If $C a {(O H)}_{2} (a q) = 0.01 \cdot m o l \cdot L^{- 1}$ $Ca(OH)_2(aq)=0.01molL^-1$ , what is $p H$ $pH$ of the solution?

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If $C a {(O H)}_{2} (a q) = 0.01 \cdot m o l \cdot L^{- 1}$ $Ca(OH)_2(aq)=0.01molL^-1$ , what is $p H$ $pH$ of the solution?

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Answer 1 · 2016-07-10T09:25:18

$p H ≅ 12$ $pH~=12$

Explanation:

We know (or should know) that in aqueous solution $p H + p O H = 14$ $pH+pOH=14$ .

Thus, if we calculate $p O H$ $pOH$ , $p H$ $pH$ is directly available.

$p O H = - {log}_{10} ([H O^{-}])$ $pOH=-log_10([HO^-])$ $= - {log}_{10} (0.02)$ $=-log_10(0.02)$ (why $0.02$ $0.02$ here and not $0.01$ $0.01$ ?) $=$ $=$ $- (- 1.70)$ $-(-1.70)$ $=$ $=$ $1.70$ $1.70$

And from the above, $p H = 14 - 1.70$ $pH=14-1.70$ $=$ $=$ $? ?$ $??$

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If $C a {(O H)}_{2} (a q) = 0.01 \cdot m o l \cdot L^{- 1}$ $Ca(OH)_2(aq)=0.01molL^-1$ , what is $p H$ $pH$ of the solution?

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Explanation:

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If Ca(OH)2(aq)=0.01⋅mol⋅L−1Ca(OH)_2(aq)=0.01*mol*L^-1, what is pHpH of the solution?

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If $C a {(O H)}_{2} (a q) = 0.01 \cdot m o l \cdot L^{- 1}$ $Ca(OH)_2(aq)=0.01molL^-1$ , what is $p H$ $pH$ of the solution?