# If Ca(OH)_2(aq)=0.01*mol*L^-1, what is pH of the solution?

Jul 10, 2016

$p H \cong 12$

#### Explanation:

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

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

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

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