# What is the pH of a 6.49 * 10^-3 M KOH solution?

May 28, 2016

$p H \cong 11.8$

#### Explanation:

It is a fact that in aqueous solution under standard conditions,

$14 = p H + p O H$

So we can directly calculate $p O H$, and then approach $p H$.

$p O H$ $=$ $- {\log}_{10} \left[H {O}^{-}\right]$ $=$ $- {\log}_{10} \left[6.49 \times {10}^{-} 3\right]$ $=$ $2.19$

Thus $p H$ $=$ $14.00 - 2.19$ $=$ ??