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What will be the effect on the solubility of $0.1\ \mathrm{mol}$ of $\ce{CH3COOAg}$ in presence of $1\ \mathrm{l}$ of $0.1~\mathrm{M}\ \ce{HNO3}$ solution?

$$K_\mathrm{sp} (\ce{CH3COOAg}) = 10^{-8}\\ K_\mathrm{a} (\ce{CH3COOH}) = 10^{-5}$$

I know that silver acetate will react with nitric acid to form silver nitrate and acetic acid. So the solubility of $\ce{CH3COOAg}$ will increase. But in the end, I am not able to find out the final solubility (as concentration of $\ce{H+}$ and $\ce{NO3-}$ is coming different).

Maybe I have an error somewhere?

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The process you are trying to investigate is:

$$\ce{AgOOCCH3(s) + HNO3(aq) <=> AgNO3 (aq) + CH3COOH (Aq)}$$

with the net ionic equation:

$$\ce{AgOOCCH3(s) + H+(aq) <=> Ag+(aq) + CH3COOH(aq)}$$

The equilibrium constant is

$$K_c = \dfrac{[\ce{Ag+}][\ce{CH3COOH}]}{[\ce{H+}]}$$

Note that this is law of mass action can be derived from the $K_a$ and the K_{sp}$ you have:

$$\begin{aligned} K_{a}&=\dfrac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}\\ K_{sp}&=[\ce{Ag+}][\ce{CH3COO-}]\\ K_c &= \dfrac{K_{sp}}{K_a}=\dfrac{[\ce{Ag+}][\ce{CH3COOH}]}{[\ce{H+}]} \end{aligned}$$

The problem is now an ICE (Initial-change-equilibrium) problem.

$$\begin{array}{|c|c|c|c|} \hline \ & [\ce{H+}] & [\ce{Ag+}] & [\ce{CH3COOH}]\\ \hline I & 0.1\ \mathrm{M} & 0\ \mathrm{M} & 0\ \mathrm{M}\\ C & -x & +x & +x \\ E & 0.1 -x \ \mathrm{M} & x \ \mathrm{M} & x \ \mathrm{M}\\ \hline \end{array}$$

Plug values into the law of mass action, and solve for $x$, the concentration of $\ce{Ag+}$, and then you can determine the solubility.

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  • $\begingroup$ Just one thing, i got x=10^-2. So solubility is 10^-6 or 10^-2 ? $\endgroup$ – Ava Jun 15 '16 at 10:07
  • $\begingroup$ Solubility is the amount that dissolved per unit volume. $\endgroup$ – Ben Norris Jun 15 '16 at 10:57

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