The `K_a` of `C_(6)H_(5)` COOH is `6.46 xx 10^(-5)` and `K_(sp)` for `C_(6)H_(5), COO^(-) Ag^(+)` is `2.5 xx 10^(-13)`. How many times the silver benzoate more soluble in a buffer of pH = 3.19 compared to its solubility in pure water?

Answer

3.2

Explanation

`{:(C_6H_5COOAg(s)" "hArrC_6H_5COO^(-)+Ag^(+),K_1=K_(sp)),(C_6H_5COOH^(-)+H^(+)hArrC_6H_5COOH," "K_2=1/K_a):}/(ul(C_6H_5COOAg(s)+H^(+)hArrC_6H_5COOH+Ag^(+),K_3=K_(sp)/(K_a)))`
`K_3=([C_6H_5COOH)[Ag^(+)])/([H^(+)])=(s.s)/([H^(+)])=s^2/([H^(+)])`
`K_(sp)/K_a`
where, s is the solution, solubility of `C_6H_5COOAg`
In a buffer of pH = 3.19
`[H^(+)] =` antilog `bar(4).81 = 6.46 xx10^(-4)`
`s^2/([H^(+]])=(K_(sp))/(K_a) ` or `s^2 = (K_(sp) xx [H^(+)])/(K_a)`
`s = sqrt((2.5 xx 10^(-13) xx 6.46 xx106(-4))/(6.46 xx 10^(-5)))`
`s= sqrt(2.5 xx10^(-13) xx10)`
`s = 1.6 xx10^(-6) ` M (in buffer)
In aqueous solution , solubility of `C_(6)H_(5)COOAg`
`s = sqrt(K_(sp)) = sqrt(2.5xx10^(-13)) = 5 xx10^(-7) M`
`(s_((C_6H_5COOAg))" in buffere")/(s_((C_6H_5COOAg))" in aqueous solution")= (1.6 xx10^(-6))/(5.0 xx10^(-7)) = 3.2`
`C_6H_5 COOAg ` is 3.2 times more soluble in buffer than in pure water.