# What volume of a 0.855 M KOH solution is required to make a 3.55 L solution at a pH of 12.4?

Dec 27, 2016

$\text{104 mL}$

#### Explanation:

Start by using the $\text{pOH}$ of the solution to figure out the concentration of hydroxide anions, ${\text{OH}}^{-}$. You should know that

color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))

To solve for the concentration of hydroxide anions, rearrange the equation as

log(["OH"^(-)]) = - "pOH"

This can be written as

10^log(["OH"^(-)]) = 10^(-"pOH")

which is equivalent to

$\left[\text{OH"^(-)] = 10^(-"pOH}\right)$

Now, an aqueous solution at room temperature has

$\textcolor{red}{\underline{\textcolor{b l a c k}{\text{pH " + " pOH} = 14}}}$

This means that the $\text{pOH}$ of the target solution is equal to

$\text{pOH} = 14 - 12.4 = 1.6$

Consequently, the target solution must have

["OH"^(-)] = 10^(-1.6) = 2.51 * 10^(-2)"M"

As you know, molarity is defined as the number of moles of solute present in $\text{1 L}$ of solution. This means that the number of moles of potassium hydroxide needed to ensure that concentration of hydroxide anions is equal to

3.55 color(red)(cancel(color(black)("L solution"))) * (2.51 * 10^(-2)"moles OH"^(-))/(1color(red)(cancel(color(black)("L solution")))) = 8.91 * 10^(-2)"moles OH"^(-)

All you have to do now is figure out what volume of the stock solution contains the needed number of moles of hydroxide anions

8.91 * 10^(-2) color(red)(cancel(color(black)("moles OH"^(-)))) * "1 L solution"/(0.855color(red)(cancel(color(black)("moles OH"^(-))))) = "0.104 L"

Expressed in milliliters, the answer will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{V}_{\text{stock KOH" = "104 mL}}}}}$

I'll leave the answer rounded to three sig figs.

So, in order to prepare this solution, take $\text{104 mL}$ of $\text{0.855 M}$ stock potassium hydroxide solution and add enough water until the final volume of the solution is equal to $\text{3.55 L}$.