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InterviewSolution
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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. |
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Answer» `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)` |
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