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?

  • $\begingroup$ Interestingly, Ag+ is much more strongly positive than H+, and -COO- more strongly negative than NO3-. Unless acetic acid splits out of solution (unlikely), what you will have is AcAg solvated to some degree in aqueous solution. When reacting metallic silver with HNO3, some acid oxidizes silver 0 to silver +1, hence AgNO3. $\endgroup$ Oct 15, 2022 at 23:38

1 Answer 1


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.

  • $\begingroup$ Just one thing, i got x=10^-2. So solubility is 10^-6 or 10^-2 ? $\endgroup$
    – Ava
    Jun 15, 2016 at 10:07
  • $\begingroup$ Solubility is the amount that dissolved per unit volume. $\endgroup$
    – Ben Norris
    Jun 15, 2016 at 10:57

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