Ionic Equilibrium | JEE Main Chemistry Chapter Wise Question Papers

Question

$K_{a}$ for ${CH}_{3} COOH$ is $1.8 \times 10^{- 5}$ and $K_{b}$ for ${NH}_{4} OH$ is $1.8 \times 10^{- 5} .$ The pH of ammonium acetate solution will be _______ . [2024]

Smart Achievers · Accepted Answer

(7)

$As K_{a} of {CH}_{3} COOH = K_{b} of {NH}_{4} OH$

$So, {pK}_{a} of {CH}_{3} COOH = {pK}_{b} of {NH}_{4} OH$

$And p K_{a} - p K_{b} = 0$

For salt of weak acid and weak base:

$pH = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b})$

$pH = \frac{1}{2} (14 + 0) = 7$

Solution

Q.
$K_{a}$ for ${CH}_{3} COOH$ is $1.8 \times 10^{- 5}$ and $K_{b}$ for ${NH}_{4} OH$ is $1.8 \times 10^{- 5} .$ The pH of ammonium acetate solution will be _______ . [2024]

Ans.
(7)

$As K_{a} of {CH}_{3} COOH = K_{b} of {NH}_{4} OH$

$So, {pK}_{a} of {CH}_{3} COOH = {pK}_{b} of {NH}_{4} OH$

$And p K_{a} - p K_{b} = 0$

For salt of weak acid and weak base:

$pH = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b})$

$pH = \frac{1}{2} (14 + 0) = 7$

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Solution

Q. Ka for CH3COOH is 1.8×10-5 and Kb for NH4OH is 1.8×10-5. The pH of ammonium acetate solution will be _______ . [2024]

Ans. (7) As Ka of CH3COOH=Kb of NH4OH So, pKa of CH3COOH=pKb of NH4OH And pKa-pKb=0 For salt of weak acid and weak base: pH=12(pKw+pKa-pKb) pH=12(14+0)=7

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Q.
$K_{a}$ for ${CH}_{3} COOH$ is $1.8 \times 10^{- 5}$ and $K_{b}$ for ${NH}_{4} OH$ is $1.8 \times 10^{- 5} .$ The pH of ammonium acetate solution will be _______ . [2024]

Ans.
(7)

$As K_{a} of {CH}_{3} COOH = K_{b} of {NH}_{4} OH$

$So, {pK}_{a} of {CH}_{3} COOH = {pK}_{b} of {NH}_{4} OH$

$And p K_{a} - p K_{b} = 0$

For salt of weak acid and weak base:

$pH = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b})$

$pH = \frac{1}{2} (14 + 0) = 7$