Question

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 $CN ^{-}+ CH _{3} COOH \rightleftharpoons HCN + CH _{3} COO ^{-}$ would be :

• A. $ 3.0 \times 10^{5} $
• B. $ 3.0 \times 10^{-5} $
• C. $ 3.0 \times 10^{-4} $
• D. $ 3.0 \times 10^{4} $

Answer 1

The Correct Option is D

To find the equilibrium constant for the given equilibrium reaction, we need to understand the relationship between the dissociation constants of the acids involved and the desired equilibrium constant.

The given equilibrium is: CN^{-} + CH_{3}COOH \rightleftharpoons HCN + CH_{3}COO^{-}.

This reaction represents the conversion of CN^{-} ion into HCN and the conversion of CH_{3}COOH into CH_{3}COO^{-}. The reaction essentially involves the transfer of a proton from acetic acid (substrate) to the cyanide ion (nucleophile).

The dissociation constants provided are:

• Dissociation constant for acetic acid, K_{a, CH_{3}COOH} = 1.5 \times 10^{-5}
• Dissociation constant for HCN, K_{a, HCN} = 4.5 \times 10^{-10}

The equilibrium constant K for the given equilibrium can be expressed in terms of the dissociation constants of the acids as follows:

K = \frac{K_{a, CH_{3}COOH}}{K_{a, HCN}}

Substitute the given values:

K = \frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}}

Perform the division:

K = \frac{1.5}{4.5} \times 10^{5}

Calculate the division:

\frac{1.5}{4.5} = \frac{1}{3} = 0.3333

Then:

K = 0.3333 \times 10^{5} = 3.333 \times 10^{4}

For rounding off:

The equilibrium constant closest to 3.333 \times 10^{4} is 3.0 \times 10^{4}.

Thus, the correct answer is:

3.0 \times 10^{4}