Question bd69c

Aug 23, 2016

$\text{0.0901 M}$

Explanation:

Titration problems, much like any stoichiometry problem, are all about the mole ratio that exists between the acid and the base.

In this particular case, hydrofluoric acid, $\text{HF}$, a weak acid, will react with potassium hydroxide, $\text{KOH}$, a strong base, in a $1 : 1$ mole ratio.

${\text{HF"_ ((aq)) + "KOH"_ ((aq)) -> "KF"_ ((aq)) + "H"_ 2"O}}_{\left(l\right)}$

This means that in order to get to the equivalence point, you must mix equal numbers of moles of weak acid and strong base.

Notice that the volume of weak acid is bigger than the volume of strong base. This should automatically tell you that the concentration of the hydrofluoric acid solution will be lower than that of the potassium hydroxide.

That is the case because you need equal numbers of moles, but have a bigger volume of weak acid solution.

So, use the molarity of the potassium hydroxide solution to calculate how many moles of solute it contains -- do not forget to convert the volume to liters!

22.15 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * overbrace("0.122 moles KOH"/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)("= 0.122 M")) = 2.7023 * 10^(-3)"moles KOH"

Since you know that the reaction consumed equal numbers of moles of weak acid and strong base to get to the equivalence point, you can say that the hydrofluoric acid solution must have contained exactly $2.7023 \cdot {10}^{- 3}$ moles of solute.

Now all you have to do is divide this number of moles to the volume of the hydrofluoric acid solution to find its molarity

c_"HF" = (2.7023 * color(red)(cancel(color(black)(10^(-3))))"moles")/(30.0 * color(red)(cancel(color(black)(10^(-3))))"L") = color(green)(|bar(ul(color(white)(a/a)color(black)("0.0901 M")color(white)(a/a)|)))#

The answer is rounded to three sig figs.

As predicted, the concentration of the weak acid turned out to be lower than the concentration of the strong base.