Question

In this problem, we calculate the pH of the intermediate form of a diprotic acid, taking activities into account.

(a) Including activity coefficients, derive Equation 10 - 11 for potassium hydrogen phthalate $\mathbf{(}{\mathbf{K}}^{\mathbf{+}}{\mathbf{HP}}^{\mathbf{-}}$ in the example following Equation 10 - 12 ).

(b) Calculate the pH of $\mathbf{0}\mathbf{.}\mathbf{050}\mathbf{MKHP}$ , using the results in part (a). Assume that the sizes of both ${\mathbf{HP}}^{\mathbf{-}}$ and ${\mathbf{P}}^{\mathbf{2}\mathbf{-}}$ are 600pm . For comparison, Equation 10 - 11 gives $\mathbf{pH}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{18}$.

Short answer

a. Equation 10 -11 is derived as follows:

First, write the following:

Charge balance: $\left[{\mathrm{K}}^{+}\right]+\left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{OH}}^{-}\right]+\left[{\mathrm{HP}}^{-}\right]+2\left[{\mathrm{P}}^{2-}\right]$

Mass balance: $\left[{\mathrm{K}}^{+}\right]=\left[{\mathrm{H}}_2\mathrm{P}\right]+\left[{\mathrm{HP}}^{-}\right]+\left[{\mathrm{P}}^{2-}\right]$

Equate the mass balance value of $\left[{\mathrm{K}}^{+}\right]$with the charge balance equation which would give us the following:

$\left[{\mathrm{H}}_2\mathrm{P}\right]+\left[{\mathrm{HP}}^{-}\right]+\left[{\mathrm{P}}^{2-}\right]+\left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{OH}}^{-}\right]+\left[{\mathrm{HP}}^{-}\right]+2\left[{\mathrm{P}}^{2-}\right]$

$\left[{\mathrm{H}}_2\mathrm{P}\right]+\left[{\mathrm{H}}^{+}\right]-\left[{\mathrm{OH}}^{-}\right]-\left[{\mathrm{P}}^{2-}\right]=0$

Next, create the equilibrium constants equation:

$$\begin{gathered} K_1=\frac{\left[H^{+}\right]\gamma \left(H^{+}\right)\left[H{P}^{-}\right]\gamma \left(H{P}^{-}\right)}{\left[H_2{P}^{-}\right]\gamma \left(H_2{P}^{-}\right)} \\ {K}_2=\frac{\left[H^{+}\right]\gamma \left(H^{+}\right)\left[P^{2-}\right]\gamma \left(P^{2-}\right)}{\left[H{P}^{-}\right]\gamma \left(H{P}^{-}\right)} \\ {K}_w=\left[H^{+}\right]\gamma \left(H^{+}\right)\left[O{H}^{-}\right]\gamma \left(O{H}^{-}\right) \end{gathered}$$

Make the following substitutions in equations for equilibrium constants to get the following result: $\frac{\left[H^{+}\right]\gamma \left(H^{+}\right)\left[H{P}^{-}\right]\gamma \left(H{P}^{-}\right)}{\left(K_1\gamma \right(H_{2}P)}+\left[H^{+}\right]-\frac{K_2\left[H{P}^{-}\right]\gamma \left(H{P}^{-}\right)}{\left[H^{+}\right]\gamma \left(H^{+}\right)\gamma \left(P^{2-}\right)}-\frac{K_w}{\left[H^{+}\right]\gamma \left(H^{+}\right)\gamma \left(O{H}^{-}\right)}=0$

b) The pH of 0.50MKHP is 4.03 .

Step-by-step solution

Monosodium oxalate

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

Derive Equation 10 - 11 for potassium hydrogen phthalate

a)

First, write the following:

Charge balance:$\left[{\mathrm{K}}^{+}\right]+\left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{OH}}^{-}\right]+\left[{\mathrm{HP}}^{-}\right]+2\left[{\mathrm{P}}^{2-}\right]$

Mass balance: $\left[K^{+}\right]=\left[H_{2}P\right]+\left[H{P}^{-}\right]+\left[P^{2-}\right]$

Equate the mass balance value of$\left[K^{+}\right]$ with the charge balance equation which would give us the following:

$\left[H_{2}P\right]+\left[H{P}^{-}\right]+\left[P^{2-}\right]+\left[H^{+}\right]=\left[O{H}^{-}\right]+\left[H{P}^{-}\right]+2\left[P^{2-}\right]$

$\left[H_{2}P\right]+\left[H^{+}\right]-\left[O{H}^{-}\right]-\left[P^{2-}\right]=0$

Next, create the equilibrium constants equation:

$$\begin{gathered} K_1=\frac{\left[H^{+}\right]\gamma \left(H^{+}\right)\left[H{P}^{-}\right]\gamma \left(H{P}^{-}\right)}{\left[H_2{P}^{-}\right]\gamma \left(H_2{P}^{-}\right)} \\ {K}_2=\frac{\left[H^{+}\right]\gamma \left(H^{+}\right)\left[P^{2-}\right]\gamma \left(P^{2-}\right)}{\left[H{P}^{-}\right]\gamma \left(H{P}^{-}\right)} \\ {K}_w=\left[H^{+}\right]\gamma \left(H^{+}\right)\left[O{H}^{-}\right]\gamma \left(O{H}^{-}\right) \end{gathered}$$

Make the following substitutions in equations for equilibrium constants to get the following result:$\frac{\left[H^{+}\right]\gamma \left(H^{+}\right)\left[H{P}^{-}\right]\gamma \left(H{P}^{-}\right)}{\left(K_1\gamma \right(H_{2}P)}+\left[H^{+}\right]-\frac{K_2\left[H{P}^{-}\right]\gamma \left(H{P}^{-}\right)}{\left[H^{+}\right]\gamma \left(H^{+}\right)\gamma \left(P^{2-}\right)}-\frac{K_w}{\left[H^{+}\right]\gamma \left(H^{+}\right)\gamma \left(O{H}^{-}\right)}=0$

The pH of 0.050MKHP

b)

Calculate the pH of 0.050MKHP using the results in part (a) as follows:

Assume that the size of both ${\mathrm{HP}}^{-}$ and ${\mathrm{P}}^{2-}$is 600pm each.

Consider the following values:

$\mathrm{\mu }\left(\mathrm{KHP}\right)=0.05\mathrm{M}$; So, $\left[\mathrm{KHP}\right]=0.05\mathrm{M}$

And,

$$\begin{gathered} \gamma \left(H{P}^{-}\right)=0.835 \\ \gamma \left(P^{2-}\right)=0.485 \\ \gamma \left(H_{2}P\right)=1.00 \\ \gamma \left(H^{+}\right)=0.860 \\ \gamma \left(O{H}^{-}\right)=0.810 \end{gathered}$$

Next, use the above values in the following equation from a) in order to get the concentration of.

$\frac{\left[H^{+}\right]\gamma \left(H^{+}\right)\left[H{P}^{-}\right]\gamma \left(H{P}^{-}\right)}{\left(K_1\gamma \right(H_{2}P)}+\left[H^{+}\right]-\frac{K_2\left[H{P}^{-}\right]\gamma \left(H{P}^{-}\right)}{\left[H^{+}\right]\gamma \left(H^{+}\right)\gamma \left(P^{2-}\right)}-\frac{K_w}{\left[H^{+}\right]\gamma \left(H^{+}\right)\gamma \left(O{H}^{-}\right)}=0$

It gives us the concentration of

Then, calculate the pH as follows:

$\begin{array}{l}\mathrm{pH}=\log \left[{\mathrm{H}}^{+}\right]\\ \mathrm{pH}=-\log \left(1.09\times {10}^{-4}\right)\\ =403\end{array}$