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Calculate the percentage hydrolysis of decinormal solution of ammonium acetate given that `k_(a) = 1.75 xx 10^(-5), K_(b) =1.80 xx 10^(-5) and K_(w) = 1.0 xx 10^(-14)` |
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Answer» (i). Calculation of hydrolysis constant `(K_(h)).` Since ammonium acetate is salt of weak acid and weak base, `K_(h) =(K_(w))/(K_(a).K_(b)) = (1.0xx10^(-14))/(1.75xx10^(-5)xx1.80xx10^(-5))=3.175 xx10^(-5)` (II). Calculatation of degree of hydrolysis (h). `h=sqrt(K_(h))= (3.175 xx 10^(-5))^(1//2) =(31.75 xx 10^(-6))^(1//2) =5.63 xx 10^(-3).` |
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