If a metal hydroxide with the molecular formu | vedclass

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(D) The dissociation reaction is $M(OH)_4 \rightleftharpoons M^{4+} + 4OH^-$. Given concentration $C = 0.0025 \ M$ and degree of ionization $\alpha =

Vedclass · Accepted Answer

$11.7$

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(D) The dissociation reaction is $M(OH)_4 ightleftharpoons M^{4+} + 4OH^-$. Given concentration $C = 0.0025 \ M$ and degree of ionization $\alpha = 0.50$ $(50\%)$. The concentration of $OH^-$ ions is given by $[OH^-] = n 	imes C 	imes \alpha$,where $n$ is the number of $OH^-$ ions produced per molecule. $[OH^-] = 4 	imes 0.0025 	imes 0.50 = 0.005 \ M = 5 	imes 10^{-3} \ M$. Now,$pOH = -\log[OH^-] = -\log(5 	imes 10^{-3}) = 3 - \log 5 = 3 - 0.699 = 2.301$. Since $pH + pOH = 14$,we have $pH = 14 - 2.301 = 11.699 \approx 11.7$.

If a metal hydroxide with the molecular formula $M(OH)_4$ is $50\%$ ionized,then the $pH$ of its $0.0025 \ M$ solution will be:

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