Question:
Calculate the molar sulfate concentration in a solution that is $0.200$ M in $\ce{Ag+}$ and saturated with $\ce{Ag2SO4}$.
My attempt:
Looking through old notes and googling lead me to this webpage. Example #2 looked similar to my problem, and my notes from general chemistry involved something called an ICE table to deal with stoichiometry and concentrations of ions. This is what I did:
- Write the dissociation equation $\ce{Ag2SO4 + <=> 2Ag+ + SO4^2-}$
- Write the solubility constant formula $$K_\mathrm{sp} = \frac{\text{products}}{\text{reactants}}$$
I was given the value of $K_\mathrm{sp}$ as $1.6 \times 10^{-5}$ and I assume these values are listed in an appendix in my text or online.
Also $\ce{Ag2SO4}$ is supposed to be a solid, I think? Because the question states the solution is "saturated" with $\ce{Ag2SO4}$ and does "saturation" not imply that the $\ce{Ag2SO4}$ is going in and out of solution as a solid and as an ion? And therefore since it is a solid we do not consider it for the equation (not sure why, but this is similar to what I have in my notes)?
$$1.6 \times 10^{-5} = \text{products}$$
$$1.6 \times 10^{-5} = \ce{[Ag+]^2 * [SO4^2- ]}$$
Here is an image if that is acceptable of the ICE table I used (note: $\text{I+C=E}$).
$$\begin{array}{|c|c|c|}\hline &\ce{Ag+}&\ce{SO4^2-}\\\hline \text{I}&\pu{0.200M}&x\\\hline \text{C}&\frac x2&0\\\hline \text{E}&\pu{0.200M}+\frac x2&x\\\hline \end{array}$$
Now I substitute the values I obtained for concentrations from the ICE table into my formula:
$$1.6 \times 10^{-5} = [0.200~\text{M} + \frac{1}{2}x]^2 \times [x]$$
Now since $K_\mathrm{sp}$ is less than $1 \times 10^{-5}$ we can ignore the $\frac{1}{2}x$? I guess because it is very small and therefore negligible?
$$1.6 \times 10^{-5} = [0.200~\text{M}]^2 \times [x]$$
$$ \frac{1.6 \times 10^{-5}}{[0.200~\text{M}]^2} = x = 0.0004$$
$$ x = 4.0 \times 10^{-4}~\text{M}$$
Therefore the concentration of $\ce{SO4^2-}$ is $ 4.0 \times 10^{-4}~\text{M}$?