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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? |
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Answer» (I) .Calculation of solubility in water `C_(6)H_(5)COOAg to C_(6)H_(5)COO^(-)+ Ag^(+)` Suppose solubility in water =x mol `L^(-1)` `[C_(6)H_(5)COO^(-)] =[Ag^(+)] =x mol L^(-1)` `x^(2) =K_(sp) " or " x(K_(sp))^(1//2) =(2.5 xx 10^(-13))^(1//2) = 5 xx 10^(-7) mol L^(-1)` (II). Calculation of solubiltiy in buffer of pH =3.19 `pH 3.19 " means " - log [H^(+)] =3.19` ` " or " " " log [H^(+)] =- 3.19 =bar(4).81 " or " [H^(+)] =6.457 xx 10^(-4)` `C_(6)H_(5)COO^(-)` ions now combine with the `H^(+)` ions to form benzoic acid but `[H^(+)]` remains almost constant because we have buffer solution. Now `C_(6)H_(5)COOH hArr C_(6)H_(5)COO^(-) +H^(+)` `:. =[[C_(6)H_(5)COO^(-)][H^(+)]]/[[C_(6)H_(5)COOH]] " or "[[C_(6)H_(5)COO^(-)]]/[[C_(6)H_(5)COOH]] =[[H^(+)]]/(K_(a)) =(6.457xx10^(-4))/(6.46xx10^(-5))=10` Suppose solubility in the buffer solution is y mol `L^(-1)`. Then as most of the benzoate ions are converted into benzoic acid molecules (which remain almost ionized), we have `y= [Ag^(+)] =[C_(6)H_(5)COO^(-)]+[C_(6)H_(5)COOH]` `=[C_(6)H_(5)COO^(-)]+ 10 [C_(6)H_(5)COO^(-)]=[C_(6)H_(5)COO^(-)]` `:. " "[C_(6)H_(5)COO^(-)]=(y)/(11)` `K_(sp) =[C_(6)H_(5)COO^(-)][Ag^(+)]` `i.e., " " 2.5 xx 10^(-13) =(y)/(1)) xx y " or "y^(2) =11 xx 2.5 xx 10^(-13)` `:. " " y^(2) =2.75 xx 10^(-12) " or " y = (2.75 xx 10)^(1//2)=1.66 xx 10^(-6) mol L^(-1)` `:. " "(y)/(x) =((1.66 xx 10^(-6) "mol " L^(-1)))/((5 xx 10^(-7) "mol " L^(-1))) = 3.32` |
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