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NEET CHEMISTRYMedium

The dissociation constants for acetic acid and HCN at 25 °C are 1.5×1051.5 \times 10^{-5} and 4.5×10104.5 \times 10^{-10}, respectively. The equilibrium constant for the equilibrium, CN+CH3COOHHCN+CH3COOCN^- + CH_3COOH \rightleftharpoons HCN + CH_3COO^- would be:

A

3.0×1053.0 \times 10^5

B

3.0×1053.0 \times 10^{-5}

C

3.0×1043.0 \times 10^{-4}

D

3.0×1043.0 \times 10^4

Step-by-Step Solution

The given dissociation reactions and their equilibrium constants are: (1) CH3COOHCH3COO+H+CH_3COOH \rightleftharpoons CH_3COO^- + H^+; Ka1=1.5×105K_{a1} = 1.5 \times 10^{-5} (2) HCNCN+H+HCN \rightleftharpoons CN^- + H^+; Ka2=4.5×1010K_{a2} = 4.5 \times 10^{-10}

The target reaction is: CN+CH3COOHHCN+CH3COOCN^- + CH_3COOH \rightleftharpoons HCN + CH_3COO^- This reaction can be obtained by subtracting equation (2) from equation (1) (or adding the reverse of equation 2 to equation 1). The equilibrium constant for the target reaction will be: Keq=Ka1Ka2=1.5×1054.5×1010=13×105=3.33×1043.0×104K_{eq} = \frac{K_{a1}}{K_{a2}} = \frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}} = \frac{1}{3} \times 10^5 = 3.33 \times 10^4 \approx 3.0 \times 10^4. Hence, the closest correct option is 3.0×1043.0 \times 10^4.

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