The solubility product of $\ce{AgCl}$ in water is given as $1.8\cdot10^{-10}$ Determine the solubility of $\ce{AgCl}$ in pure water and in a $0.25~\mathrm{M}$ $\ce{MgSO4}$ solution.

I solve to the $K_\mathrm{{sp}}$ and got $1.34\cdot10^{-10}$ for Part B I was told to use the activity and am not sure if I did it right, my new $K_\mathrm{{sp}}$ turn out to be $1.123\cdot10^{11}$ could someone help me with part B?

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    – Asker123
    Commented Mar 12, 2015 at 0:50

1 Answer 1


In pure water, as you calculated, for the reaction
$\ce{AgCl -> Ag+ + Cl-}$ with $K_\mathrm{sp} = 1.8\cdot10^{-10}$
the concentration of $\ce{Ag+}$, i.e. $\ce{[Ag+]}$ will be $1.34\cdot10^{-5}~\mathrm{M}$.

In a solution of $0.25~\mathrm{M}$ $\ce{MgSO4}$ (which dissociates almost completely), there's an additional equilibrium here:
$\ce{Ag2SO4 -> 2Ag+ + SO4^2-}$ with $K_\mathrm{sp} = 1.4\cdot10^{-5}$

Looking at the $K_\mathrm{sp}$ of both the reactions, we can see that $\ce{AgCl}$, having the lesser solubility, decides the $\ce{[Ag+]}$. Here we can even check if the concentration of $\ce{[Ag+]} = 1.34\cdot10^{-5}~\mathrm{M}$ and $\ce{[SO4^{2-}]} = 0.25~\mathrm{M}$ exceeds the solubility limit of $\ce{Ag2SO4}$. It doesn't. $(1.34\cdot10^{-5})^2 \cdot 0.25 = 4.5\cdot10^{-11} < (K_\mathrm{sp})_{\ce{Ag2SO4}}$

The final composition of the solution will be $\ce{[Ag+]} = 1.34\cdot10^{-5}~\mathrm{M}$, $\ce{[Cl- ]} = 1.34\cdot10^{-5}~\mathrm{M}$, $\ce{[Mg^2+]} = 0.25~\mathrm{M}$ and $\ce{[SO4^2- ]} = 0.25~\mathrm{M}$.

In conclusion, the $\ce{[Ag+]}$ is the same for both the cases.

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