# 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?

Apr 26, 2017

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...

$\left(1\right)$ Use an ICE Table to organize your work.

${\text{0.0120 L" xx "3.10 M NaOH" = "0.0372 mols OH}}^{-}$ added

$\text{HA"(aq) " "" "+" "" " "OH"^(-)(aq) -> "A"^(-)(aq) + "H"_2"O} \left(l\right)$

$\text{I"" ""0.27 mols"" "" "" ""0.0372 mols"" "" ""0 mols"" "" } -$
$\text{C"" " - x" "" "" "" "-"0.0372 mols"" "" } + x$
$\text{E"" "(0.27 - x) "mols"" ""0 mols"" "" "" "" "(x) "mols"" } -$

Therefore, you form the following base to acid ratio:

$\left({\left[{\text{A"^(-)]_(buffer))/(["HA"]_(buffer)) = (["A"^(-)]_(eq))/(["HA"]_i - ["A}}^{-}\right]}_{e q}\right)$

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

$= \frac{x}{0.27 - x}$

$\left(2\right)$ Recognize what this ratio is.

This is the ratio in the Henderson-Hasselbalch equation:

$\text{pH" = "pKa} + \log \left(\frac{x}{0.27 - x}\right)$

Hence, the $\text{pH}$ you measured of $3.80$ gives you the $\text{pKa}$ directly:

$3.80 - \log \left(\frac{x}{0.27 - x}\right) = \text{pKa}$

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

$\textcolor{b l u e}{\text{pKa}} = 3.80 - \log \left(\frac{0.0372}{0.27 - 0.0372}\right)$

$= \textcolor{b l u e}{4.60}$

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.