The degree of dissociation of a 0.1 M monobasic acid is 0.4%. Its diss

The degree of dissociation of a 0.1 M monobasic acid is 0.4%. Its dissociation constant is :(a) 0.4 × 10-4(b) 4.0 × 10-4(c) 1.6 × 10-6(d) 0.8 × 10-5

Answer»

(C) \(1.6 × 10^{-6}\)

\(K_a=\frac{[B^+][\overset{-}{o}{H}]}{[BOH]}\)

\(K_a=\frac{C^2\alpha^2}{C(1-\alpha)}\)

\(\alpha=\frac{0.4}{100}=0.004\)

\(\therefore K_a=\frac{C\alpha ^2}{(1-\alpha)}\)

\(K_a= C\alpha ^2\)

\(\because \alpha <<1\)

\(K_a=0.1\times(4\times10^{-3})^2\)

= \(0.1\times16\times10^{-6}\)

= \(16\times10^{-7}\) or \(1.6\times10^{-6}\)

Option : (c) 1.6 × 10^-6