The number of moles of Ca(OH)_2 required to p | vedclass

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(C) Given $pH = 14$. Since $pH + pOH = 14$,we have $pOH = 14 - 14 = 0$. $pOH = -\log[OH^-] = 0$,which implies $[OH^-] = 10^0 = 1 \ M$. $Ca(OH)_2$ unde

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$0.125$

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(C) Given $pH = 14$. Since $pH + pOH = 14$,we have $pOH = 14 - 14 = 0$. $pOH = -\log[OH^-] = 0$,which implies $[OH^-] = 10^0 = 1 \ M$. $Ca(OH)_2$ undergoes complete ionization as $Ca(OH)_2 ightarrow Ca^{2+} + 2OH^-$. Thus,$1 \ mole$ of $Ca(OH)_2$ produces $2 \ moles$ of $OH^-$. Therefore,the concentration of $Ca(OH)_2$ required is $[Ca(OH)_2] = \frac{[OH^-]}{2} = \frac{1}{2} = 0.5 \ M$. For $250 \ mL$ $(0.25 \ L)$ of solution,the number of moles required is $n = M 	imes V(L) = 0.5 \ mol \ L^{-1} 	imes 0.25 \ L = 0.125 \ mole$.

The number of moles of $Ca(OH)_2$ required to prepare $250 \ mL$ of solution with $pH$ $14$ (assuming complete ionization) is :-

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