The $K_{s p}$ of $Ag X_{2} CrO X_{4}$ is $1.5 \times 1 0^{- 12}$ . Calculate its molar solubility in pure water, expressed in units of $1 0^{- 5} mol L^{- 1}$ (two decimals).

Numerical answer type, answer in x 10^-5 mol/L, tolerance 0.02.

Answer: 7.21 x 10^-5 mol/L

K_{s p} = 4 s^{3}

, so

s = (1.5 \times 1 0^{- 12} /4)^{1/3} = 7.2 \times 1 0^{- 5} M

, i.e. 7.21 in units of

1 0^{- 5} M