Sulfuric acid, ${\mathrm{H}}_2{\mathrm{SO}}_4,\text{readily dissociates into}{\mathrm{H}}^{+}\text{and}{\mathrm{HSO}}_4^{-}{\mathrm{H}}^{+}\text{and}{\mathrm{HSO}}_4^{-}ions$

${\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{H}}^{+}+{\mathrm{HSO}}_4^{-}$

The hydrogen sulfate ion, in turn, can dissociate again:

${\mathrm{HSO}}_4^{-}⟷{\mathrm{H}}^{+}+{\mathrm{SO}}_4^{2-}$

The equilibrium constants for these reactions, in aqueous solutions at 298 K, are approximately 10 and 10*, respectively. (For dissociation of acids it is usually more convenient to look up K than $∆G°$. By the way, the negative base-10 logarithm of K for such a reaction is called pK, in analogy to pH. So for the first reaction pK = -2, while for the second reaction pK = 1.9.)

(a) Argue that the first reaction tends so strongly to the right that we might as well consider it to have gone to completion, in any solution that could possibly be considered dilute. At what pH values would a significant fraction of the sulfuric acid not be dissociated?

(b) In industrialized regions where lots of coal is burned, the concentration of sulfate in rainwater is typically 5 x 10 mol/kg. The sulfate can take any of the chemical forms mentioned above. Show that, at this concentration, the second reaction will also have gone essentially to completion, so all the sulfate is in the form of SOg. What is the pH of this rainwater?

(c) Explain why you can neglect dissociation of water into H* and OH in answering the previous question. (d) At what pH would dissolved sulfate be equally distributed between HSO and SO2-?

(a) We have two equations:

$$\begin{gathered} {\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{H}}^{+}+{\mathrm{HSO}}_4^{-}\left(1\right) \\ {\mathrm{HSO}}_4^{-}\leftrightarrow {\mathrm{H}}^{+}+{\mathrm{SO}}_4^{2-}\left(2\right) \end{gathered}$$

The equilibrium constant for the first reaction is:

role="math" localid="1647214128324" $$\begin{gathered} K=\frac{m_{{\mathrm{H}}^{+}}\times m_{{\mathrm{HSO}}_4^{-}}}{m_{{\mathrm{H}}_2{\mathrm{SO}}_4}} \\ \text{since}K\text{is large we can replace}{m}_{{\mathrm{H}}^{+}}\text{with}{m}_{{\mathrm{HSO}}_4^{-}}\text{, so:} \\ K=\frac{m_{{\mathrm{HSO}}_4^{-}}^2}{m_{{\mathrm{H}}_2{\mathrm{SO}}_4}} \end{gathered}$$

The equilibrium constant is $K=100$

$m_{{\mathrm{HSO}}_4^{-}}=10\sqrt{m_{{\mathrm{H}}_2{\mathrm{SO}}_4}}$

When the morality of H₂SO₄ equals 1, then the morality of HSO₄^-, is 10, which implies that the morality of H⁺ is 10. Then the pH of the solution is:

$$\begin{gathered} \mathrm{pH}=-{\log }_{10}\left(m_{{\mathrm{H}}^{+}}\right) \\ =-{\log }_{10}\left(10\right) \\ =1 \end{gathered}$$

This indicates that the solution is highly acidic; the first equation's left side is acidic, implying that the reaction will move to the right. As a result, all of the sulfuric acid will be separated.

(b)The equilibrium constant for first reaction is:

$K=\frac{M_{{\mathrm{H}}^{+}}\times M_{{\mathrm{SO}}_4^{2-}}}{M_{{\mathrm{HSO}}_4^{-}}}$

The equilibrium constant can be written in terms of the pK of the reaction, as:

$$\begin{gathered} \mathrm{pK}=-{\log }_{10}\left(K\right) \\ K={10}^{-\mathrm{pK}} \end{gathered}$$

Therefore,

${10}^{-\mathrm{pK}}=\frac{M_{{\mathrm{H}}^{+}}\times M_{{\mathrm{SO}}_4^{2-}}}{M_{{\mathrm{HSO}}_4^{-}}}$

Each molecule of sulfuric acid gives two molecules of HT, thus the molality of the HT is twice the molality of the sulfuric acid, therefore we can write:

$$\begin{gathered} {10}^{-\mathrm{pK}}=\frac{2M_{{\mathrm{SO}}_4^{2-}}\times M_{{\mathrm{SO}}_4^{2-}}}{M_{{\mathrm{HSO}}_4^{-}}} \\ {10}^{-\mathrm{pK}}=\frac{2M_{{\mathrm{SO}}_4^{2-}}^2}{M_{{\mathrm{HSO}}_4^{-}}} \\ {M}_{{\mathrm{HSO}}_4^{-}}=\frac{2M_{{\mathrm{SO}}_4^{2-}}^2}{{10}^{-\mathrm{pK}}} \end{gathered}$$

If the concentration of the sulfate in rain water is $M_{{\mathrm{SO}}_4^{2-}}=5.0\times {10}^{-5}\mathrm{mol}/\mathrm{kg}$, substitute this value

$$\begin{gathered} M_{{\mathrm{HSO}}_4^{-}}=\frac{2{\left(5.0\times {10}^{-5}\mathrm{mol}/\mathrm{kg}\right)}^2}{{10}^{-1.9}} \\ {M}_{{\mathrm{HSO}}_4^{-}}=3.9716\times {10}^{-7}\mathrm{mol}/\mathrm{kg} \end{gathered}$$

The concentration of $H^{+}$is

$M_{{\mathrm{H}}^{+}}=2M_{{\mathrm{SO}}_4^{-2}}=2\left(5.0\times {10}^{-5}\mathrm{mol}/\mathrm{kg}\right)=10\times {10}^{-5}\mathrm{mol}/\mathrm{kg}$

pH of rainwater is

$$\begin{gathered} \mathrm{pH}=-{\log }_{10}\left(M_{{\mathrm{H}}^{+}}\right) \\ =-{\log }_{10}\left(10\times {10}^{-5}\mathrm{mol}/\mathrm{kg}\right) \\ =4 \end{gathered}$$

(c)The concentrations of the OH^- and H⁺

$$\begin{gathered} M_{{\mathrm{H}}^{+}}·M_{{\mathrm{OH}}^{-}}={10}^{-14} \\ {M}_{{\mathrm{OH}}^{-}}=\frac{{10}^{-14}}{M_{{\mathrm{H}}^{+}}} \end{gathered}$$

Substitute the value of ${\mathrm{M}}_{{\mathrm{H}}^{+}}$

role="math" localid="1647244030613" $$\begin{gathered} M_{{\mathrm{OH}}^{-}}=\frac{{10}^{-14}}{10\times {10}^{-5}\mathrm{mol}/\mathrm{kg}} \\ =1.0\times {10}^{-10}\mathrm{mol}/\mathrm{kg} \end{gathered}$$

(d)When the dissolved sulfate is equally distrbuted between ${\mathrm{HSO}}_4^{-}\text{and}{\mathrm{SO}}_4^{-2}$we can set $M_{{\mathrm{SO}}_4^{2-}}$equals to $M_{{\mathrm{HSO}}_4^{-}}$

role="math" localid="1647244228696" $$\begin{gathered} {10}^{-\mathrm{pK}}=\frac{M_{{\mathrm{H}}^{+}}\times M_{{\mathrm{SO}}_4^{2-}}}{M_{{\mathrm{HSO}}_4^{-}}^{-}} \\ {M}_{{\mathrm{H}}^{+}}={10}^{-\mathrm{pK}} \\ {M}_{{\mathrm{H}}^{+}}={10}^{-1.9} \end{gathered}$$

Hence, pH of the solution is:

$$\begin{gathered} \mathrm{pH}=-{\log }_{10}\left(M_{{\mathrm{H}}^{+}}\right) \\ =-{\log }_{10}\left({10}^{-1.9}\mathrm{mol}/\mathrm{kg}\right) \\ \mathrm{pH}=1.9 \end{gathered}$$