Question

What is the solubility of CaCO₃ in g/mL if the K_sp = 9.9 x 10^-9

Answer 1

CaCO₃ $\rightarrow$ Ca²⁺ + CO₃^2-

Solubility (s) = [Ca²⁺] = [CO₃^2-]

K_sp= [Ca²⁺][CO₃^2-]

So, K_sp = s²

s = √K_sp

= √ 9.9 x 10^-9

= 9.94 x 10^-5 mol/L

= 9.94 x 10^-8 mol/ml

Molas mass of CaCO₃ = 100g/mol

s = (100 g/mol) x (9.94 x 10^-8 mol/ml)

= 9.94 x 10^-6 g/ml

So the amount of CaCO₃ dissolved in water is 9.94 x 10^-6 g/ml.