Calculate the percentage hydrolysis and the pH of 0.02 M CH3COONH4 : Kb(NH3) = 1.6 × 10–5, Ka(CH3COOH) = 1.6 × 10–5.
Dear student,
For hydrolysis of salt of weak acid and weak base,
kh = kw /( ka * kb)
kw = 10-14 ,, kb = ka = 1.6 * 10-5
kh = 10-14 /(1.6 * 10-5)2
= 0.39 * 10-4
kh = h2/1-h2
0.39 * 10-4 = h2/1-h2
0.39 * 10-4 - 0.39 * 10-4 h2 = h2
1.39 * 10-4 h2 = 0.39 * 10-4
h2 =0.28
h = 0.52
% hydrolysis = 0.52 * 100
= 52 %
For pH,
pH = (pKw + pKa - pKb) / 2
since
pKa = pKb
pH = pKw / 2
-log Kw/2
= 14/2
pH = 7
Regards
For hydrolysis of salt of weak acid and weak base,
kh = kw /( ka * kb)
kw = 10-14 ,, kb = ka = 1.6 * 10-5
kh = 10-14 /(1.6 * 10-5)2
= 0.39 * 10-4
kh = h2/1-h2
0.39 * 10-4 = h2/1-h2
0.39 * 10-4 - 0.39 * 10-4 h2 = h2
1.39 * 10-4 h2 = 0.39 * 10-4
h2 =0.28
h = 0.52
% hydrolysis = 0.52 * 100
= 52 %
For pH,
pH = (pKw + pKa - pKb) / 2
since
pKa = pKb
pH = pKw / 2
-log Kw/2
= 14/2
pH = 7
Regards
