The pH of rain water is approximately:

Pure water at $$25^{\circ}\text{C}$$ is neutral, so its hydrogen-ion concentration is $$[ \text{H}^{+} ] = 1.0 \times 10^{-7}\,\text{mol L}^{-1}$$ and the corresponding pH is

$$\text{pH} = -\log\!\bigl([ \text{H}^{+} ]\bigr) = -\log\!\bigl(1.0 \times 10^{-7}\bigr) = 7.0.$$

However, rain water is not absolutely pure. Atmospheric carbon dioxide dissolves in the falling drops and makes the water very slightly acidic. We start by writing the chemical steps that introduce extra $$\text{H}^{+}$$ ions.

The first step is dissolution of carbon dioxide, described by Henry’s law. The law states

$$C = k_{\mathrm H}\,P,$$

where $$C$$ is the concentration (in $$\text{mol L}^{-1}$$) of dissolved gas, $$P$$ is its partial pressure (in atmospheres) and $$k_{\mathrm H}$$ is the Henry’s-law constant. For carbon dioxide in water at $$25^{\circ}\text{C}$$ we use $$k_{\mathrm H} = 3.3 \times 10^{-2}\,\text{mol L}^{-1}\,\text{atm}^{-1}$$ and the atmospheric partial pressure $$P_{\mathrm{CO_2}} = 0.00040\,\text{atm}.$$

Substituting these numbers,

$$ C_{\mathrm{CO_2}} = 3.3 \times 10^{-2}\,\text{mol L}^{-1}\,\text{atm}^{-1} \times 0.00040\,\text{atm} = 1.32 \times 10^{-5}\,\text{mol L}^{-1}. $$

Once dissolved, the gas is hydrated:

$$\mathrm{CO_2(aq)} + \mathrm{H_2O} \;\rightleftharpoons\; \mathrm{H_2CO_3(aq)}.$$

For the pH calculation we treat the whole dissolved amount as carbonic acid $$\mathrm{H_2CO_3}$$ with the initial concentration

$$[\mathrm{H_2CO_3}]_0 = 1.32 \times 10^{-5}\,\text{mol L}^{-1}.$$

Carbonic acid is a weak diprotic acid, but its first dissociation is the principal source of $$\text{H}^{+}$$ ions:

$$\mathrm{H_2CO_3} \;\rightleftharpoons\; \mathrm{H^{+}} + \mathrm{HCO_3^{-}}.$$

The first dissociation constant is

$$K_a = 4.3 \times 10^{-7}.$$

For a weak acid with initial concentration $$C$$ and degree of dissociation $$\alpha$$, equilibrium gives

$$K_a = \dfrac{C\alpha^2}{1 - \alpha}\;.$$

Because the acid is very weak, $$\alpha \ll 1$$, so $$1 - \alpha \approx 1,$$ and the expression reduces to

$$K_a \approx C\alpha^{2}.$$

Solving for $$\alpha$$:

$$\alpha \approx \sqrt{\dfrac{K_a}{C}}.$$

Substituting the numerical values,

$$ \alpha \approx \sqrt{\dfrac{4.3 \times 10^{-7}}{1.32 \times 10^{-5}}} = \sqrt{3.26 \times 10^{-2}} = 1.81 \times 10^{-1}. $$

The hydrogen-ion concentration produced is

$$ [\text{H}^{+}]_{\text{acid}} = C\alpha = 1.32 \times 10^{-5}\,\text{mol L}^{-1} \times 1.81 \times 10^{-1} = 2.39 \times 10^{-6}\,\text{mol L}^{-1}. $$

The total $$[\text{H}^{+}]$$ in rain water combines this amount with the neutral-water value $$1.0 \times 10^{-7}\,\text{mol L}^{-1}$$, but the acid contribution is much larger, so

$$ [\text{H}^{+}]_{\text{total}} \approx 2.39 \times 10^{-6}\,\text{mol L}^{-1}. $$

The pH is therefore

$$ \text{pH} = -\log\!\bigl(2.39 \times 10^{-6}\bigr) = -\Bigl(\log 2.39 + \log 10^{-6}\Bigr) = -\bigl(0.378 + (-6)\bigr) = 5.62 \;(\text{approximately}). $$

This theoretical value, $$\text{pH} \approx 5.6$$, matches the experimentally observed average pH of natural rain water.

Hence, the correct answer is Option A.