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NEET CHEMISTRYMedium

The dissociation constants for acetic acid and HCN at 25∘C25^\circ \text{C} are 1.5×10−51.5 \times 10^{-5} and 4.5×10−104.5 \times 10^{-10}, respectively. The equilibrium constant for the equilibrium, CN−+CH3COOH⇌HCN+CH3COO−CN^- + CH_3COOH \rightleftharpoons HCN + CH_3COO^- would be:

A

3.0×1053.0 \times 10^5

B

3.0×10−53.0 \times 10^{-5}

C

3.0×10−43.0 \times 10^{-4}

D

3.0×1043.0 \times 10^4

Step-by-Step Solution

For the given reaction: CN−+CH3COOH⇌HCN+CH3COO−CN^- + CH_3COOH \rightleftharpoons HCN + CH_3COO^- The equilibrium constant KK is given by: K=[HCN][CH3COO−][CN−][CH3COOH]K = \frac{[HCN][CH_3COO^-]}{[CN^-][CH_3COOH]} Multiplying the numerator and denominator by [H+][H^+], we get: K=[CH3COO−][H+][CH3COOH]×[HCN][CN−][H+]K = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} \times \frac{[HCN]}{[CN^-][H^+]} K=Ka(CH3COOH)Ka(HCN)K = \frac{K_a(CH_3COOH)}{K_a(HCN)} Given Ka(CH3COOH)=1.5×10−5K_a(CH_3COOH) = 1.5 \times 10^{-5} and Ka(HCN)=4.5×10−10K_a(HCN) = 4.5 \times 10^{-10}, K=1.5×10−54.5×10−10=13×105=0.333×105=3.33×104K = \frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}} = \frac{1}{3} \times 10^5 = 0.333 \times 10^5 = 3.33 \times 10^4 The closest value among the given options is 3.0×1043.0 \times 10^4.

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