] =

K

sp

[ IO

3

−

]

2

=

6. 5 ∞ 1 0

− 10

( 0. 024 )

2

= 1. 1 ∞ 1 0

− 6

M

1. 1 ∞ 1 0

− 6

mol

100 0 mL

∞

137340 mg B a

2+

1 mol

=

= 1.5× 10

mg / mL Ba

2+

1973 D

The molar solubility of silver bromide is diminished

by the addition of a small amount of solid potassium

bromide to a saturated solution. However, the molar

solubility of silver bromide is increased by the

addition of solid potassium nitrate, a salt whose ions

are not common to those of silver bromide.

Explain these experimental observations in terms of

the principles involved.

Answer:

AgBr (s) ↔ Ag

(aq) + Br

(aq) ; As KBr dissolves, the

concentration of Br

ions increase and force the

equilibrium to shift to the left (LeChatelier’s principle)

where the concentrations of the ions in solution

decrease and less can dissolve.

The diverse (“uncommon”) ion effect – “the salt

effect”. As the total ionic concentration of a solution

increases, interionic attractions become more

important. Activities become smaller than the

stoichiometric or measured concentrations. For the

ions involved in the solution process this means that a

higher concentration must appear in solution before

equilibrium is established. - the solubility must

increase.

1977 D

The solubility of Zn(OH) 2

is not the same in the

following solutions as it is in pure water. In each case

state whether the solubility is greater or less than that

in water and briefly account for the change in

solubility.

(a) 1-molar HCl (c) 1-molar NaOH

(b) 1-molar Zn(NO 3

2

(d) 1-molar NH 3

Answer:

• Magnesium
• Sodium
• Silver

The following reagents are used for identifying the

metals.

• Pure water
• A solution of 1.0 molar HCl
• A solution of concentrated HNO 3

(a) Which metal can be easily identified because it is

much softer than the other two? Describe a

chemical test that distinguishes this metal from

the other two, using only one of the reagents

above. Write a balanced chemical equation for the

reaction that occurs.

(b) One of the other two metals reacts readily with the

HCl solution. Identify the metal and write the

balanced chemical equation for the reaction that

occurs when this metal is added to the HCl

solution. Use the table of standard reduction

potentials (attached) to account for the fact that

this metal reacts with HCl while the other does

not.

(c) The one remaining metal reacts with the

concentrated HNO 3

solution. Write a balanced

chemical equation for the reaction that occurs.

(d) The solution obtained in (c) is diluted and a few

drops of 1 M HCl is added. Describe what would

be observed. Write a balanced chemical equation

for the reaction that occurs.

Answer:

(a) Sodium is softest of the three.

Na added to water → gas and base

Na + H 2

O → H

2

• NaOH

(b) Mg reacts with HCl. Mg + 2 H

→ Mg

2+

+ H

2

Reduction potentials, E°: Mg = -2.37v; Ag = +

0.80v. Mg, not Ag, reacts with HCl

(c) Ag + 4 H

+ NO

3

→ 3 Ag

+ NO + 2 H

2

O OR

Ag + 2 H

+ NO 3

→ Ag

+ NO 2 + H 2 O

(d) A white precipitate forms: Ag

• Cl

→ AgCl (s)

1990 A

The solubility of iron(II) hydroxide, Fe(OH) 2

, is

1.43× 10

gram per litre at 25°C.

(a) Write a balanced equation for the solubility

equilibrium.

(b) Write the expression for the solubility product

constant, Ksp, and calculate its value.

(c) Calculate the pH of a saturated solution of

Fe(OH) 2

at 25°C.

(d) A 50.0 millilitre sample of 3.00× 10

molar FeSO 4

solution is added to 50.0 millilitres of 4.00× 10

molar NaOH solution. Does a precipitate of

Fe(OH) 2

form? Explain and show calculations to

support your answer.

Answer:

(a) Fe(OH) 2

→ Fe

2+

+ 2 OH

(b)

×

−

3

g

L

×

1 mol

89. 9 g

×

−

5

mol

L

Fe(OH)

2

= 1.59× 10

M = [Fe

2+

]

= 3.18× 10

M = [OH

]

K

sp

= [Fe

2+

][OH

]

2

= (1.59× 10

)(3.18× 10

2

= 1.61× 10

(c)

[ H

] =

1. 0 ∞ 1 0

− 14

[ OH

−

]

=

1. 0 ∞ 1 0

− 14

3. 18

∞ 1 0

− 5

= 3. 14 ∞ 1 0

− 10

M

pH = -log[H

] = 9.

OR

pOH = -log[OH

] = -log(3.18× 10

pH = 14 - pOH = 9.

(d) 50.0 mL of 3.00× 10

M Fe

2+

diluted of 100.0 mL

= 1.50× 10

M Fe

2+

50.0 mL of 4.00× 10

M OH

diluted of 100.0 mL

= 2.00× 10

M OH

Q = [Fe

2+

][OH

]

2

= (1.50× 10

)(2.00× 10

2

= 6.00× 10

Precipitate will NOT form since Q < K sp

1995 D (repeated in the thermo section)

Lead iodide is a dense, golden yellow, slightly soluble

solid. At 25°C, lead iodide dissolves in water forming

a system represented by the following equation.

PbI 2 (s) ↔ Pb

2+

+ 2 I

∆ H = +46.5 kilojoules

(a) How does the entropy of the system PbI 2

(s) +

H

2

O (l) change as PbI 2

(s) dissolves in water at 25°C?

Explain

(b) If the temperature of the system were lowered

from 25°C to 15°C, what would be the effect on

the value of K sp

? Explain.

(c) If additional solid PbI 2

were added to the system

at equilibrium, what would be the effect on the

concentration of I

in the solution? Explain.

(d) At equilibrium, ∆ G = 0. What is the initial effect

on the value of ∆ G of adding a small amount of

Pb(NO 3

2

to the system at equilibrium? Explain.

Answer:

(a) Entropy increases. At the same temperature,

liquids and solids have a much lower entropy than

do aqueous ions. Ions in solutions have much

greater “degrees of freedom and randomness”.

(b) K sp

value decreases. K sp

= [Pb

2+

][I

]

2

. As the

temperature is decreased, the rate of the forward

(endothermic) reaction decreases resulting in a net

decrease in ion concentration which produces a

smaller K sp

value.

(c) No effect. The addition of more solid PbI 2

does

not change the concentration of the PbI 2

which is

a constant (at constant temperature), therefore,

neither the rate of the forward nor reverse reaction

is affected and the concentration of iodide ions

remains the same.

(d) ∆ G increases. Increasing the concentration of Pb

2+

ions causes a spontaneous increase in the reverse

reaction rate (a “shift left” according to

LeChatelier’s Principle). A reverse reaction is

spontaneous when the ∆ G >0.

1998 A (Required)

Solve the following problem related to the solubility

equilibria of some metal hydroxides in aqueous

solution.

(a) The solubility of Cu(OH) 2

(s) is 1.72× 10

gram per

100. milliliters of solution at 25°C.

(i) Write the balanced chemical equation for the

dissociation of Cu(OH) 2

(s) in aqueous

solution.

(ii) Calculate the solubility (in moles per liter) of

Cu(OH) 2

at 25°C.

(iii) Calculate the value of the solubility-product

constant, Ksp , for Cu(OH) 2 at 25°C.

(b) The value of the solubility-product constant, K sp

for Zn(OH) 2

is 7.7× 10

at 25°C.

(i) Calculate the solubility (in moles per liter) of

Zn(OH) 2

at 25°C in a solution with a pH of

(ii) At 25 °C, 50.0 milliliters of 0.100-molar

Zn(NO 3

2

is mixed with 50.0 milliliters of

0.300-molar NaOH. Calculate the molar

concentration of Zn

2+

(aq) in the resulting

solution once equilibrium has been

established. Assume that volumes are

additive.

Answer

(a)i Cu(OH) 2

→ Cu

2+

+ 2 OH

ii

1. 72 ∞ 10

− 6

g

1. 10 0 L

∞

1 mol

97.5 g

= 1. 76 ∞ 10

− 7

mol

L

iii K sp

= [Cu

2+

][OH

]

2

= [1.76× 10

][3.53× 10

]

2

= 2.20× 10

(b)i Zn(OH) 2

→ Zn

2+

+ 2 OH

Ksp = [Zn

2+

][OH

]

2

pH 9.35 = pOH 4.65; [OH

] = 10

-pOH

[OH

] = 10

-4.

= 2.24× 10

M

[Zn

2+

] = solubility of Zn(OH) 2

in

mol

/ L

[Zn

2 +

]

=

K

sp

[OH

]

2

=

7. 7

∞ 10

− 1 7

( 2. 24

∞ 10

− 5

)

2

=

1. 5

∞ 10

− 7

M

ii [Zn

2+

]init = 0.100 M ×

50 mL

100 mL

= 0.0500 M

[OH

]

init

= 0.300 M ×

50 mL

100 mL

= 0.150 M

X = conc. loss to get to equilibrium

K

sp

= 7.7× 10

= (0.0500 - X )(0.150 - 2 X )

2

[Zn

2+

] = 0.0500 - X = 3.1× 10

M

2001 A Required

Answer the following questions relating to the

solubility of the chlorides of silver and lead.

(a) At 10°C, 8.9 × 10

g of AgCl (s) will dissolve in

100. mL of water.

(i) Write the equation for the dissociation of

AgCl (s) in water.

(ii) Calculate the solubility, in mol L

, of AgCl (s)

in water at 10°C.

(iii) Calculate the value of the solubility-product

constant, K sp

for AgCl (s) at 10°C.

(b) At 25°C, the value of K sp

for PbCl 2

(s) is 1.6 × 10

and the value of K sp

for AgCl (s) is 1.8 × 10

(i) If 60.0 mL of 0.0400 M NaCl (aq) is added to

60.0 mL of 0.0300 M Pb(NO 3

2

(aq) , will a

precipitate form? Assume that volumes are

additive. Show calculations to support your

answer.

volume. As the volume evaporates and becomes

half, half of the Ag

will precipitate out.

FROM OTHER PACKET:

1985 A

At 25ºC the solubility product constant, Ksp, for

strontium sulfate, SrSO 4

, is 7.6× 10

• . The solubility

product constant for strontium fluoride, SrF 2

, is

7.9× 10

(a) What is the molar solubility of SrSO 4

in pure

water at 25ºC?

(b) What is the molar solubility of SrF 2

in pure water

at 25ºC?

(c) An aqueous solution of Sr(NO 3

2

is added slowly

to 1.0 litre of a well-stirred solution containing

0.020 mole F

and 0.10 mole SO 4

2-

at 25ºC. (You

may assume that the added Sr(NO 3

2

solution does

not materially affect the total volume of the

system.)

1. Which salt precipitates first?
2. What is the concentration of strontium ion,

Sr

2+

, in the solution when the first precipitate

begins to form?

(d) As more Sr(NO 3

2

is added to the mixture in (c) a

second precipitate begins to form. At that stage,

what percent of the anion of the first precipitate

remains in solution?

Answer:

(a) SrSO 4

(s) ↔ Sr

2+

(aq) + SO 4

2-

(aq)

At equilibrium: [Sr

2+

] = X M = [SO

4

2-

]

X

2

= K

sp

= 7.6× 10

X = 8.7× 10

mol/L, solubility of SrSO 4

(b) SrF 2

(s) ↔ Sr

2+

(aq) + 2 F

(aq)

At equilibrium: [Sr

2+

] = X M = [F

] = 2 X M

KSP = [Sr

2+

][F

]

2

= ( X )(2 X )

2

= 7.9× 10

X = 5.8× 10

mol/L, solubility of SrF 2

(c) Solve for [Sr

2+

] required for precipitation of each

salt.

K

sp

= [Sr

2+

][F

]

2

= 7.9× 10

= ( x )

( 0. 020 mol )

2

1. 0 L

⊇

⊄

ℑ

_

↓

= 7. 9 ∞ 1 0

− 10

; x = 2. 0 ∞ 1 0

− 6

M

K

sp

= [Sr

2+

][SO

4

2-

] = 7.6× 10

= ( y )

1. 10 mol )

2. 0 L

⊇

⊄

ℑ

_

↓

= 7. 6 ∞ 1 0

− 7

; y = 7. 6 ∞ 1 0

− 6

M

Since 2.0× 10

M < 7.6× 10

M, SrF 2

must

precipitate first.

When SrF 2

precipitates, [Sr

2+

] = 2.0× 10

M

(d) The second precipitate to form is SrSO 4

, which

appears when [Sr

2+

] = 7.6× 10

M (based on

calculations in Part c.)

When [Sr

2+

] = 7.6× 10

M, [F

] is determined as

follows:

K

sp

= [Sr

2+

][F

]

2

= 7.9× 10

= (7.6× 10

)(z)

2

= 7.9× 10

; z = 1.0× 10

M

% F

−

still in solution

=

∞ 10

2. 0

∞ 10

− 2

∞ 100

= 50 .%

1994 A

MgF 2

(s) ↔ Mg

2+

(aq) + 2 F

(aq)

In a saturated solution of MgF 2

at 18ºC, the

concentration of Mg

2+

is 1.21× 10

molar. The

equilibrium is represented by the equation above.

(a) Write the expression for the solubility-product

constant, Ksp, and calculate its value at 18ºC.

(b) Calculate the equilibrium concentration of Mg

2+

in

1.000 liter of saturated MgF 2

solution at 18ºC to

which 0.100 mole of solid KF has been added.

The KF dissolves completely. Assume the volume

change is negligible.

(c) Predict whether a precipitate of MgF 2

will form

when 100.0 milliliters of a 3.00× 10

-molar

Mg(NO 3

2

solution is mixed with 200.0 milliliters

of a 2.00× l

-molar NaF solution at 18ºC.

Calculations to support your prediction must be

shown.

(d) At 27ºC the concentration of Mg

2+

in a saturated

solution of MgF 2

is 1.17× 10

molar. Is the

dissolving of MgF 2

in water an endothermic or an

exothermic process? Give an explanation to

support your conclusion.

Answer:

(a) K sp

= [Mg

2+

][F

]

2

= (1.21× 10

)(2.42× 10

2

= 7.09× 10

(b) X = concentration loss by Mg

2+

ion

2 X = concentration loss by F

ion

[Mg

2+

] = (1.21× 10

- X ) M

[F

] = (0.100 + 2.42× 10

- 2 X ) M

since X is a small number then (0.100 + 2.42× 10

- 2 X ) ≈ 0.

K

sp

= 7.09× 10

= (1.21× 10

- X )(0.100)

2

X = 1.2092914× 10

[Mg

2+

]= 1.21× 10

-1.20929× 10

= 7.09× 10

M

(c) [Mg

2+

] = 3.00× 10

M × 100.0 mL/300.0 mL =

1.00× 10

M

[F

] = 2.00× 10

M × 200.0 mL/300.0 mL =

1.33× 10

M

trial K sp

= (1.00× 10

)(1.33× 10

2

= 1.78× 10

trial K sp

< = 7.09× 10

, ∴no ppt.

(d) @ 18ºC, 1.21× 10

M MgF 2

dissolves

@ 27ºC, 1.17× 10

M MgF 2

dissolves

MgF 2

↔ Mg

2+

+ 2 F

• heat

dissolving is exothermic; if heat is increased it

forces the equilibrium to shift left (according to

LeChatelier’s Principle) and less MgF 2

will

dissolve.