# How do you calculate the pH of 0.075M KOH?

May 17, 2017

$p H = 12.88$

#### Explanation:

We know $\left(i\right)$ that $p H + p O H = 14$ in aqueous solution under standard conditions...........

We also know $\left(i i\right)$ that $p O H = - {\log}_{10} \left[H {O}^{-}\right]$

$= - {\log}_{10} \left(0.075\right) = - \left(- 1.12\right) = + 1.12$.

And so $p H = 14 - p O H = 12.88$