The average concentration of `SO_(2)` in the atmosphere over a city on a cetrain day is 10 ppm, when the average temperature is 298 K. Given that the solubility of `SO_(2)` in water at 298 K is `1.3653` mol `litre^(-1)` and the `pK_(a)` of `H_(2)SO_(3)` is `1.92`, estimate the pH of rain on that day.

Explanation

To find the pH of rain on a certain day given the concentration of `SO2` in the atmosphere, we can follow these steps:

### Step 1: Convert the concentration of SO2 from ppm to mol/L
The concentration of `SO2` is given as 10 ppm. To convert this to molarity (mol/L), we use the following formula:

\[ \text{Concentration (mol/L)} = \frac{\text{ppm}}{10^6} \times \frac{1 \text{ g}}{\text{molar mass of SO2 (g/mol)}} \]

The molar mass of `SO2` is approximately 64.07 g/mol. Thus,

\[ \text{Concentration of SO2} = \frac{10 \text{ ppm}}{10^6} \times \frac{1 \text{ g}}{64.07 \text{ g/mol}} \approx 1.56 \times 10^{-7} \text{ mol/L} \]

### Step 2: Determine the concentration of H2SO3
Given that the solubility of `SO2` in water at 298 K is 1.3653 mol/L, we can assume that the concentration of `H2SO3` formed from the dissolved `SO2` will also be approximately equal to the solubility of `SO2`.

\[ [\text{H2SO3}] \approx 1.3653 \text{ mol/L} \]

### Step 3: Set up the equilibrium expression
The dissociation of `H2SO3` can be represented as follows:

\[ \text{H2SO3} \rightleftharpoons \text{H}^+ + \text{HSO3}^- \]

Let \( C \) be the initial concentration of `H2SO3`, and let \( A \) be the amount that dissociates. At equilibrium, we have:

- Concentration of `H2SO3` = \( C - A \)
- Concentration of `H^+` = \( A \)
- Concentration of `HSO3^-` = \( A \)

### Step 4: Use the Ka expression
The dissociation constant \( K_a \) for `H2SO3` can be calculated from the given \( pK_a \):

\[ K_a = 10^{-pK_a} = 10^{-1.92} \approx 0.0120 \]

The equilibrium expression is:

\[ K_a = \frac{[\text{H}^+][\text{HSO3}^-]}{[\text{H2SO3}]} = \frac{A \cdot A}{C - A} \approx \frac{A^2}{C} \]

Since \( A \) is very small compared to \( C \), we can neglect \( A \) in the denominator.

### Step 5: Substitute values into the equation
Substituting the values we have:

\[ 0.0120 = \frac{A^2}{1.3653} \]

### Step 6: Solve for A
Rearranging gives:

\[ A^2 = 0.0120 \times 1.3653 \]
\[ A^2 = 0.0164 \]
\[ A = \sqrt{0.0164} \approx 0.128 \]

### Step 7: Calculate pH
The concentration of \( H^+ \) ions is equal to \( A \):

\[ [\text{H}^+] \approx 0.128 \text{ mol/L} \]

Now, we can calculate the pH:

\[ \text{pH} = -\log(0.128) \approx 0.89 \]

### Final Answer
The estimated pH of rain on that day is approximately **0.89**.

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