How do you calculate the equilibrium concentration of H_3O+ in a 0.20 M solution of oxalic acid?

Jul 17, 2018

["H"_3"O"^(+)] = 0.0803 color(white)(l) "mol" * "dm"^(-3)

Explanation:

The oxalic acid is a dicarboxylic acid. Each molecule of the acid partially disassociates to produce up to two hydronium ions when dissolved in water. Calculation of ${\text{H"_3"O}}^{+}$ for the $0.20 \textcolor{w h i t e}{l} \text{M}$ solution requires knowledge of both first and the second acid disassociation constants of the oxalic acid.

• $p {K}_{a} = 1.27$ for the partial disassociation of ${\text{C"_2"H"_2"O}}_{4}$
• $p {K}_{b} = 4.27$ for the partial disassociation of ${\text{C"_2"H" "O}}_{4}^{-}$

Construct a rice table (also in $\text{M}$) for each of the process. Let the final increase in hydronium concentration due to the disassociation of ${\text{C"_2"H"_2"O}}_{4}$ be $x$.

R "C"_2"H"_2"O"_4 (aq) + "H"_2"O"(l) rightleftharpoons "C"_2"H""O"_4^(-)(aq) + "H"_3"O"^(+)(aq)
I color(white)(I)0.20 color(white)(."O"_4 (aq) + "H"_2"O"(l) rightleftharpoons) 0 color(white)("."_2"H""O"_4^(-)(aq) + ) 0
C -x color(white)("l"_2"O"_4 (aq) + "H"_2"O"(l)rightleftharpoons) +x color(white)("O"_4^(-)(aq) + ) +x
E 0.20 - x color(white)((aq) + "H"_2"O"(l) rightleftharpoons) color(grey)(cancel(x)) color(white)("."_2"H""O"_4^(-)(aq) + ) color(grey)(cancel(x))

The disassociation of ${\text{C"_2"H"_2"O}}_{4}$ accounts for all ${\text{H"_3"O}}^{+}$ and ${\text{C"_2"H" "O}}_{4}^{-}$ in the system before the onset of the second process. Concentrations for these two species in the initial conditions of the second RICE table for the disassociation of ${\text{C"_2"H" "O}}_{4}^{-}$ are therefore identical to those in the final condition of the first process. Let the increase in hydronium concentration due to the disassociation of ${\text{C"_2"H" "O}}_{4}^{-}$ be $y$.

R "C"_2"H""O"_4^(-) (aq) + "H"_2"O"(l) rightleftharpoons "C"_2"O"_4^(2-)(aq) + "H"_3"O"^(+)(aq)
I color(white)(I)x color(white)("c"_2"H""O"_4 (aq) + "H"_2"O"(l) rightleftharpoons) 0 color(white)("."_2"O"_4^(2-)(aq) + ) x
C -y color(white)("l"_2"O"_4 (aq) + "H"_2"O"(l)rightleftharpoons) +y color(white)("."_4^(2-)(aq) + ) +y
E  color(purple)(x - y) color(white)("C"_4 (aq) + "H"_2"O"(l) rightleftharpoons) y color(white)("."_2"l""O"_4^(-)(aq) + ) color(purple)(x + y)

The solution to the two RICE table shall satisfy the equilibrium condition for both processes simultaneously. However, both equations are supposed to take the equilibrium concentration found in the second and the last table in case the concentration of a species (e.g., ${\text{H"_3"O}}^{+}$ is changed in both processes. That is:

• $\frac{\left(\textcolor{p u r p \le}{x - y}\right) \cdot \left(\textcolor{p u r p \le}{x + y}\right)}{0.20 - x} = {10}^{- 1.27}$ for the first process and
• $\frac{\left(\textcolor{p u r p \le}{x + y}\right) \cdot \left(y\right)}{\textcolor{p u r p \le}{x - y}} = {10}^{- 4.27}$ for the second process.

Solving the quadratic system (preferably with a Computer Algebra System) yields a collection of sets of values for $x$ and $y$. Both $x$ and $y$ are supposed to be greater than zero given that concentrations of species such as ${\text{H"_3"O}}^{+}$ and ${\text{C"_2"O}}_{4}^{-}$ always take non-negative values. Screening the result with the condition $x \ge 0$, $y \ge 0$ help narrows down to a single possible outcome:

• $x = 8.02 \times {10}^{- 2}$
• $y = 5.36 \times {10}^{- 4}$

The second RICE table gives the equilibrium concentration for ${\text{H"_3"O}}^{+}$:

["H"_3"O"^(+)] = color(purple)(x + y) ~~ 0.0803 color(white)(l) "mol" * "dm"^(-3)