Question

Use the "x is small" approximation to find the concentration of the products in the following reaction which initially contains only 0.85 M HCN. Decide whether using the approximation was valid or invalid.HCN(aq) ⇌ H+(aq) + CN−(aq), Kc= 6.17 x 10−10

Answer 1

Answer:

Concentration of product at equilibrium ;

$[H^+]=0.0000229 M$

$[CN^-]=0.0000229 M$

Explanation:

$HCN(aq)\rightleftharpoons H^+(aq) + CN^-(aq)$

initially

0.85 M        0    0

(0.85-x)M    x      x

The equilibrium constant of reaction = $K_c= 6.17\times 10^{-10}$

The expression of an equilibrium cannot can be written as:

$K_c=\frac{[H^+][CN^-]}{[HCN]}$

$6.17\times 10^{-10}=\frac{x\times x}{(0.85-x)}$

Solving for x:

x = 0.0000229

Concentration of product at equilibrium ;

$[H^+]=0.0000229 M$

$[CN^-]=0.0000229 M$