Question

if the solubility of ZnCO3 in water at 25 degrees celsius is 1.54*10^-3 g/L, calculate the solubility-product constant of ZnCO3, assuming complete dissociaition of ZnCO3 that has dissolved.

Answer 1

Molar mass of ZnCO3 = 1*MM(Zn) + 1*MM(C) + 3*MM(O)

= 1*65.38 + 1*12.01 + 3*16.0

= 125.39 g/mol

Molar mass of ZnCO3= 125.39 g/mol

s = 1.54*10^-3 g/L

To covert it to mol/L, divide it by molar mass

s = 1.54*10^-3 g/L / 125.39 g/mol

s = 1.228*10^-5 g/mol

The salt dissolves as:

ZnCO3 <----> Zn2+ + CO32-

   s s

Ksp = [Zn2+][CO32-]

Ksp = (s)*(s)

Ksp = 1(s)^2

Ksp = 1(1.228*10^-5)^2

Ksp = 1.508*10^-10

Answer: 1.51*10^-10