Question

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

Answer 1

`"104 mL"`

Explanation:

Start by using the `"pOH"` of the solution to figure out the concentration of hydroxide anions, `"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

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

Now, an aqueous solution at room temperature has

`color(red)(ul(color(black)("pH " + " pOH" = 14)))`

This means that the `"pOH"` of the target solution is equal to

`"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 `"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

`color(darkgreen)(ul(color(black)(V_"stock KOH" = "104 mL")))`

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

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