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

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

#H_2A + H_2O rightleftharpoons HA^(-) + H_3^(+)O#, #-> "pKa"_1 = 4.19#

#HA^(-) + H_2O rightleftharpoons A^(2-) + H_3^(+)O#, #-> "pKa"_2 = 5.57#

Once again, the pH is bigger than #"pKa"_1# and smaller than #"pKa"_2#, which means that the dominant form of the acid will be #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_2A] + [HA^(-)] = 2.35# **(1)**

#pH = pKa_1 + log(([HA^(-)])/([H_2A]))# **(2)**

Solve equation **(2)** first.

#5.10 = 4.19 + log(([HA^(-)])/([H_2A])) => log(([HA^(-)])/([H_2A])) = 0.91#

#([HA^(-)])/([H_2A]) = 8.13#

Since you're interested in determining the value of #[HA^(-)]#, plug #[H_2A] = ([HA^(-)])/8.13# into equation **(1)**

#[HA^(-)]/8.13 + [HA^(-)] = 2.35 => [HA^(-)] + 8.13 * [HA^(-)] = 19.1#

#[HA^(-)] = 19.1/9.13 = "2.09 mmol/L"#