Question

The ionization constant of benzoic acid is 6.46 × 10^–5 and K_sp 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,

`["H"_3"O"^+]  = 6.46  xx 10^(-4) "M"`

\[\ce{C_6H_5COOH  + H_2O <-> C_6H_5COO- + H_3O}\]

`"K"_"a" = (["C"_6"H"_5"COO"^(-)]["H"_3"O"^(+)])/["C"_6"H"_5"COOH"]`

`["C"_6"H"_5"COOH"]/["C"_6"H"_5"COO"^(-)] = ["H"_3"O"^+]/["K"_"a"]`

`= (6.46 xx 10^(-4))/(6.46 xx 10^(-5)) = 10`

Let the solubility of C₆H₅COOAg be x mol/L.

Then,

`["Ag"^(+)] = x`

`["C"_6"H"_5"COOH"] + ["C"_6"H"_5"COO"^(-)] = x`

`10["C"_6"H"_5"COO"^(-)] + ["C"_6"H"_5"COO"^(-)] = x`

`["C"_6"H"_5"COO"^(-)] = x/11`

`"K"_("sp")["Ag"^(+)]["C"_6"H"_5 "COO"^(-)]`

`2.5 xx 10^(-13) = x(x/11)`

`x = 1.66 xx 10^(-6) "mol"/"L"`

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 `["CH"_3"COO"^(-)] =  x'"M"`

`"K"_("sp") = ["Ag"^+] ["CH"_3"COO"^-]`

`"K"_("sp") = (x')^2`

`x' = sqrt("K"_("sp")) = sqrt(2.5 xx 10^(-13)) = 5xx 10^(-7) "mol"/"L"`

`therefore x/(x') = (1.66 xx 10^(-6))/5xx10^(-7) = 3.32`

Hence, C₆H₅COOAg is approximately 3.317 times more soluble in a low pH solution.