Answer
$ K_{sp}$ (Silver Arsenate) $= (3.1 \times 10^{-22})$
Work Step by Step
1. Calculate the molar mass:
107.87* 3 + 74.92* 1 + 16* 4 = 462.53g/mol
2. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 8.5\times 10^{- 7}}{ 462.53}$
$n(moles) = 1.838\times 10^{- 9}$
3. Find the concentration in mol/L:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 1.838\times 10^{- 9}}{ 0.001} $
$C(mol/L) = 1.838\times 10^{- 6}$
4. Write the $K_{sp}$ expression:
$ Ag_3AsO_4(s) \lt -- \gt 3Ag^+(aq) + 1As{O_4}^{3-}(aq)$
$ K_{sp} = [Ag^+]^ 3[As{O_4}^{3-}]^ 1$
5. Determine the ions concentrations:
$[Ag^+] = [Ag_3AsO_4] * 3 = [1.838 \times 10^{-6}] * 3 = 5.513 \times 10^{-6}$
$[As{O_4}^{3-}] = [Ag_3AsO_4] * 1 = 1.838 \times 10^{-6}$
6. Calculate the $K_{sp}$:
$ K_{sp} = (5.513 \times 10^{-6})^ 3 \times (1.838 \times 10^{-6})^ 1$
$ K_{sp} = (1.675 \times 10^{-16}) \times (1.838 \times 10^{-6})$
$ K_{sp} = (3.079 \times 10^{-22})$
