What is the $[O H^{-}]$ $[OH^-]$ of a $4.0 \times 10^{- 4}$ $4.0 times 10^-4$ $M$ $M$ solution of $C a (O H_{2})$ $Ca(OH_2)$ ?

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What is the $[O H^{-}]$ $[OH^-]$ of a $4.0 \times 10^{- 4}$ $4.0 times 10^-4$ $M$ $M$ solution of $C a (O H_{2})$ $Ca(OH_2)$ ?

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Answer 1 · 2017-04-07T18:43:08

$[H O^{-}] = 8.0 \times 10^{- 4} \cdot m o l \cdot L^{- 1}$ $[HO^-]=8.0xx10^-4*mol*L^-1$ ............

Explanation:

Calcium hydroxide speciates in aqueous solution according to the following reaction:

$C a {(O H)}_{2} (s) H_{2} O - - \to C a^{2 +} + 2 H O^{-}$ $Ca(OH)_2(s) stackrel(H_2O)rarrCa^(2+) + 2HO^-$

If the the concentration is $4.0 \times 10^{- 4} \cdot m o l \cdot L^{- 1}$ $4.0xx10^-4*mol*L^-1$ with respect to $C a^{2 +}$ $Ca^(2+)$ , stoichiometry, i.e. the composition of the salt, demands the solution is TWICE this concentrated with respect to $^{-} O H$ $""^(-)OH$ .

And thus $[H O^{-}] = 2 \times [C a^{2 +}] = 2 \times 4.0 \times 10^{- 4} \cdot m o l \cdot L^{- 1} = ? ? \cdot m o l \cdot L^{- 1} .$ $[HO^-]=2xx[Ca^(2+)]=2xx4.0xx10^-4*mol*L^-1=??*mol*L^-1.$

What are $p O H$ $pOH$ and $p H$ $pH$ of this solution?

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What is the $[O H^{-}]$ $[OH^-]$ of a $4.0 \times 10^{- 4}$ $4.0 times 10^-4$ $M$ $M$ solution of $C a (O H_{2})$ $Ca(OH_2)$ ?

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Explanation:

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What is the $[O H^{-}]$ $[OH^-]$ of a $4.0 \times 10^{- 4}$ $4.0 times 10^-4$ $M$ $M$ solution of $C a (O H_{2})$ $Ca(OH_2)$ ?