Question

Calculate the extent of hydrolisis and the pH of 0.1M ammonium acetate. Given that Ka=Kb=1.8×10^-5​

Answer 1

Answer:

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

Answer 2

5) Gl Culate the pH of a solution which results from the mixing of 5o.oml of 0.3M HCl which 50mL of 0.4M NH3