# Question cbd61

Jan 30, 2015

The concentration of the dominant form of the acid at pH = 5.10"# is $\text{2.09 mmol/L}$.

This time you're dealing with a diprotic acid. The general forms of the equations look like this:

${H}_{2} A + {H}_{2} O r i g h t \le f t h a r p \infty n s H {A}^{-} + {H}_{3}^{+} O$, $\to {\text{pKa}}_{1} = 4.19$
$H {A}^{-} + {H}_{2} O r i g h t \le f t h a r p \infty n s {A}^{2 -} + {H}_{3}^{+} O$, $\to {\text{pKa}}_{2} = 5.57$

Once again, the pH is bigger than ${\text{pKa}}_{1}$ and smaller than ${\text{pKa}}_{2}$, which means that the dominant form of the acid will be $H {A}^{-}$. You need to set up two equations with two unknows - the concentrations of the acid forms. One will the Henderson - Hasselbalch equation, and the other the total given concentration for both forms of the acid.

$\left[{H}_{2} A\right] + \left[H {A}^{-}\right] = 2.35$ (1)
$p H = p K {a}_{1} + \log \left(\frac{\left[H {A}^{-}\right]}{\left[{H}_{2} A\right]}\right)$ (2)

Solve equation (2) first.

$5.10 = 4.19 + \log \left(\frac{\left[H {A}^{-}\right]}{\left[{H}_{2} A\right]}\right) \implies \log \left(\frac{\left[H {A}^{-}\right]}{\left[{H}_{2} A\right]}\right) = 0.91$

$\frac{\left[H {A}^{-}\right]}{\left[{H}_{2} A\right]} = 8.13$

Since you're interested in determining the value of $\left[H {A}^{-}\right]$, plug $\left[{H}_{2} A\right] = \frac{\left[H {A}^{-}\right]}{8.13}$ into equation (1)

$\frac{H {A}^{-}}{8.13} + \left[H {A}^{-}\right] = 2.35 \implies \left[H {A}^{-}\right] + 8.13 \cdot \left[H {A}^{-}\right] = 19.1$

$\left[H {A}^{-}\right] = \frac{19.1}{9.13} = \text{2.09 mmol/L}$