Millimoles of calcium hydroxide required to p | vedclass

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Question

(A) Given $pH = 12$,we know that $pOH = 14 - pH = 14 - 12 = 2$. $[OH^-] = 10^{-pOH} = 10^{-2} \ M$. Since $Ca(OH)_2$ dissociates completely as $Ca(OH)

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$5$

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(A) Given $pH = 12$,we know that $pOH = 14 - pH = 14 - 12 = 2$. $[OH^-] = 10^{-pOH} = 10^{-2} \ M$. Since $Ca(OH)_2$ dissociates completely as $Ca(OH)_2 ightarrow Ca^{2+} + 2OH^-$,the concentration of $Ca(OH)_2$ is $[Ca(OH)_2] = \frac{[OH^-]}{2} = \frac{10^{-2}}{2} = 0.5 	imes 10^{-2} = 5 	imes 10^{-3} \ M$. Number of millimoles $= 	ext{Molarity} 	imes 	ext{Volume in } mL = (5 	imes 10^{-3} \ M) 	imes (100 \ mL) = 0.5 \ 	ext{millimoles}$. This can be written as $5 	imes 10^{-1} \ 	ext{millimoles}$. Comparing with $x 	imes 10^{-1}$,we get $x = 5$.

Millimoles of calcium hydroxide required to produce $100 \ mL$ of the aqueous solution of $pH$ $12$ is $x \times 10^{-1}$. The value of $x$ is $..............$. (Nearest integer). Assume complete dissociation.

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