The standard hydrolysis constant of anilinium acetate at 25- C is KaCH3COOH-1-8 - 10-5 M- KbAniline - 4-6 - 10-10

Given: nbsp; (K_a(CH_3COOH)=1.8 × 10^ 5 M, K_b(Aniline) = 4.6 × 10^ 10 ) PhNH_3^+CH_3COO^ + H_2O ⇌ PhNH_2 + CH_3COOH + H_2O PhNH_2 + nbsp; H_2O ⇌ nbsp;Ph ...

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Standard XI

Chemistry

Acidic Buffer Action

The standard ...

Question

The standard hydrolysis constant of anilinium acetate at $25^{\circ} C$ is $(K_{a} (C H_{3} C O O H) = 1.8 \times 10^{- 5} M, K_{b} (Aniline) = 4.6 \times 10^{- 10})$

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1.21

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1.35

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Solution

The correct option is B 1.21
Given: $(K_{a} (C H_{3} C O O H) = 1.8 \times 10^{- 5} M, K_{b} (Aniline) = 4.6 \times 10^{- 10})$
$P h N H_{3}^{+} C H_{3} C O O^{-} + H_{2} O ⇌ P h N H_{2} + C H_{3} C O O H + H_{2} O$
$P h N H_{2} + H_{2} O ⇌ P h N H_{3}^{+} + O H^{-}$
$C H_{3} C O O H + H_{2} O ⇌ C H_{3} C O O^{-} + H_{3} O^{+}$
From the hydrolysis of salt, we know,
$K_{h} = \frac{K_{w}}{K_{a} K_{b}}$
$K_{h} = \frac{1.0 \times 10^{- 14}}{(1.8 \times 10^{- 5}) (4.6 \times 10^{- 10})} = 1.21$

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Similar questions

Q. The standard hydrolysis constant of anilinium acetate at

25^{\circ} C

(K_{a} (C H_{3} C O O H) = 1.8 \times 10^{- 5} M, K_{b} (Aniline) = 4.6 \times 10^{- 10})

Q. The standard hydrolysis constant of anilinium acetate at

25^{\circ} C

(K_{a} (C H_{3} C O O H) = 1.8 \times 10^{- 5} M, K_{b} (Aniline) = 4.6 \times 10^{- 10})

Q. Hydrolysis constant

(K_{h})

of ammonium acetate is :
(Given that

K_{a} =

1.8 \times

10^{- 5}

K_{b} =

2 \times

10^{- 5}

K_{w} = 10^{- 14}

)

Q. 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

1.8 \times 10^{- 5}

K_{b}

for aniline is

4.6 \times 10^{- 10}

Q. If

K_{a} = 1.8 \times 10^{- 5}

, calculate the degree of hydrolysis of 0

.1 M

solution of sodium acetate at

25^{\circ} C

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Acidic Buffer Action

Standard XI Chemistry

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The standard hydrolysis constant of anilinium acetate at 25∘ C is (Ka(CH3COOH)=1.8×10−5 M, Kb(Aniline)=4.6×10−10 )

The correct option is B 1.21Given: (Ka(CH3COOH)=1.8×10−5 M, Kb(Aniline)=4.6×10−10 ) PhNH+3CH3COO−+H2O⇌PhNH2+CH3COOH+H2O PhNH2+H2O⇌PhNH+3+OH− CH3COOH+H2O⇌CH3COO−+H3O+ From the hydrolysis of salt, we know, Kh=KwKaKb Kh=1.0×10−14(1.8×10−5)(4.6×10−10)=1.21

The standard hydrolysis constant of anilinium acetate at $25^{\circ} C$ is $(K_{a} (C H_{3} C O O H) = 1.8 \times 10^{- 5} M, K_{b} (Aniline) = 4.6 \times 10^{- 10})$