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: $$\begin{align} \frac{1.6\times10^{-10}}{4.00\times10^{-3}} &= 4.00\times10^{-14}\\\\ \sqrt{4.00\times10^{-14}} &= 2.0\times10^{-7} \end{align}$$

Is this correct?

  • 2
    $\begingroup$ You've got a few problems: 1. The concentration of chloride in a solution of $\ce{CaCl2}$ is twice that of the salt. 2. A common calculator error when dividing using scientific notation. (The upper fraction does not equal $4\times 10^{-14}$, it is $4\times 10^{-8}$ as written.) 3. The dissociation equilibrium for silver chloride doesn't contain any squares, so there's no reason to use a square root in the solution. $\endgroup$ Commented Jul 9, 2015 at 13:54

1 Answer 1


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 $$\ce{CaCl2<=>Ca^2+ + 2Cl-}$$

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 $400000:1$).


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