Question #cbd61

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Answer 1 · 2015-01-30T00:08:45

The concentration of the dominant form of the acid at pH $= 5.10$ $= 5.10"$ is $2.09 mmol/L$ $"2.09 mmol/L"$ .

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

$H_{2} A + H_{2} O ⇌ H A^{-} + H_{3}^{+} O$ $H_2A + H_2O rightleftharpoons HA^(-) + H_3^(+)O$ , $\to {pKa}_{1} = 4.19$ $-> "pKa"_1 = 4.19$
$H A^{-} + H_{2} O ⇌ A^{2 -} + H_{3}^{+} O$ $HA^(-) + H_2O rightleftharpoons A^(2-) + H_3^(+)O$ , $\to {pKa}_{2} = 5.57$ $-> "pKa"_2 = 5.57$

Once again, the pH is bigger than ${pKa}_{1}$ $"pKa"_1$ and smaller than ${pKa}_{2}$ $"pKa"_2$ , which means that the dominant form of the acid will be $H A^{-}$ $HA^(-)$ . 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.

$[H_{2} A] + [H A^{-}] = 2.35$ $[H_2A] + [HA^(-)] = 2.35$ (1)
$p H = p K a_{1} + log (\frac{[H A^{-}]}{[H_{2} A]})$ $pH = pKa_1 + log(([HA^(-)])/([H_2A]))$ (2)

Solve equation (2) first.

$5.10 = 4.19 + log (\frac{[H A^{-}]}{[H_{2} A]}) \Rightarrow log (\frac{[H A^{-}]}{[H_{2} A]}) = 0.91$ $5.10 = 4.19 + log(([HA^(-)])/([H_2A])) => log(([HA^(-)])/([H_2A])) = 0.91$

$\frac{[H A^{-}]}{[H_{2} A]} = 8.13$ $([HA^(-)])/([H_2A]) = 8.13$

Since you're interested in determining the value of $[H A^{-}]$ $[HA^(-)]$ , plug $[H_{2} A] = \frac{[H A^{-}]}{8.13}$ $[H_2A] = ([HA^(-)])/8.13$ into equation (1)

$\frac{H A^{-}}{8.13} + [H A^{-}] = 2.35 \Rightarrow [H A^{-}] + 8.13 \cdot [H A^{-}] = 19.1$ $[HA^(-)]/8.13 + [HA^(-)] = 2.35 => [HA^(-)] + 8.13 * [HA^(-)] = 19.1$

$[H A^{-}] = \frac{19.1}{9.13} = 2.09 mmol/L$ $[HA^(-)] = 19.1/9.13 = "2.09 mmol/L"$

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