BotanyThe dissociation constants for acetic acid and HCN at 250C are 1.5 ×  10-5 and 4.5 ×  10-10, respectively. The equilibrium constant for the equilibriumCN- + CH3COOH  ⇌  HCN  +  CH3COO-would be

The dissociation constants for acetic acid and HCN at 250C are 1.5 ×  10-5 and 4.5 ×  10-10, respectively. The equilibrium constant for the equilibrium

CN- + CH3COOH    HCN  +  CH3COO-

would be

  1. A

    3.0 ×  105

  2. B

    3.0 ×  10-5

  3. C

    3.0 ×  10-4

  4. D

    3.0 ×  104

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    Solution:

    given,  CH3COOH    CH3COO- + H+;

    Ka1 = 1.5 ×  10-5                                     …..(i)

    HCN    H+  +  CN-;

    Ka2 = 4.5 ×  10-10                                   …..(ii)

    CN- + CH3COOH    HCN  +  CH3COO-      K = ?

    On subtracting Eq. (ii) from Eq.(i), we get 

    CH3COOH + CN-    HCN  +  CH3COO-

    K = Ka1Ka2 = 1.5 × 10-54.5 × 10-10 = 1053 = 3.33 × 104

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