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?

Explanation

(I) .Calculation of solubility in water
`C_(6)H_(5)COOAg to C_(6)H_(5)COO^(-)+ Ag^(+)`
Suppose solubility in water =x mol `L^(-1)`
`[C_(6)H_(5)COO^(-)] =[Ag^(+)] =x mol L^(-1)`
`x^(2) =K_(sp) " or " x(K_(sp))^(1//2) =(2.5 xx 10^(-13))^(1//2) = 5 xx 10^(-7) mol L^(-1)`
(II). Calculation of solubiltiy 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)`
`C_(6)H_(5)COO^(-)` ions now combine with the `H^(+)` ions to form benzoic acid but `[H^(+)]` remains almost constant because we have buffer solution. Now
`C_(6)H_(5)COOH hArr C_(6)H_(5)COO^(-) +H^(+)`
`:. =[[C_(6)H_(5)COO^(-)][H^(+)]]/[[C_(6)H_(5)COOH]] " or "[[C_(6)H_(5)COO^(-)]]/[[C_(6)H_(5)COOH]] =[[H^(+)]]/(K_(a)) =(6.457xx10^(-4))/(6.46xx10^(-5))=10`
Suppose solubility in the buffer solution is y mol `L^(-1)`. Then as most of the benzoate ions are converted into benzoic acid molecules (which remain almost ionized), we have
`y= [Ag^(+)] =[C_(6)H_(5)COO^(-)]+[C_(6)H_(5)COOH]`
`=[C_(6)H_(5)COO^(-)]+ 10 [C_(6)H_(5)COO^(-)]=[C_(6)H_(5)COO^(-)]`
`:. " "[C_(6)H_(5)COO^(-)]=(y)/(11)`
`K_(sp) =[C_(6)H_(5)COO^(-)][Ag^(+)]`
`i.e., " " 2.5 xx 10^(-13) =(y)/(1)) xx y " or "y^(2) =11 xx 2.5 xx 10^(-13)`
`:. " " y^(2) =2.75 xx 10^(-12) " or " y = (2.75 xx 10)^(1//2)=1.66 xx 10^(-6) mol L^(-1)`
`:. " "(y)/(x) =((1.66 xx 10^(-6) "mol " L^(-1)))/((5 xx 10^(-7) "mol " L^(-1))) = 3.32`