Answer to Question #250597 in General Chemistry for cash

Question

What is the solubility of calcite (CaCO₃) in grams per milliliter at a temperature at which its K_sp is 9.89×10^-9?

Answer 1

CaCO3 (s) ------> Ca2+ (aq) + CO32- (aq)

Ksp = [Ca2+][CO32-] =9.9 x 10^-9

Let s = solubility of CaCO3

[Ca2+] = s

[CO32-] = s

Ksp = s^2

s^2 = 9.9 x 10^-9

s = 9.95x 10^-5 mol/L

The molar mass of CaCO3 = 100.1 g/mol

9.95 x 10^-5 mol/L x (100.1 g / 1 mol) = 0.00995 g/L