Calculate the percentage hydrolysis & the pH 0.02 M ${C H}_{3} {C O O N H}_{4} . K_{b} ({N H}_{3}) = 1.6 \times 10^{- 5}, K_{a} ({C H}_{3} C O O H) = 1.6 \times 10^{- 5} .$

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Solution

For details :-
$K_{h} = \frac{K w}{k a . k b}$
$= \frac{10^{- 14}}{1.6 \times 10^{- 5} \times 1.6 \times 10^{- 5}}$
$k_{h} = 3.906 \times 10^{- 4}$
As $k_{h}$ is very small
$∴ K_{h} = h^{2}$
$h^{2} = \sqrt{K_{h}}$ (h = hydrolysis)
$= \sqrt{3.906 \times 10^{- 4}}$
$h = 1.976 \times 10^{- 2}$
$% h = \frac{1.976 \times 10^{- 2}}{0.02} \times 100$
hydrolysis percentage $= 98.8 %$
$p H = \frac{1}{2} (P^{k_{w}} + P^{k_{o}} - P^{k_{b}})$
$= 7 + \frac{1}{2} (l o g 1.6 \times 10^{- 5} - l o g 1.6 \times 10^{- 5})$
$= 7 + \frac{1}{2} (l o g \frac{1.6 \times 10^{- 5}}{1.6 \times 10^{5}})$
$= 7 + \frac{1}{2} (0)$
$p H = 7$

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Calculate the percentage hydrolysis & the pH 0.02 M CH3COONH4. Kb(NH3) = 1.6 × 10−5, Ka(CH3COOH) = 1.6 × 10−5.

Calculate the percentage hydrolysis & the pH 0.02 M ${C H}_{3} {C O O N H}_{4} . K_{b} ({N H}_{3}) = 1.6 \times 10^{- 5}, K_{a} ({C H}_{3} C O O H) = 1.6 \times 10^{- 5} .$