The pH of ammonium acetate can be calculated by using the expression-A- p H-1-2-p Kw-p Ka-p Kb-B- p H-1-2-p Kw-p Ka-p Kb- p H-1-2-p Kw-p Ka-p Kb-D- p H-1-2-p Kw-p Ka-p Kb-

The correct option is A [CH_3COO^⊖][H^⊕]/[CH_3COOH]=K_a; [NH_4^⊕][OH^⊖]/[NH_4OH]=K_b……(1) and K_w=[H^⊕][OH^⊖] CH_3COO^⊖+NH_4^⊕+H_2O⇌ CH_3COOH+NH_4OH K ...

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Byju's Answer

Standard XII

Chemistry

Salt of Weak Acid and Strong Base

The pH of amm...

Question

The pH of ammonium acetate can be calculated by using the expression:

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Solution

The correct option is A

$\frac{[C H_{3} C O O^{⊖}] [H^{⨁}]}{[C H_{3} C O O H]} = K_{a}; \frac{[N H_{4}^{⨁}] [O H^{⊖}]}{[N H_{4} O H]} = K_{b} \dots \dots (1) a n d K_{w} = [H^{⨁}] [O H^{⊖}] C H_{3} C O O^{⊖} + N H_{4}^{⨁} + H_{2} O ⇌ C H_{3} C O O H + N H_{4} O H K_{H} = \frac{[C H_{3} C O O H] [N H_{4} O H]}{[C H_{3} C O O^{⊖}] [N H_{4}^{⨁}]} \dots \dots (2)$
$\approx K_{H} = {[\frac{C H_{3} C O O H}{C H_{3} C O O^{⊖}}]}^{2} [\frac{C H_{3} C O O H}{C H_{3} C O O^{⊖}}] \approx \frac{N H_{4} O H}{[N H_{4}^{⨁}]} (o r) \sqrt{K_{n}} = \frac{C H_{3} C O O H}{C H_{3} C O O^{⊖}}$
$B u t K_{H} = \frac{K_{w}}{K_{a} + K_{b}} \to i n t h e d e n o m i n a t o r c h a n g e t o (K_{a}) \times (K_{b})$
$∴ \sqrt{\frac{K_{w}}{K_{a} + K_{b}}} = \frac{[H^{+}]}{K_{a}} (f r o m (1)) (M a k e t h e s a m e c h a n g e h e r e)$
$∴ [H^{+}] = \sqrt{\frac{K_{w} \times K_{a}}{K_{b}}} (o r) p H = \frac{1}{2} l o g K_{w} + - \frac{1}{2} l o g K_{a} + \frac{1}{2} l o g K_{b} p H = \frac{1}{2} [p K_{w} + p K_{a} - p K_{b}]$

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Similar questions

Q. Prove that

p H = \frac{1}{2} [p K w - p K b - l o g_{10} C]

where

h < 5 %

Q. Match the List-I (solution of salts) with List-II (pH of the solution) and select the correct answer using the codes given below the lists:

List-I	List-II
A. Weak acid and strong base	1. $1 / 2 p K_{w}$
B. Strong acid and weak acid	2. $1 / 2 [p K_{w} - p K_{b} + l o g c]$
C. Weak acid and weak base	3. $1 / 2 [p K_{w} - p K_{b} - l o g c]$
D. Strong acid and strong base	4. $1 / 2 [p K_{w} + p K_{a} + l o g c]$