Question

A certain aqueous solution contains `"0.27 mols"` of weak acid `"HA"`. If `"12.0 mL"` of `"3.10 M"` `"NaOH"` was added to it, resulting in a `"pH"` of `3.80`, what is the `"pK"_a` of the weak acid?

Answer 1

You added less mols of strong base to a weak acid. Hence, you form a buffer by neutralizing some weak acid to generate some weak base. Here's how I would do it...

`(1)` Use an ICE Table to organize your work.

`"0.0120 L" xx "3.10 M NaOH" = "0.0372 mols OH"^(-)` added

`"HA"(aq) " "" "+" "" " "OH"^(-)(aq) -> "A"^(-)(aq) + "H"_2"O"(l)`

`"I"" ""0.27 mols"" "" "" ""0.0372 mols"" "" ""0 mols"" "" "-`
`"C"" " - x" "" "" "" "-"0.0372 mols"" "" "+x`
`"E"" "(0.27 - x) "mols"" ""0 mols"" "" "" "" "(x) "mols"" "-`

Therefore, you form the following base to acid ratio:

`(["A"^(-)]_(buffer))/(["HA"]_(buffer)) = (["A"^(-)]_(eq))/(["HA"]_i - ["A"^(-)]_(eq))`

`= (n_("A"^(-),eq))/(n_(HA,i) - n_(A^(-),eq)`

`= x/(0.27 - x)`

`(2)` Recognize what this ratio is.

This is the ratio in the Henderson-Hasselbalch equation:

`"pH" = "pKa" + log(x/(0.27 - x))`

Hence, the `"pH"` you measured of `3.80` gives you the `"pKa"` directly:

`3.80 - log(x/(0.27 - x)) = "pKa"`

But we know that `x = 0.0372`, so:

`color(blue)("pKa") = 3.80 - log(0.0372/(0.27 - 0.0372))`

`= color(blue)(4.60)`

Remember that we don't have to care about the total volume in a buffer. It all cancels out in the Henderson-Hasselbalch ratio anyway, because the solution is shared.