Weight of Ca(OH)_2 needed to prepare 250 mL | vedclass

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(A) Given $pH = 13$,we have $pOH = 14 - 13 = 1$. Therefore,the concentration of hydroxide ions is $[OH^-] = 10^{-pOH} = 10^{-1} = 0.1 \ M$. Since $Ca(

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$0.925$

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(A) Given $pH = 13$,we have $pOH = 14 - 13 = 1$. Therefore,the concentration of hydroxide ions is $[OH^-] = 10^{-pOH} = 10^{-1} = 0.1 \ M$. Since $Ca(OH)_2$ is a diacidic base,its molarity $M$ is related to normality $N$ by $N = 2 	imes M$. For $[OH^-] = 0.1 \ M$,the molarity of $Ca(OH)_2$ is $0.05 \ M$. The molar mass of $Ca(OH)_2$ is $40 + 2 	imes (16 + 1) = 74 \ g/mol$. Weight $w = M 	imes 	ext{Molar Mass} 	imes 	ext{Volume in Liters} = 0.05 	imes 74 	imes 0.250 = 0.925 \ g$.

Weight of $Ca(OH)_2$ needed to prepare $250 \ mL$ of solution with $pH = 13$. (in $g$)

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