The percentage hydrolysis of 0.15 M solution of ammonium acetate, Ka for CH3COOH is 1.8 × 10-5 and Kb for NH3 is 1.8 × 10-5

Question

The percentage hydrolysis of 0.15 M solution of ammonium acetate, Ka for CH3COOH is 1.8 × 10-5 and Kb for NH3 is 1.8 × 10-5 (A) 0.556 (B) 4.72 (C) 9.38 (D) 5.56

Tardigrade Admin · Accepted Answer

α = √(KW/Ka × Kb) = √(1×10-14/1.8 ×10-5 × 1.8 ×10-5) = 0.55

Q. The percentage hydrolysis of $0.15 M$ solution of ammonium acetate, $K_{a}$ for $C H_{3} COO H$ is $1.8 \times 1 0^{- 5}$ and $K_{b}$ for $N H_{3}$ is $1.8 \times 1 0^{- 5}$

0.556

4.72

9.38

5.56

$α = \frac{K _{W}}{K _{a} \times K _{b}}$
$= \frac{1 \times 1 0 ^{- 14}}{1.8 \times 1 0 ^{- 5} \times 1.8 \times 1 0 ^{- 5}} = 0.55$

Q. The percentage hydrolysis of 0.15M solution of ammonium acetate, Ka​ for CH3​COOH is 1.8×10−5 and Kb​ for NH3​ is 1.8×10−5