0.001 mol of a strong electrolyte M(OH)_2 is | vedclass

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Question

(A) The strong electrolyte $M(OH)_2$ dissociates completely as: $M(OH)_2 \rightarrow M^{2+} + 2OH^-$. The number of moles of $M(OH)_2 = 0.001 \ mol$.

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(A) The strong electrolyte $M(OH)_2$ dissociates completely as: $M(OH)_2 ightarrow M^{2+} + 2OH^-$. The number of moles of $M(OH)_2 = 0.001 \ mol$. The concentration of $M(OH)_2 = \frac{0.001 \ mol}{20 \ mL} 	imes 1000 \ mL/L = 0.05 \ M$. Since $1 \ mol$ of $M(OH)_2$ gives $2 \ mol$ of $OH^-$,the concentration of $[OH^-] = 2 	imes 0.05 \ M = 0.1 \ M$. $pOH = -\log[OH^-] = -\log(0.1) = 1$. $pH = 14 - pOH = 14 - 1 = 13$.

$0.001 \ mol$ of a strong electrolyte $M(OH)_2$ is dissolved to make $20 \ mL$ of a saturated solution. Calculate its $pH$ $(K_w = 1 \times 10^{-14})$.

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