The ionisation constant of benzoic acid `(PhCOOH)` is `6.46 xx 10^(-5)` and `K_(sp)` for silver benzoate is `2.5 xx 10^(-3)`. How many times is silver benzoate more soluble in a buffer of `pH 3.19` compared to its solubility is pure water?

Answer

Silver benozate is `3.317` times more soluble at lower `pH`

Explanation

Since `pH = 3.19`
`[H_(3)O^(+)] = 6.46 xx 10^(-4) M`
`C_(6)H_(5)COOH + H_(2)O harr C_(6)H_(5)COO^(-) + H_(3)O`
`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 `C_(6)H_(5)COOAg` 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`
This ionic product exceeds the Ksp of Zns and CdS. Therefore, precipitation will occur in `CdCl_(2)` and `ZnCl_(2)` solution is `1.66 xx 10^(-6) "mol"//L`
Now let the solubility of `C_(6)H_(5)COOAg` 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)) = 5 xx 10^(-7) "mol"//L`
`:. (x)/(x') = (1.66 xx 10^(-6))/(5 xx 10^(-7)) = 3.32`
Hence, `C_(6)H_(5)COOAg` is approximately 3.317 times more soluble in a low pH solution.