x mg of Mg(OH)_2 (molar mass = 58 g/mol) is | vedclass

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(A) Given $pH = 10.0$,so $pOH = 14 - 10 = 4.0$. The concentration of hydroxide ions is $[OH^-] = 10^{-pOH} = 10^{-4} \ M$. Since $Mg(OH)_2$ dissociate

Vedclass · Accepted Answer

$3$

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(A) Given $pH = 10.0$,so $pOH = 14 - 10 = 4.0$. The concentration of hydroxide ions is $[OH^-] = 10^{-pOH} = 10^{-4} \ M$. Since $Mg(OH)_2$ dissociates as $Mg(OH)_2 ightarrow Mg^{2+} + 2OH^-$,the concentration of $Mg(OH)_2$ required is $[Mg(OH)_2] = \frac{[OH^-]}{2} = \frac{10^{-4}}{2} = 5 	imes 10^{-5} \ M$. For $1.0 \ L$ of solution,the number of moles of $Mg(OH)_2 = 5 	imes 10^{-5} \ mol$. Mass of $Mg(OH)_2 = 	ext{moles} 	imes 	ext{molar mass} = 5 	imes 10^{-5} \ mol 	imes 58 \ g/mol = 290 	imes 10^{-5} \ g = 2.9 	imes 10^{-3} \ g$. Converting to $mg$: $2.9 	imes 10^{-3} \ g 	imes 1000 \ mg/g = 2.9 \ mg$. The nearest integer value of $x$ is $3$.

$x \ mg$ of $Mg(OH)_2$ (molar mass $= 58 \ g/mol$) is required to be dissolved in $1.0 \ L$ of water to produce a $pH$ of $10.0$ at $298 \ K$. The value of $x$ is $........... mg$ ($Nearest$ $integer$). ($Given$: $Mg(OH)_2$ is assumed to dissociate completely in $H_2O$)

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