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. 3.317
B. 9.5
C. 1000
D. 7.5

Answer

A

Explanation

In pure water : `C_6H_5COOAg hArr underset((s-x))(C_6H_5COO^(-)) + underset(s)(Ag^+)`
`underset((s-x))(C_6H_5COO^(-)) + H_2O hArr underset(x)(C_6H_5COOH)+underset(x)(OH^(-))`
`s(s-x)=2.5xx10^(-13)`….(1)
`x^2/((s-x))=10^(-14)/((6.46xx10^(-5)))`...(2)
Calculate s.
In buffer : `C_6H_5COOAg hArr underset((s'-x'))(C_6H_5COO^(-)) + underset(s')(Ag^+)`
`underset(s'(s'-x')=2.5xx10^(-13))(C_6H_5COO^(-) + H^+ hArr) C_6H_5COOH` .....(3)
`((s'-x')xx10^(-3.19))/x'=6.46xx10^(-5)`....(4) solve for s'.
Calculate `(s')/s`