Question

The ionization constant of benzoic acid is 6.46 × 10–5 and Ksp for silver benzoate is 2.5 × 10–13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

Answer 1

Since pH = 3.19,

Let the solubility of C6H5COOAg be x mol/L.
Then,

Thus, the solubility of silver benzoate in a pH 3.19 solution is 1.66 × 10-6 mol/L. Now, let the solubility of C6H5COOAg be x' mol/L.
Then [Ag+]=x'M and [CH3COO-]=x'M
$K_{sp}=[Ag^+][CH3COO^-]$
$K_{sp}=(x')^2$
$x'=\sqrt{K_{sp}}=\sqrt{2.5\times10^{-13}}=5\times10^{-7}\,mol/L$
$\therefore \,\frac{x}{x'}=\frac{1.66\times10^{-6}}{5\times10^{-7}}=3.32$
Hence, C6H5COOAg is approximately 3.317 times more soluble in a low-pH solution.