Question

What is the equilibrium Mn²⁺ concentration when 1.96 L of a 0.199 M manganese(II) acetate solution are mixed with 2.04 L of a 0.356 M potassium carbonate solution?

[Mn²⁺] = ? M

Answer 1

Initial moles of Mn(CH3COO)2 = 1.96*0.199 = 0.39004 moles

Initial moles of K2CO3 = 2.04* 0.356 = 0.72624

Mn(CH3COO)2  + K2CO3 = MnCO3+ 2KCH3COO

I 0.39004 0.72624

C - x - x +x +2x

E 0.39004-x 0.72624-x +x +2x

The solubility product, Ksp, for manganese (II) carbonate is 5x10^-10

Ksp = [MnCO3] [K CH3COO]² / [K2CO3] [Mn(CH3COO)2]

5*10^-10 = x*(2x)² / (0.72624-x)*(0.39004-x)

x = 3.28*10^-4

Moles of Mn2+ = moles of MnCO3 = 3.28*10^-4 moles

[Mn2+] = 3.28*10^-4 / (1.96+2.04) = 8.21*10^-5 M