The dissociation constants for acetic acid and HCN at 25oC are 1-5 - 10-5 and 4-5 - 10-10- respectively- The equilibrium constant for the equilibrium CN- - CH3COOH- HCN - CH3COO-would be-

The correct option is A 3.0 × 10^4 Given, #160; C H _ 3 COOH⇋ C H _ 3 CO O ^ + H ^ + K _ a =1.5× 10 ^ 5 .....(1) HCN⇋ H ^ + +C N ^ ; ...

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Standard XII

Chemistry

Heat of Formation

The dissociat...

Question

The dissociation constants for acetic acid and HCN at $25^{o} C$ are $1.5 \times 10^{- 5}$ and $4.5 \times 10^{- 10}$ , respectively. The equilibrium constant for the equilibrium $C N^{-} + C H_{3} C O O H ⇌ H C N + C H_{3} C O O^{-}$
would be:

3.0 \times 10^{4}

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3.0 \times 10^{5}

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3.0 \times 10^{- 5}

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3.0 \times 10^{- 4}

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Solution

The correct option is A $3.0 \times 10^{4}$
Given,
$C H_{3} C O O H ⇋ C H_{3} C O O^{-} + H^{+}$
$K_{a} = 1.5 \times 10^{- 5} . . . . . (1)$
$H C N ⇋ H^{+} + C N^{-}; K_{a} = 4.5 \times 1 0^{- 10} . . . . (2)$
$C N^{-} + C H_{3} C O O H ⇋ H C N + C H_{3} C O O^{-}$
On substracting equation (2) from equation(1) we get
$C H_{3} C O O H + C N^{-} ⇋ H C N + C H_{3} C O O H;$
$K = \frac{K_{a}}{K_{a 1}} = \frac{1.5 \times 1 0^{- 5}}{4.5 \times 1 0^{- 10}} = \frac{10^{5}}{3} = 3.33 \times 1 0^{4}$

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Similar questions

Q. The dissociation constants for acetic acid and HCN at

25^{\circ} C

are

1.5 \times 10^{- 5}

and

4.5 \times 10^{- 10}

, respectively. The equilibrium constant for the equilibrium,

C N^{-} + C H_{3} C O O H ⇌ H C N + C H_{3} C O O^{-}

would be?

Q. Given that

K_{a}

for

H C l O

and

H C N

are

3.0 \times 10^{- 8}

and

4.8 \times 10^{- 10}

, respectively, the equilibrium constant for the reaction

H C l O (a q) + C N^{-} (a q) ⇌ C l O^{-} (a q) + H C N (a q)

, is:

K_{a}

for

H C N

and

{C H}_{3} C O O H

are

4.9 \times 10^{- 10}

and

1.8 \times 10^{- 5}

respectively. Calculate the equilibrium constant for the reaction:

{C H}_{3} C O O H + N a C N ⇌ {C H}_{3} C O O N a + H C N

Q. The dissociation constants of two acids

H A_{1}

and

H A_{2}

are

3.0 \times 10^{- 4}

and

1.8 \times 10^{- 5}

respectively. The relative strengths of the acids will be:

Q. Which of the following are true regarding the aqueous dissociation of HCN,

K_{a}

= 4.9 x

10^{- 10}

25^{o} C

?
i. At equilibrium,

[H^{+}] = [C N^{-}]

ii. At equilibrium,

[H^{+}]

= [HCN]
iii.

H C N

is a strong acid

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The dissociation constants for acetic acid and HCN at 25oC are 1.5×10−5 and 4.5×10−10, respectively. The equilibrium constant for the equilibrium CN−+CH3COOH⇌HCN+CH3COO−would be:

The correct option is A 3.0×104Given, CH3COOH⇋CH3COO−+H+Ka=1.5×10−5.....(1)HCN⇋H++CN−;Ka=4.5×10−10....(2)CN−+CH3COOH⇋HCN+CH3COO−On substracting equation (2) from equation(1) we getCH3COOH+CN−⇋HCN+CH3COOH;K=KaKa1=1.5×10−54.5×10−10=1053=3.33×104

The dissociation constants for acetic acid and HCN at $25^{o} C$ are $1.5 \times 10^{- 5}$ and $4.5 \times 10^{- 10}$ , respectively. The equilibrium constant for the equilibrium $C N^{-} + C H_{3} C O O H ⇌ H C N + C H_{3} C O O^{-}$
would be: