A saturated solution of silver benzoate, `AgOCOC_(6)H_(5)` has a `pH` of `8.63, K_(a)` for benzoic acid is `6.5 xx 10^(-5)`. Estimate `K_(sp)` for silver benzoate.

Explanation

`pH = 8.63, pOH = 5.37, [overset(Θ)OH] = 4.3 xx 10^(-6)`
`{:(,C_(6)H_(5)COO^(Θ)+H_(2)Orarr,C_(6)H_(5)COOH+,overset(Θ)OH,),("Initially",C,0,0,),("Final",C(1-h),Ch,Ch,):}`
`K_(h) = ([C_(6)H_(5)COOH][overset(Θ)OH])/([C_(6)H_(5)COO^(Θ)] ),Ch = 4.3 xx 10^(-6)`
`:. [C_(^)H_(5)COO^(Θ)] = ((4.3xx10^(-6))(4.3xx10^(-6)))/(K_(h)) ...(i)`
`K_(h) = (K_(w))/(K_(a)) = (10^(-14))/(6.5 xx 10^(-5))`
Substituting the value of `K_(h)` in equaiton (i)
`:. [C_(6)H_(5)COO^(Θ)] = ((4.3xx10^(-6))^(2)xx(6.5xx10^(-5)))/(10^(-14))`
`C_(6)H_(5)COOAg hArr C_(6)H_(5)COo^(Θ) + Ag^(o+)`
`K_(sp) = [Ag^(o+)] [C_(6)H_(5)COO^(Θ)] = (0.12)^(2) = 1.4 xx 10^(-2)`
Second method:
Use formula for the `pH` of salt of `S_(B)//W_(A))`
`pH = (1)/(2) (pK_(w) + pK_(a) + logC)`
`8.63 = (1)/(2) (14 + 4.187 + logC)`
`:. C = 0.12`
`K_(sp) = [Ag^(o+)] [C_(6)H_(5)COO^(Θ)] = (0.12)^(2) = 1.4 xx 10^(-2)`