The degree of dissociation of a 0.05 M solut | vedclass

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(B) For a basic solution,$pH + pOH = 14$. Given $pH = 11$,so $pOH = 14 - 11 = 3$. Thus,$[OH^-] = 10^{-pOH} = 10^{-3} \ M$. For a weak base $NH_3$,the

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$2 	imes 10^{-2}$

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(B) For a basic solution,$pH + pOH = 14$. Given $pH = 11$,so $pOH = 14 - 11 = 3$. Thus,$[OH^-] = 10^{-pOH} = 10^{-3} \ M$. For a weak base $NH_3$,the dissociation is $NH_3 + H_2O ightleftharpoons NH_4^+ + OH^-$. The concentration of $OH^-$ is given by $[OH^-] = C 	imes \alpha$,where $C$ is the concentration of the base and $\alpha$ is the degree of dissociation. Given $C = 0.05 \ M$ and $[OH^-] = 10^{-3} \ M$. $\alpha = \frac{[OH^-]}{C} = \frac{10^{-3}}{0.05} = \frac{10^{-3}}{5 	imes 10^{-2}} = 0.2 	imes 10^{-1} = 0.02 = 2 	imes 10^{-2}$.

The degree of dissociation of a $0.05 \ M$ solution of $NH_3$ having $pH = 11$ at $25 \ ^oC$ is ...............

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