Answer
$ K_{sp} (CaCrO_4) = (7.07 \times 10^{-4})$
Work Step by Step
1. Calculate the molar mass $(CaCrO_4)$:
40.08* 1 + 52* 1 + 16* 4 = 156.08g/mol
2. Calculate the number of moles $(CaCrO_4)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 4.15}{ 156.08}$
$n(moles) = 0.02659$
3. Find the concentration in mol/L $(CaCrO_4)$:
$0.02659$ mol in 1L: $0.02659 M (CaCrO_4)$
4. Write the $K_{sp}$ expression:
$ CaCrO_4(s) \lt -- \gt 1Ca^{2+}(aq) + 1CrO_4^{2-}(aq)$
$ K_{sp} = [Ca^{2+}]^ 1[CrO_4^{2-}]^ 1$
5. Determine the ion concentrations:
$[Ca^{2+}] = [CaCrO_4] * 1 = [0.02659] * 1 = 0.02659$
$[CrO_4^{2-}] = [CaCrO_4] * 1 = 0.02659$
6. Calculate the $K_{sp}$:
$ K_{sp} = (0.02659) \times (0.02659)$
$ K_{sp} = (7.07 \times 10^{-4})$
