Question

Part A)

Silver chromate is sparingly soluble in aqueous solutions. The Ksp of Ag, CrO, is 1.12 x 10-12 M'. What is the solubility (in moles per liter) of silver chromate in a 1.30 M potassium chromate aqueous solution? solubility: What is the solubility (in moles per liter) of silver chromate in a 1.30 M silver nitrate aqueous solution? solubility: M . What is the solubility (in moles per liter) of silver chromate in pure water? solubility: solubility:

Part B)

Answer 1

Part A

i) Molar solubility of Ag2CrO4 in 1.30M potassium chromate

Solubility equilibrium of Ag2CrO4 is

Ag2CrO4(s) <------> 2Ag⁺(aq) + CrO4^2-(aq)

Ksp = [Ag⁺]²[CrO4^2-] = 1.12 ×10^-12M³

Initial concentration

[Ag⁺] = 0

[CrO4^2-] = 1.30

change in concentration

[Ag⁺] = + 2x

[CrO4^2-] = + x

Equilibrium concentration

[Ag⁺] = 2x

[CrO4^2-] = 1.30 + x

so,

(2x)²(1.30 + x) = 1.12 ×10^-12

solving for x

x = 4.64×10^-7

Therefore

molar solubility of Ag2CrO4 in 1.30M potassium chromate = 4.64 ×10^-7M

ii) Molar solubility Ag2CrO4 in 1.30M silver nitrate

[Ag⁺]²[CrO4^2-] = 1.12 ×10^-12

Initial concentration

[Ag⁺] = 1.30

[CrO4^2-] = 0

change in concentration

[Ag⁺] = + 2x

[CrO4^2-] = + x

equilibrium concentration

[Ag⁺] = 1.30 + 2x

[CrO4^2-] = x

so,

(1.30 +2x)²( x) = 1.12 ×10^-12

solving for x

x = 6.63 × 10^-13

Therefore,

molar solubility of Ag2CrO4 in 1.30M AgNO3 = 6.63 ×10^-13M

iii)

Ksp = [Ag⁺]²[CrO4^2-] = 1.12 ×10^-13

if solubility of Ag2CrO4 is represented by S

at saturate solution

[Ag⁺] = 2S

[CrO4^2-] = 1S

(2S)²(1S) = 1.12 ×10^-12 M³

4S³ = 1.12 ×10^-12 M³

S³ = 2.80 ×10^-13M³

S = 6.54 × 10^-5M

Therefore,

molar solubility of Ag2CrO4 in pure water = 6.54 ×10^-5 M

Part B

pOH = 14 - pH

pOH = 14 - 9.70

pOH = 4.30

pOH = -log[OH^-]

-log[OH^-] = 4.30

[OH^-] = 5.01×10^-5M

Solubility equilibrium of Fe(OH)2 is

Fe(OH)2(s) <------> 2Fe²⁺(eq) + 2OH^-(aq)

Ksp = [Fe²⁺][OH^-]² = 4.87×10^-17

substitute the OH^- concentration in Ksp expression

[Fe²⁺] ×( 5.01 × 10^-5M) = 4.87 ×10^-17M²

[Fe²⁺] = 1.94 × 10^-8M

Therefore,

Fe(OH)2 will precipitate above 1.94×10^-8 M of Fe²⁺