If the pH of an aqueous solution of #"0.15 M"# #"HF"# is #2#, what is the percent ionization of #"HF"#?
2 Answers
The percent "ionization" is the same as the percent dissociation. Since you know the pH is
#"HF" rightleftharpoons "H"^(+) + "F"^(-)#
We can use the Henderson-Hasselbalch equation.
#pH = pKa + log \frac{["A"^(-)]}{["HA"]}#
where
#2 = 3.15 + log \frac{["A"^(-)]}{["HA"]}#
#-1.15 = log \frac{["A"^(-)]}{["HA"]}#
#10^(-1.15) = \frac{["A"^(-)]}{["HA"]}#
But we know that as
Additionally, both species are within the same solution, so their concentrations are determined by the same total volume. When we compare concentrations, we simply get:
Furthermore, we are looking for a percent. If we let
#10^(-1.15) = x/(1-x)#
#10^(-1.15) - 10^(-1.15)x = x#
#10^(-1.15) = (1 + 10^(-1.15))x#
#x = (10^(-1.15))/(1 + 10^(-1.15))#
#= 0.066 => color(blue)("6.6% dissociated")#
Explanation:
Here's an alternative approach you can use to find the percent ionization of the weak acid.
When placed in aqueous solution, hydrofluoric acid,
#"HF"_text((aq]) rightleftharpoons "H"_text((aq])^(+) + "F"_text((aq])^(-)#
Notice that you have
Simply put, one molecule of hydrofluoric acid that ionizes will split into one hydrogen cation and one fluoride anion.
Use the pH of the solution to determine the concentration of hydrogen cations
#["H"^(+)] = 10^(-"pH")#
#["H"^(+)] = 10^(-2) = "0.010 M"#
So, if the ionization of the acid produced
This means that out of the initial
Percent ionization is defined as
#color(blue)("% ionization" = "amount that ionized"/"initial amount" xx 100)#
In this case, you would have
#"% ionization" = (0.010 color(red)(cancel(color(black)("M"))))/(0.15color(red)(cancel(color(black)("M")))) xx 100 = 6.667% ~~ color(green)(6.7%)#