## Chemistry: The Molecular Nature of Matter and Change 7th Edition

$K_{sp} = (2.8 \times 10^{-11})$
1. Calculate the molar mass $(Ag_2Cr_2O_7)$: 107.87* 2 + 52* 2 + 16* 7 = 431.74g/mol 2. Calculate the number of moles $(Ag_2Cr_2O_7)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{8.3 \times 10^{-3}}{ 431.74}$ $n(moles) = 1.9\times 10^{- 5}$ 3. Find the concentration in mol/L $(Ag_2Cr_2O_7)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 1.9\times 10^{- 5}}{ 0.1}$ $C(mol/L) = 1.9\times 10^{- 4}$ 4. Write the $K_{sp}$ expression: $Ag_2Cr_2O_7(s) \lt -- \gt 2Ag^{+}(aq) + 1Cr_2O{_7}^{2-}(aq)$ $K_{sp} = [Ag^{+}]^ 2[Cr_2O{_7}^{2-}]^ 1$ 5. Determine the ion concentrations: $[Ag^{+}] = [Ag_2Cr_2O_7] * 2 = [1.9 \times 10^{-4}] * 2 = 3.8 \times 10^{-4}$ $[Cr_2O{_7}^{2-}] = [Ag_2Cr_2O_7] * 1 = 1.9 \times 10^{-4}$ 6. Calculate the $K_{sp}$: $K_{sp} = (3.8 \times 10^{-4})^ 2 \times (1.9 \times 10^{-4})^ 1$ $K_{sp} = (1.5 \times 10^{-7}) \times (1.9 \times 10^{-4})$ $K_{sp} = (2.8 \times 10^{-11})$