# What is the pH of a 0.003 M KOH solution?

Apr 17, 2018

$p H \approx 11.5$

#### Explanation:

When put in water, the $K O H$ dissolves into ${K}^{+}$ and $O {H}^{-}$ ions, with the latter increasing the $p H$ of the solution.
As hydroxides can be considered strong bases, it will completely dissociate into the water solution, and form an equal amount of moles of hydroxide ions:

0.003 mol of $O {H}^{-}$.

Now,

$p O H = - \log \left[O {H}^{-}\right]$
$p O H = 2.522$

And if we assume this is done under standard conditions:

$p O H + p H = 14$
$p H = 14 - 2.522$
$p H \approx 11.5$