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 will be- CN-CH3COOH - HCN-CH3COO-

The correct option is D 3.0× 10^4 CH_3COOH = HCN + CH_3COO^ K_a = 1.5 × 10^ 5 #160; #160; #160; #160; #160; #160; #160; ...

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Standard XII

Chemistry

Characteristics of Equilibrium Constant

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 will be: $C N^{-} + C H_{3} C O O H ⇌ H C N + C H_{3} C O O^{-}$

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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3.0 \times 10^{4}

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Solution

The correct option is D $3.0 \times 10^{4}$
$C H_{3} C O O H = H C N + C H_{3} C O O^{-}$
$K_{a} = 1.5 \times 10^{-} 5$ .....(1)
$H C N = H^{+} + C N^{-}; K_{a 1} = 4.5 \times 10^{-} 10$ ......(2)
$C N^{-} + C H_{3} C O O H = H C N + C H_{3} C O O H^{-}$
K = ?
On subtracting equation (2) from equation (1) we get $K = K_{a} / K_{a 1} = 1.5 \times \frac{10^{- 5}}{4.5} \times 10^{- 10}$ $= \frac{10^{5}}{3}$ $= 3.33 \times 10^{4}$ So option D is correct.

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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:

Find the true statements regarding the aqueous dissociation of HCN,

K_{a} = 4.9 \times 10^{- 10}

25^{o}

C.
I. At equilibrium,

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

II. At equilibrium,

[H^{+}] > [H C N]

III.

H C N (a q)

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 will be: CN−+CH3COOH⇌HCN+CH3COO−

The correct option is D 3.0×104CH3COOH=HCN+CH3COO−Ka=1.5×10−5 .....(1)HCN=H++CN−;Ka1=4.5×10−10 ......(2)CN−+CH3COOH=HCN+CH3COOH−K = ?On subtracting equation (2) from equation (1) we get K=Ka/Ka1=1.5×10−54.5×10−10 =1053 =3.33×104 So option D is correct.

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 will be: $C N^{-} + C H_{3} C O O H ⇌ H C N + C H_{3} C O O^{-}$