# What is the "pH" of "0.1 N" "Ca"("OH")_2? Thanks.

Jul 1, 2017

$p H = 13.0$

#### Explanation:

We work out (i) the $p O H$........

We have $0.1 \cdot N$ with respect to $H {O}^{-}$. This is $0.05 \cdot m o l \cdot {L}^{-} 1$ with respect to $C a {\left(O H\right)}_{2} \left(a q\right)$. Agreed?

Now.................

$p O H = - {\log}_{10} \left[H {O}^{-}\right] = - {\log}_{10} \left(0.1\right) = - \left(- 1\right) = + 1$.

And we can now (ii) address the $p H$.

Because $p O H + p H = 14$, and thus $p H = 14 - 1.00 = 13.0$.

Is this solution acidic or basic?

See here for a similar problem.

Thanks to Truong-Son who pointed out an error on my part......