Answer
The molar solubility of $AgCl$ in that solution is equal to $8.9 \times 10^{-10}$ M.
Work Step by Step
1. Calculate the molar mass $(CaCl_2)$:
40.08* 1 + 35.45* 2 = 110.98g/mol
2. Calculate the number of moles $(CaCl_2)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 10.0}{ 110.98}$
$n(moles) = 0.0901$
3. Find the concentration in mol/L $(CaCl_2)$:
$0.0901$ mol in 1L: $0.0901 M (CaCl_2)$
- Therefore, since each $CaCl_2$ has 2 chloride ions, $[Cl^-] = [CaCl_2]*2 = 0.0901*2 = 0.1802$ M
4. Write the $K_{sp}$ expression:
$ AgCl(s) \lt -- \gt 1Cl^-(aq) + 1Ag^{+}(aq)$
$1.6 \times 10^{-10} = [Cl^-]^ 1[Ag^{+}]^ 1$
$1.6 \times 10^{-10} = (0.1802 + S)^ 1( 1S)^ 1$
5. Find the molar solubility.
Since 'S' has a very small value, we can approximate: $[Cl^-] = 0.1802$
$1.6 \times 10^{-10}= (0.1802)^ 1 \times ( 1S)^ 1$
$ \frac{1.6 \times 10^{-10}}{0.1802} = ( 1S)^ 1$
$8.9 \times 10^{-10} = S$
