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

Let the solubility of `C_6H_5COO^-` Ag in water in x mol `L^(-1)`
`C_6H_5COOAg hArr C_6H_5COO+Ag^+`
`[C_6H_5COO^-]`=x mol `L^(-1), [Ag^+]= "x mol" L^(-1)`
`K_(sp)=[C_6H_5COO^-][Ag^+]`
`K_(sp)=x xx x =x^2`
`therefore x^2= K_(sp)` or `x=sqrtK_(sp)`
`therefore x=sqrt(2.5xx10^(-13))`
`=5.0xx10^(-7) " mol L"^(-1)`
`therefore` Solubility of silver benzoate in water , x =`5.0xx10^(-7)"mol L"^(-1)`
pH of buffer =3.19
or -log `[H^+]` =3.19
or log `[H^+]`=-3.19 =4.81
`[H^+]`=antilog (4.81)=`6.456xx10^(-4)`
`C_6H_5COO^-` ions now combine with `H^+` ions to form `C_6H_6COOH` but `[H^+]` remains almost constant because, we have buffer solution.
Benzoic acid ionize as :
`C_6H_5COOH hArr C_6H_5COO^(-) + H^+`
`K_a=([C_6H_5COO^-][H^+])/([C_6H_5COOH])`
or `([C_6H_5COO])/([C_6H_5COO^-])=([H^+])/K_a=(6.456xx10^(-4))/(6.46xx10^(-5))=10`
Suppose solubility in buffer solution is y mol `L^(-1)`. Then most of the benzoate ions are converted into benzoic acid molecules (which remain almost ionized).
`y=[Ag^+]=[C_6H_5COO^-]+[C_6H_5COOH]`
`=[C_6H_5COO^-]+10[C_6H_5COO^-]`
`=11[C_6H_5COO^-]`
`therefore [C_6H_5COO^-]=y/11`
But `K_(sp)=[C_6H_5COO^-][Ag^+]`
`2.5xx10^(-13)=y/11xxy`
or `y^2=2.75xx10^(-12)`
`therefore y=(2.75xx10^(-12))^(1/2)=1.66xx10^(-6)`
Thus, `y/x =(1.66xx10^(-6))/(5.0xx10^(-7))`=3.32
`therefore` Silver benzoate is 3.32 times more soluble in buffer.