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

1 Answer
Dec 27, 2016

Answer:

#"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"#.