# Determine the pH of a 0.09 M KOH solution. Your answer must be within ± 0.1%?

May 4, 2018

$\text{pH} = 12.95$

#### Explanation:

$\text{KOH}$ is a strong base, meaning that it dissociates completely. Therefore, the (balanced) equation is:
${\text{KOH (aq)" -> "K"^+ "(aq)" + "OH}}^{-}$(aq)"

This balanced equation tells us that every mole of $\text{KOH}$ produces one mole of ${\text{OH}}^{-}$.
Therefore, we know that ["KOH"] = ["OH"^-] = 0.09 " M"

Now, to find the $\text{pOH}$, we use the formula:
"pOH" = -log["OH"^-]

So:
"pOH" = -log[0.09 " M"] ~~ 1.05

However, the question wants the $\text{pH}$, not the $\text{pOH}$. To find this, we use:
$\text{pH" = 14 - "pOH}$

So:
$\text{pH} = 14 - 1.05 = 12.95$

The exact solution to $8$ decimal places is $12.95424251$. This is within 0.1%:

|(12.95 - 12.95424251)/(12.95424251)| xx 100% ~~ 0.03%

Hope this helps!