Question

What is the pH of a solution that is 0.20 M in HF and 0.40 M in NaF? [Ka = 7.2*10^-4] I know the answer is 3.44 but could someone please explain how they got it? Thank you so much!!!! (:

Answer 1

You get it like this:

Explanation:

HF is a weak acid which dissociates:

`sf(HFrightleftharpoonsH^(+)+F^-)`

`sf(K_a=([H^+][F^-])/([HF])=7.2xx10^(-4))`

Rearranging:

`sf([H^+]=K_axx([HF])/([F^-]))`

Putting in the numbers:

`sf([H^+]=7.2xx10^(-4)xx0.2/(0.4)=3.6xx10^(-4)color(white)(x)"mol/l")`

`sf(pH=-log[H^+]=-log(3.6xx10^(-4))=3.44)`

Answer 2

When I read this, I interpreted it as having initial concentrations of `"HF"` and `"NaF"`. From that, we would consider the following dissociation (units of `"M"`):

`"HF"(aq) rightleftharpoons "H"^(+)(aq) + "F"^(-)(aq)`

`"I"" "0.20" "" "" "0" "" "" "0.40`
`"C"" "-x" "" "+x" "" "" "+x`
`"E"" "0.20-x" "x" "" "" "0.40+x`

This would give:

`K_a = 7.2 xx 10^(-4) = (["H"^(+)]_(eq)["F"^(-)]_(eq))/(["HF"]_(eq))`

`= (x(0.40+x))/(0.20-x)`

Taking the `-log` of both sides, this can be made to look like the Henderson-Hasselbalch equation for buffers:

`-log(K_a) = -logx - log((0.40+x)/(0.20-x))`

Since `x = ["H"^(+)]`, this is really:

`"pK"_a = "pH" - log ((0.40+x)/(0.20-x))`

Or, we can then get the Henderson-Hasselbalch equation like so:

`"pH" = "pK"_a + log ((0.40+x)/(0.20-x))`

In this case, we have a `K_a` that is fairly small, so we approximate that `0.40 + x ~~ 0.40` and `0.20 - x ~~ 0.20`:

`color(blue)("pH") ~~ -log(7.2 xx 10^(-4)) + log(0.40/0.20)`

`= color(blue)(3.44)`

In homework problems and exams, you can just use the Henderson-Hasselbalch equation to begin with, as long as you know that it only works on buffers containing a weak acid and its conjugate weak base.