# What is the [OH^-] of a 4.0 times 10^-4 M solution of Ca(OH_2)?

Apr 7, 2017

$\left[H {O}^{-}\right] = 8.0 \times {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1$............

#### Explanation:

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

$C a {\left(O H\right)}_{2} \left(s\right) \stackrel{{H}_{2} O}{\rightarrow} C {a}^{2 +} + 2 H {O}^{-}$

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

And thus [HO^-]=2xx[Ca^(2+)]=2xx4.0xx10^-4*mol*L^-1=??*mol*L^-1.

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