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?

Options

A. 4
B. 3.32
C. 3.01
D. 2.5

Answer

B

Explanation

Suppose S is the molar solubility of silver benzoate in water, then
`C_(6)H_(5)COOAg_((s))hArrC_(6)H_(5)COO_((aq))^(-)+Ag_((aq))^(+)`
`K_(sp)=S^(2):.S=sqrt(2.5xx10^(-13))=5.0xx10^(-7)M`
If the solubility of salt of weak acid of ionization constant `K_(a)` is `S'`, Then `K_(sp),K_(a)andS'` are related to each other at `pH=3.19`.
`:." "[H^(+)]=6.46xx10^(-4)M" "(becausepH=3.19)`
`K_(sp)=S^('2)[(K_(a))/(K_(a)+[H^(+)])]`
`S'={(2.5xx10^(-13))/([(6.46xx10^(-5))/(6.46xx10^(-5)+6.46xx10^(-4))])}^(1//2)`
`S'={(2.5xx10^(-13)xx7.0.6xx10^(-4))/(6.46xx10^(-5))}^(1//2)`
`=(2.75xx10^(-12))^(1//2)=1.658xx10^(-6)M`
`:.` The ration of `(S')/(S)=(1.658xx10^(-6))/(5.0xx10^(-7))=3.32`
Silver benzoate is 3.32 times more soluble in buffer of `pH=3.19` than in pure water.