For a 10^{-2} M Ca(OH)_2 solution with a de | vedclass

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(C) The concentration of $Ca(OH)_2$ is $C = 10^{-2} \ M$. Since $Ca(OH)_2$ provides $2 \ OH^-$ ions per molecule,the total concentration of $OH^-$ ion

Vedclass · Accepted Answer

$2.7, 11.3$

Vedclass · Answer

(C) The concentration of $Ca(OH)_2$ is $C = 10^{-2} \ M$. Since $Ca(OH)_2$ provides $2 \ OH^-$ ions per molecule,the total concentration of $OH^-$ ions is given by $[OH^-] = 2 	imes C 	imes \alpha$. Given $\alpha = 10\% = 0.1$,we have $[OH^-] = 2 	imes 10^{-2} 	imes 0.1 = 2 	imes 10^{-3} \ M$. Now,calculate $pOH$: $pOH = -\log[OH^-] = -\log(2 	imes 10^{-3}) = -(\log 2 + \log 10^{-3}) = -(0.301 - 3) = 2.699 \approx 2.7$. Finally,calculate $pH$: $pH = 14 - pOH = 14 - 2.7 = 11.3$. Thus,the values are $pOH = 2.7$ and $pH = 11.3$.

For a $10^{-2} \ M \ Ca(OH)_2$ solution with a degree of dissociation $\alpha = 10\%$,what are the values of $pOH$ and $pH$ respectively?

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