Solubility when multiple salts are involved Calculate the solubility of $\ce{AgCl}$ ($K_{\mathrm{sp}} = 1.6\times10^{-10}$) in $100.0~\mathrm{mL}$ of $4.00\times10^{-3}~\mathrm{M}$ calcium chloride.


The stated answer is: $2.0\times10^{-8}~\mathrm{M}$ 

My way of solving was: 
$$\begin{align}
\frac{1.6\times10^{-10}}{4.00\times10^{-3}} &amp;= 4.00\times10^{-14}\\\\
\sqrt{4.00\times10^{-14}} &amp;= 2.0\times10^{-7}
\end{align}$$

Is this correct?
I'm not sure what your train of thought was on this problem. Maybe annotating the problem would help see where you've gone wrong.

We're given that $K_{\mathrm{sp}}$ = $[\ce{Ag+}][\ce{Cl-}] = 1.6\times10^{-10}~\ce{mol^2/L^2}$. Since the solution already contains a known concentration of the chloride ion, we will solve for the amount of $[\ce{Ag+}]$ that will dissolve from the added $\ce{AgCl}$ to figure out the molar solubility of $\ce{AgCl}$.

$$[\ce{Ag+}] = \frac{K_{\mathrm{sp}}}{[\ce{Cl-}]} = \frac{1.6\times10^{-10}~\mathrm{mol^2/L^2}}{8\times10^{-3}~\mathrm{mol/L}} = 2\times10^{-8}\mathrm{mol/L}$$ 

Note that $[\ce{Cl-}]$ is double that of $\ce{[CaCl2]}$ since $$\ce{CaCl2&lt;=&gt;Ca^2+ + 2Cl-}$$

Since the dissolution of $\ce{AgCl}$ is in a one-to-one ratio with respect to the silver ion, $\ce{Ag+}$, the above answer is sufficient to show the molar solubility of $\ce{AgCl}$ in the given solution.

Any further chloride ions that dissolve will be negligible to the amount that is already in solution (according to the final solubility, only about $2\times10^{-8}~\mathrm{M}$ would be added, which is negligible to $8\times10^{-3}~\mathrm{M}$ by a factor of $400000:1$).