1998 #1 Solubility Product Equilibrium

Solve the following problem related to the solubility equilibria of some metal hydroxides in aqueous solution.

a) The solubility of Cu(OH)₂ is 1.72 x 10¯⁶ 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, K_sp, for Cu(OH)₂ at 25 °C.

b) The value of the solubility-product constant, K_sp, for Zn(OH)₂ is 7.7 x 10¯¹⁷ 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 9.35.
(ii) At 25°C, 50.0 milliliters of 0.100-molar Zn(NO₃)₂ is mixed with 50.0 milliliters of 0.300-molar NaOH. Calculate the molar concentration of Zn²⁺(aq) in the resulting solution once equilibrium has been established. Assume that volumes are additive.

1998 #1 Solubility Product Equilibrium Answers

a. (i) Cu(OH)₂(s) <===> Cu²⁺(aq) + 2 OH¯(aq) (one point)

Correct stoichiometry and charges (but not phases) necessary
No credit earned if water as a reactant or product

(ii) 1.72 x 10¯⁶ g / 97.57 g/mol = 1.763 x 10¯⁸ mol Cu(OH)₂ (one point)

1.763 x 10¯⁸ mol Cu (OH)₂ / 0.100 L = 1.76 x 10¯⁷ moles per liter (one point)

One point earned for conversion of mass to moles (need not to be computed explicitly)
One point earned for calculation of moles per liter

(iii) [ Cu²⁺] = 1.76 x 10¯⁷ M
[OH¯] = 2 x (1.76 x 10¯⁷ M ) = 3.52 x 10¯⁷ (one point)

K_sp = [Cu²⁺] [OH¯]² = (1.76 x 10¯⁷) (3.52 x 10¯⁷)² = 2.18 x 10¯²⁰ (one point)

One point earned for correct [Cu²⁺] and [OH¯]
One point for correct substitution into K_sp expression and answer
Response need not include explicit statement of [OH¯] if K_sp expression is written with correct values of [Cu²⁺] and [OH¯]

(b) (i) pH =9.35 ----> pOH = 4.65 ----> [OH¯] = 2.24 x 10¯⁵ M (one point)

[Zn²⁺] = K_sp / [OH¯]² = 7.7 x 10¯¹⁷ / (2.24 x 10¯⁵)² = 1.5 x 10¯⁷ M (one point)

One point earned for correct determintion of [OH¯]
One point for correct answer (assume [Zn²⁺] equals solubility in moles per liter)
No points earned if [OH¯] is assumed equal to twice [Zn²⁺]

(b) (ii)

	Zn^2+	+	2 OH¯	---->	Zn(OH)2
initial amount	0.0050 mol		0.0150 mol		0 mol
final amount	0 mol		0.0050 mol		0.0050 mol

OR

[OH¯] = 0.0050 mol / 0.100 L = 0.0050 M (one point)

One point earned if precipitation reaction is clearly indicated and moles or concentration of OH¯ is calculated correctly

Zn(OH)2(s) --->	Zn^2+	+	OH¯
	x		(0.050 + 2x)

K_sp = 7.7 x 10¯¹⁷ = [Zn²⁺] [OH¯]² = (x)(0.050 + 2x)² = (x)(0.050)² ----> [Zn²⁺] = 3.1 x 10¯¹⁴ M (one point)

OR

Zn(OH)2(s) --->	Zn^2+	+	2 OH¯
	(0.050-x		(0.150-2x)

K_sp = 7.7 x 10¯¹⁷ = [Zn²⁺] [OH¯]² = (0.050 - x)(0.150 - 2x)² (one point)

Solve for x, then subtract x from 0.050 M to obtain [Zn²⁺] (one point)

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