A solution has initially 0.1 M-HCOOH and 0.2 M-HCN. The value of Ka for HCOOH = 2.56 × 10−5, Ka of H

Question

A solution has initially 0.1 M-HCOOH and 0.2 M-HCN. The value of Ka for HCOOH = 2.56 × 10−5, Ka of HCN = 9.6 × 10−10. The only incorrect statement for the solution is (log 2 = 0.3)

Accepted Answer

HCN $ightleftharpoons$ H+ + CN–Equilibrium (0.2 – x) M (x + y) M x M$9.6 	imes 10^{- 10} = \frac{(x + y).x}{(0.2 - x)} \approx \frac{y.x}{0.2}$HCOOH $ightleftharpoons$ H+ + HCOO–Equilibrium (0.1 – y) M (x + y) M y M$$2.56 	imes 10^{- 5} = \frac{(x + y).y}{(0.1 - y)} \approx \frac{y.y}{0.1}$$∴ y = 1.6 × 10–3 and x = 1.2 × 10–7Option (d) may be answered without solving because solution is acidic.

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