The ionization contant of benzoic acid is `6.46xx10^(-5) and K_(sp)` for silver benzoate is `2.5xx10^(-13)`. How many times is silver benzoate more soluble in abuffer of pH=3.19 compared to its solubility in pure water ?

Explanation

` {:(C_6H_5COOAg,hArr, C_6H_5COO^(-) ,+,Ag^(+) ),(s,,s,,s):}`
solubility in water
` s= sqrt(K_(sp) ) = sqrt( 2.5 xx 10 ^(-13)) = 5.0 xx 10 ^(-7) " mol " L^(-1)`
solubility in a buffer of pH = 3.19 :
In a buffer of pH = 3.19
` - log_(10) [H^(+)] = 3.19 `
` therefore " " [H^(+) ] = " antilog"_(10) (-3.19) = 6.457 xx 10 ^(4) " mol " L^(-1)`
The `H^(+) ` ions presents in buffer solutions will combine with ` C_6H_5COO^(-)` ions to form ` C_6H_5COO^(-) ` but ` [H^(+) ]` in the reactions will remain unchanged because the solutions is a buffer solutions
` C_6H_5COOhArr C_6H_5COO^(-) + H^(+) `
` therefore " "K_a = ( [C_6H_5COO^(-) ][H^(+) ])/([C_6H_5COOH]) `
`or " " ([C_6H_5COOH])/([C_6H_5COO^(-)]) = ([H^(+) ])/(K_a) = ( 6.47 xx 10 ^(-4))/(6.46 xx 10 ^(-5)) =9.995" "...(i)`
Suppose the solubility of silver benzoate in the buffer is s. mol ` L^(-1)`
Since, almost all the `C_6H_5COO^(-)` ions get converted to bezoic acid which remains almost unionised we have
`" " s. = [Ag^(+)] = [C_6H_5COO^(-) ] +[C_6H_5COOH]`
` " " [C_6H_5COO^(-)] + 9.995 xx [C_6H_5COO^(-)]`
` " " ` [From eq. (i)]
` " " = 10 . 995 [C_6H_5COO^(-)]`
` therefore [C_6H_5COO^(-)] = (s.)/(10.995)`
` because " " K_(sp) = [C_6H_5COO^(-)] [Ag^(+)]`
`or " " 2.5 xx 10 ^(-1) = (s.)/( 10.995)`
` or " " s.^(2) = 2.5 xx 10 ^(-13) xx 10.995`
` or " " s. = sqrt(2.5 xx 10 ^(-13) xx 10.995)`
` " " = 1.657 xx 10 ^(_6) "mol " L^(-1)`
` therefore " " (s.)/(s) = ( 1.657 xx10^(-6))/(5.0 xx 10 ^(-7)) = 3.314`
Hence, silver benzoate is 3.314 times more soluble in the beffer.