91Ó°ÊÓ

A diprotic acid dissociates in water in two stages as shown below:

${\mathrm{H}}_2\mathrm{X}\left(\mathrm{aq}\right)↔{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{HX}}^{-}\left(\mathrm{aq}\right)$

${\mathrm{HX}}^{-}\left(\mathrm{aq}\right)↔{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{X}}^{2-}\left(\mathrm{aq}\right)$

In this task we have a diprotic acid with = 8.00, we need to determine the following:

a) the pH when$\left[H_{2}A\right]=\left[H{A}^{-}\right]$

b) the pH when $\left[{\mathrm{HA}}^{-}\right]=\left[{\mathrm{A}}^{2-}\right]$

c) the principal species at pH = 2.00
d) the principal species at pH = 6.00

e) the principal species at pH = 10.00

First consider the reaction of dissociation:

$$\begin{gathered} {\mathrm{H}}_2\mathrm{X}\left(\mathrm{aq}\right)↔{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{HX}}^{-}\left(\mathrm{aq}\right) \\ {\mathrm{HX}}^{-}\left(\mathrm{aq}\right)↔{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{X}}^{2-}\left(\mathrm{aq}\right) \end{gathered}$$

(a)use the reaction (1) and ${\mathrm{pk}}_1$= 4.00 via the Henderson-Hasselbach equation equation;

$$\begin{gathered} \mathrm{pH}={\mathrm{pk}}_1+\log \left(\left[{\mathrm{HA}}^{-}\right]/\left[{\mathrm{H}}_2\mathrm{A}\right]\right) \\ \mathrm{pH}=4.00+\log \left(1\right) \\ \mathrm{pH}=4.00+0 \\ \mathrm{pH}=4.00 \end{gathered}$$

(b)use the reaction (1) and ${\mathrm{pk}}_2$= 8.00 via the Henderson-Hasselbach equation;

$$\begin{gathered} \mathrm{pH}={\mathrm{pk}}_2+\log \left(\left[{\mathrm{HA}}^{2-}\right]/\left[{\mathrm{HA}}^{-}\right]\right) \\ \mathrm{pH}=8.00+\log \left(1\right) \\ \mathrm{pH}=8.00+0 \\ \mathrm{pH}=8.00 \end{gathered}$$

(c) the principal species at pH=2.00 is determined as;

$$\begin{gathered} pH=p{k}_1+log\left(\right[H{A}^{-}]/[H_{2}A\left]\right) \\ 2.00=4.00+log\left(\right[H{A}^{-}]/[H_{2}A\left]\right) \\ -2.00=log\left(\right[H{A}^{-}]/[H_{2}A\left]\right) \\ {10}^{-2.00}=\left[H{A}^{-}\right]/\left[H_{2}A\right] \\ 0.01=\left[H{A}^{-}\right]/\left[H_{2}A\right] \end{gathered}$$

Considering the quotient $H_{2}A$is the principal species

(d) the principal species at pH=6.00 is determined as;

$$\begin{gathered} \mathrm{pH}={\mathrm{pk}}_1+\log \left(\right[{\mathrm{HA}}^{-}]/[{\mathrm{H}}_2\mathrm{A}\left]\right) \\ 6.00=4.00+\log \left(\right[{\mathrm{HA}}^{-}]/[{\mathrm{H}}_2\mathrm{A}\left]\right) \\ 2.00=\log \left(\right[{\mathrm{HA}}^{-}]/[{\mathrm{H}}_2\mathrm{A}\left]\right) \\ {10}^{2.00}=\left[{\mathrm{HA}}^{-}\right]/\left[{\mathrm{H}}_2\mathrm{A}\right] \\ 100=\left[{\mathrm{HA}}^{-}\right]/\left[{\mathrm{H}}_2\mathrm{A}\right] \end{gathered}$$

Considering the quotient $H{A}^{-}$is the principal species

(e) the principal species at pH=10.00 is determined as;

$$\begin{gathered} \mathrm{pH}={\mathrm{pk}}_2+\log \left(\right[{\mathrm{A}}^{2-}]/[{\mathrm{HA}}^{-}\left]\right) \\ 10.00=8.00+\log \left(\right[{\mathrm{A}}^{2-}]/[{\mathrm{HA}}^{-}\left]\right) \\ 2.00=\log \left(\right[{\mathrm{HA}}^{-}]/[{\mathrm{HA}}^{-}\left]\right) \\ {10}^{2.00}=\left[{\mathrm{A}}^{2-}\right]/\left[{\mathrm{HA}}^{-}\right] \\ 100=\left[{\mathrm{A}}^{2-}\right]/\left[{\mathrm{HA}}^{-}\right] \end{gathered}$$

Considering the quotient $A^{2-}$is the principal species