An alkaline solution of Mg(OH)_2 has pH = 12. | vedclass

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(A) Given $pH = 12$,so $pOH = 14 - 12 = 2$. Concentration of hydroxide ions: $[OH^-] = 10^{-pOH} = 10^{-2} \ M$. The dissociation reaction is: $Mg(OH)

Vedclass · Accepted Answer

$0.58$

Vedclass · Answer

(A) Given $pH = 12$,so $pOH = 14 - 12 = 2$. Concentration of hydroxide ions: $[OH^-] = 10^{-pOH} = 10^{-2} \ M$. The dissociation reaction is: $Mg(OH)_2 ightleftharpoons Mg^{2+} + 2OH^-$. Since $1 \ mol$ of $Mg(OH)_2$ produces $2 \ mol$ of $OH^-$,the concentration of $Mg(OH)_2$ is $[Mg(OH)_2] = \frac{[OH^-]}{2} = \frac{10^{-2}}{2} = 0.5 	imes 10^{-2} \ M$. Molar mass of $Mg(OH)_2 = 24 + 2 	imes (16 + 1) = 58 \ g/mol$. Using the formula $Molarity = \frac{w_B 	imes 1000}{M_B 	imes V(mL)}$: $0.5 	imes 10^{-2} = \frac{w_B 	imes 1000}{58 	imes 200}$. $w_B = 0.5 	imes 10^{-2} 	imes 58 	imes 0.2 = 0.58 \ g$.

An alkaline solution of $Mg(OH)_2$ has $pH = 12$. What amount of $Mg(OH)_2$ (in $gm$) is dissolved in $200 \ mL$ of the final solution?

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