Use three cycles of approximation by the method of Box$\mathbf{10}\mathbf{-}\mathbf{2}$to calculate$\left[H^{+}\right]\mathbf{,}\left[H_{2}A\right]\mathbf{,}\left[{\mathrm{HA}}^{-}\right]\mathbf{,}$and$\left[A^2\right]$inrole="math" localid="1654933890973" $\mathbf{0}\mathbf{.}\mathbf{00100}\mathbf{M}$monosodium oxalate, NaHA.

Monosodium oxalate is a dicarboxylic acid that may be found in a variety of plants and vegetables. Glyoxylic acid or ascorbic acid metabolism produces it in the body.

Consider the following for ${\mathrm{H}}_2\mathrm{A}$:

$$\begin{gathered} {\mathrm{K}}_1=5.62\times {10}^{-2} \\ {\mathrm{K}}_2=5.42\times {10}^{-5} \end{gathered}$$

Next make the first approximation that localid="1654934565542" $\left[{\mathrm{HA}}^{-}\right]=0.001\mathrm{M}$so the following would be:

$$\begin{gathered} {\left[{\mathrm{H}}^{+}\right]}_1=\sqrt{\left({\mathrm{K}}_1{\mathrm{K}}_2\left(0.001\right)+{\mathrm{K}}_1{\mathrm{K}}_{\mathrm{w}}\right)/\left({\mathrm{K}}_1+0.001\right)} \\ {\left[{\mathrm{H}}^{+}\right]}_1=2.31\times {10}^{-4}\mathrm{M} \end{gathered}$$

Where the pH would be:

$$\begin{gathered} \mathrm{pH}=-\log {\left[{\mathrm{H}}^{+}\right]}_1 \\ \mathrm{pH}=-\log \left(2.31\times {10}^{-4}\right) \\ =3.64 \end{gathered}$$

Then calculate the rest of the concentrations:

$$\begin{gathered} {\left[{\mathrm{H}}_2\mathrm{A}\right]}_1=\frac{{\left[{\mathrm{H}}^{+}\right]}_1\left[{\mathrm{HA}}^{-}\right]}{{\mathrm{K}}_1} \\ =4.1\times {10}^{-6}\mathrm{M}{\left[{\mathrm{A}}^{2-}\right]}_1 \\ =\frac{{\mathrm{K}}_2\left[{\mathrm{HA}}^{-}\right]}{{\left[{\mathrm{H}}^{+}\right]}_1} \end{gathered}$$

Next make the second approximation that

$$\begin{gathered} {\left[{\mathrm{HA}}^{-}\right]}_2=0.001-{\left[{\mathrm{A}}^{2-}\right]}_1-{\left[{\mathrm{H}}_2\mathrm{A}\right]}_1 \\ =0.000761\mathrm{M} \end{gathered}$$

so the following would be:

$$\begin{gathered} {\left[{\mathrm{H}}^{+}\right]}_2=\sqrt{\left({\mathrm{K}}_1{\mathrm{K}}_2\left(0.00761\right)+{\mathrm{K}}_1{\mathrm{K}}_{\mathrm{w}}\right)/\left({\mathrm{K}}_1+0.00761\right)} \\ {\left[{\mathrm{H}}^{+}\right]}_2=2.02\times {10}^{-4}\mathrm{M} \end{gathered}$$

Where the pH would be:

$$\begin{gathered} \mathrm{pH}=-\log {\left[{\mathrm{H}}^{+}\right]}_2 \\ \mathrm{pH}=-\log \left(2.02\times {10}^{-4}\right) \\ =3.70 \end{gathered}$$

Then calculate the rest of the concentrations:

$$\begin{gathered} {\left[{\mathrm{H}}_2\mathrm{A}\right]}_2=\frac{{\left[{\mathrm{H}}^{+}\right]}_2{\left[{\mathrm{HA}}^{-}\right]}_2}{{\mathrm{K}}_1} \\ =2.73\times {10}^{-6}\mathrm{M}{\left[{\mathrm{A}}^{2-}\right]}_2 \\ =\frac{{\mathrm{K}}_2\left[{\mathrm{HA}}^{-}\right]}{{\left[{\mathrm{H}}^{+}\right]}_2} \\ =2.04\times {10}^{-4}\mathrm{M} \end{gathered}$$

Next make the third approximation that

$$\begin{gathered} {\left[{\mathrm{HA}}^{-}\right]}_3=0.001-{\left[{\mathrm{A}}^{2-}\right]}_2-{\left[{\mathrm{H}}_2\mathrm{A}\right]}_2 \\ =0.000793\mathrm{M} \end{gathered}$$

so the following would be:

$$\begin{gathered} {\left[{\mathrm{H}}^{+}\right]}_3=\sqrt{\left({\mathrm{K}}_1{\mathrm{K}}_2\left(0.00793\right)+{\mathrm{K}}_1{\mathrm{K}}_{\mathrm{w}}\right)/\left({\mathrm{K}}_1+0.00793\right)} \\ {\left[{\mathrm{H}}^{+}\right]}_3=2.06\times {10}^{-4}\mathrm{M} \end{gathered}$$

Where the pH would be:

$$\begin{gathered} \mathrm{pH}=-\log {\left[{\mathrm{H}}^{+}\right]}_3 \\ \mathrm{pH}=-\log \left(2.06\times {10}^{-4}\right) \\ =3.69 \end{gathered}$$

Then calculate the rest of the concentrations:

$$\begin{gathered} {\left[{\mathrm{H}}_2\mathrm{A}\right]}_3=\frac{{\left[{\mathrm{H}}^{+}\right]}_3\left[{\mathrm{HA}}^{-}\right]3}{{\mathrm{K}}_1} \\ =2.90\times {10}^{-6}\mathrm{M} \\ {\left[{\mathrm{A}}^{2-}\right]}_3=\frac{{\mathrm{K}}_2{\left[{\mathrm{HA}}^{-}\right]}_3}{{\left[{\mathrm{H}}^{+}\right]}_3} \\ =2.09\times {10}^{-4}\mathrm{M} \end{gathered}$$