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# The hydronium ion concentration in pure water is $1 \times 10^{-7}$ mol$L^{-1}$. The degree of dissociation of water is

(a) $1.8 \times 10^{-9}$

(b) $0.8 \times 10^{-8}$

(c) $9.8 \times 10^{-6}$

(d) $3 \times 10^{-9}$

Answer: $1.8 \times 10^{-9}$

If $\alpha$ is degree of dissociation

$H_2O \rightleftharpoons H^+ + OH^-$

Initial Concentration, [$H_2O$] = C mol$L^{-1}$
Equilibrium concentration, $[H_2O] = C(1-\alpha)$ mol $L^{-1}, [H^+] = C\alpha$ mol $L^{-1}, [OH^-] = C\alpha$ mol $L^{-1}$

$[H^+] = C\alpha \Rightarrow \alpha = \frac{[H^+]}{C} = \frac{1 \times 10^{-7}}{55.5} = 1.8 \times 10^{-9}$