Calculate the degree of hydrolysis h of a salt of aniline and acetic acid in 0-1 M solution at 25- CGiven -Ka for CH3COOH is 1-8- 10-5Kb for aniline is 4-6- 10-10

Since aniline and acetic acid are both weak electrolytes. So, K_h=K_wK_a× K_b K_h=10^ 141.8× 10^ 5× 4.6× 10^ 10= 1.2 if we see, nbsp; K_hC=1.20.1=12 Since, K ...

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Standard XI

Chemistry

Salt of Weak Acid and Weak Base

Calculate the...

Question

Calculate the degree of hydrolysis $(h)$ of a salt of aniline and acetic acid in 0.1 M solution at $25^{\circ} C$
Given :
$K_{a}$ for $C H_{3} C O O H$ is $1.8 \times 10^{- 5}$
$K_{b}$ for aniline is $4.6 \times 10^{- 10}$

0.996

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0.235

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0.85

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0.523

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Solution

The correct option is D 0.523
Since aniline and acetic acid are both weak electrolytes.
So,
$K_{h} = \frac{K_{w}}{K_{a} \times K_{b}} K_{h} = \frac{10^{- 14}}{1.8 \times 10^{- 5} \times 4.6 \times 10^{- 10}} = 1.2$

if we see, $\frac{K_{h}}{C} = \frac{1.2}{0.1} = 12$
Since, $\frac{K_{h}}{C} < 100$ ,
So we cannot take the approximation $1 - h \approx 1$ and we have to solve quadratic,
$K_{h} = \frac{h^{2}}{(1 - h)^{2}} \sqrt{K_{h}} = \frac{h}{1 - h} \sqrt{1.2} = \frac{h}{1 - h} 1.096 = \frac{h}{1 - h} h = \frac{1.096}{2.096} = 0.523$

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Calculate the degree of hydrolysis (h) of a salt of aniline and acetic acid in 0.1 M solution at 25∘C Given : Ka for CH3COOH is 1.8×10−5 Kb for aniline is 4.6×10−10

Calculate the degree of hydrolysis $(h)$ of a salt of aniline and acetic acid in 0.1 M solution at $25^{\circ} C$
Given :
$K_{a}$ for $C H_{3} C O O H$ is $1.8 \times 10^{- 5}$
$K_{b}$ for aniline is $4.6 \times 10^{- 10}$