Question

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?

Answer 1

`K_(a)` of `PhCOOAg = 6.45 xx 10^(-5)`.
`K_(sp)` of `PhCOOAg = 2.5 xx 10^(-13)`.
`pH` of buffer `= 3.19`.
`:. [H^(o+)] = "Antilog" (-3.19) = "Antilog" (-3-0.19 +1-1)`
`= "Antilog" (bar(4).81)`
`= 6.45 xx 10^(-4)M`
Let `S_("buffer")` (solubility of salt in a buffer)
`= sqrt(K_(sp)) = [1+([H^(o+)])/(K_(a))]^(1//2)`
`S_(H_(2))O = sqrt(K_(sp))`
`:.(S_("buffer"))/(S_(H_(2)O))=(sqrt cancel(K_(sp))[1+([H^(oplus)])/(K_(a))]^(1//2))/(sqrtcancel(K_(sp)))`
`= [1+(6.45xx10^(-4))/(6.45xx10^(-5))]^(1//2)`
`= [1+10]^(1//2) = (11)^(1//2) = 3.317`
Thus, silver benzote is `3.317` times more soluble at lower `pH` than in water.