Question

Need help on Solubility Equilibria and explanation how to do and steps please! :)

1a) What is the molar solubility of Iron(II) carbonate in 0.10M Sodium Carbonate? The solubility product constant (Ksp) is 2.1x10^-11

Na₂CO₃$\rightarrow$0.20M Na⁺ and 0.10M CO₃^2-

1b) What is the molar solubility of Silver phosphate (Ksp= 1.8x10^-18) in 0.20M silver nitrate?

AgNO₃$\rightarrow$0.20M Ag⁺ and 0.20M NO₃^-

Answer 1

Q1a) Let molar solubility of FeCO₃ = s

FeCO₃ dissociates as : FeCO₃ (s) $\rightarrow$ Fe²⁺ (aq) + CO₃^2- (aq)

initial concentration of Fe²⁺ = molar solubility of FeCO₃ = s

initial concentration of CO₃^2- = (molar solubility of FeCO₃) + (concentration from Na₂CO₃)

initial concentration of CO₃^2- = s + 0.10 M

K_sp FeCO₃ = [Fe²⁺][CO₃^2-]

2.1 x 10^-11 = (s) * (s + 0.10 M)

assuming s << 0.10 M

2.1 x 10^-11 = (s) * (0.10 M)

s = (2.1 x 10^-11) / (0.10 M)

s = 2.1 x 10^-10 M

molar solubility of FeCO₃ = s = 2.1 x 10^-10 M

Q1b) Let molar solubility of Ag₃PO₄ = s

Ag₃PO₄ dissociates as : Ag₃PO₄ (s) $\rightarrow$ 3 Ag⁺ (aq) + PO₄^3- (aq)

initial concentration Ag⁺ = 3 * (molar solubility of Ag₃PO₄) + (concentration from AgNO₃)

initial concentration Ag⁺ = 3 * (s) + 0.20 M

initial concentration PO₄^3- = s

K_sp Ag₃PO₄ = [Ag⁺]³[PO₄^3-]

1.8 x 10^-18 = (3s + 0.20 M)³ * (s)

assuming 3s << 0.20 M

1.8 x 10^-18 = (0.20 M)³ * (s)

s = (1.8 x 10^-18) / (0.20 M)³

s = 2.25 x 10^-16 M

molar solubility of Ag₃PO₄ = s = 2.25 x 10^-16 M