Answer
$ K_{sp} = (2.8 \times 10^{-11})$
Work Step by Step
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})$
