# What is the pH of a 0.000460 M solution of Ca(OH)_2?

Jun 21, 2016

Approx. $11$

#### Explanation:

$p O H$ $=$ $- {\log}_{10} \left[H {O}^{-}\right]$.

Thus $p O H = - {\log}_{10} \left\{0.009200\right\}$ $=$ $- \left(- 3.04\right) = 3.04$

But it is a FACT that $p H + p O H = 14$

Thus $p H \cong 11$

Why is the concentration of $H {O}^{-}$ double that of $C a {\left(O H\right)}_{2}$?

PS If you have trouble with the $\log$ function, voice your problem . Someone here will help you.