`C_(6)H_(5)CO Oag rarr C_(6)H_(5)CO O^(-) + Ag^(+)`
Solubility in water. Suppose solubility in water = x mol `L^(-1)`. Then
`[C_(6)H_(5)CO O^(-)]=[Ag^(+)] = x "mol" L^(-1) . :. X^(2)=K_(sp) or x = sqrt(K_(sp))=sqrt(2.5xx10^(-13))=5xx10^(-7)` mol ` L^(-1)`
Solubility in buffer of pH = 3.19
`pH = 3.19 ` means ` - log (H^(+)]=3.19 or log [H^(+)]=-3.19 = bar(4).81 or [H^(+)] = 6.457 xx 10^(-4)M`
`C_(6)H_(5)CO O^(-)` ions now conbine with the `H^(+)` ions to from benzoic acid but `[H^(+)]` remains almost constant because we have buffer solution . Now
`C_(6)H_(5) CO OH hArr C_(6)H_(5)CO O^(-) + H^(+)`
`:. K_(a)=([C_(6)H_(5)CO O^(-)][H^(+)])/([C_(6)H_(5)CO OH]) or ([C_(6)H_(5)CO OH ])/([C_(6)H_(5)CO O^(-)])=([H^(+)])/(K_(a))=(6.457xx10^(-4))/(6.46xx10^(-5))=10 ` ...(i)
Suppose solubility in the buffer solution is y mol `L^(-1)` . Then as most of the benzote ions are converted into bezoic acid molecules (which remain almost ionized), we have
`y = [Ag^(+)] = [ C_(6)H_(5)CO O^(-)]+[C_(6)H_(5)CO O^(-)] + [ C_(6)H_(5)CO OH]=[C_(6)H_(5)CO O^(-)] + 10 [C_(6)H_(5)CO O^(- )]=11[C_(6)H_(5)CO O^(-)] " " `(using eqn. (i) )
`:. [C_(6)H_(5)CO O^(-)]=(y)/(11)`
`K_(sp)=[C_(6)H_(5)CO O^(-)][Ag^(+)]`
i.e., `2.5xx10^(-3)=(y)/(11) xx y or y^(2)=2.75 xx 10^(-12) or y=1.66xx10^(-6) :. (y)/(x)=(1.66xx10^(-6))/(5xx10^(-7))=3.32`.
Note that in case of salts of weak acids, the solubility is more in the acidic solution than in water. The reason, in general, may be explained as follows :
Taking example of `C_(6)H_(5)CO O Ag`, we have `C_(6)H_(5)CO O Ag hArr C_(6)H_(5)CO O^(-) + Ag^(+)`
In acidic solution, the anions (`C_(6)H_(5)CO O^(-)` in the present case) undergo protonation in presence of acid . Thus, `C_(6)H_(5)CO O^(-)` ions are removed. Hence, equilibrium shifts forward producing more `Ag^(+)` ions.
Alternatively, as `C_(6)H_(5)CO O^(-)` ions are removed, `Q_(sp)` decrease. In order to maintain solubility product equilibrium `(Q_(sp) = K_(sp)), Ag^(+)` ion concentration must increase. Hence, solubility is more.
