## Introductory Chemistry (5th Edition)

$K_{sp} (CaCrO_4) = (7.07 \times 10^{-4})$
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 $K_{sp}$: $K_{sp} = (0.02659) \times (0.02659)$ $K_{sp} = (7.07 \times 10^{-4})$