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Question
Calculate the extent of hydrolysis and the pH of 0.1 M ammonium acetate Given that Ka = Kb = 1.8 × 10−5
Solution
h = `sqrt("K"_"h") = sqrt(("K"_"w")/("K"_"a" "K"_"b")) = sqrt((1 xx 10^-14)/(1.8 xx 10^-5 xx 1.8 xx 10^-5))`
= `sqrt(1/1.8 xx 10^-4)`
= 0.7453 × 10−5
pH = `1/2 "pK"_"w" + 1/2 "pK"_"a" - 1/2 "pK"_"b"`
Given that Ka = Kb = 1.8 × 10−5
if Ka = Kb, then, pKa = pKb
∴ pH = `1/2 "pK"_"w"`
= `1/2 (14)`
= 7
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