The pH of a saturated solution of Ca(OH)_{2} | vedclass

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Question

(A) The dissociation of $Ca(OH)_{2}$ is given by: $Ca(OH)_{2(s)} \rightleftharpoons Ca^{2+}_{(aq)} + 2OH^{-}_{(aq)}$Given $pH = 9$,we know $pOH = 14 -

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$0.5 	imes 10^{-15}$

Vedclass · Answer

(A) The dissociation of $Ca(OH)_{2}$ is given by: $Ca(OH)_{2(s)} ightleftharpoons Ca^{2+}_{(aq)} + 2OH^{-}_{(aq)}$Given $pH = 9$,we know $pOH = 14 - pH = 14 - 9 = 5$.Therefore,$[OH^{-}] = 10^{-pOH} = 10^{-5} \ M$.From the stoichiometry,$[OH^{-}] = 2S$,where $S$ is the solubility of $Ca(OH)_{2}$.So,$2S = 10^{-5} \implies S = 0.5 	imes 10^{-5} \ M$.The solubility product expression is $K_{sp} = [Ca^{2+}][OH^{-}]^{2} = (S)(2S)^{2} = 4S^{3}$.Substituting the value of $S$: $K_{sp} = 4 	imes (0.5 	imes 10^{-5})^{3} = 4 	imes 0.125 	imes 10^{-15} = 0.5 	imes 10^{-15}$.

$S.No.$	$Formula$	$K_{sp}$
$1$	$PQ$	$4.0 \times 10^{-20}$
$2$	$PQ_2$	$3.2 \times 10^{-14}$
$3$	$PQ_3$	$2.7 \times 10^{-35}$

The $pH$ of a saturated solution of $Ca(OH)_{2}$ is $9$. The solubility product $(K_{sp})$ of $Ca(OH)_{2}$ is:

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