Acid rain takes place dur to combination of acidic oxides with water and it is an environmental concern all over the world. Assuming rain water is uncontaminated with `HNO_(3)` or `H_(2)SO_(4)` and is in equilibrium with `1.25 xx 10^(-4) atm CO_(2)`. The Henry's law constant `(K_(H))` is `1.25 xx 10^(6) "torr". K_(a_(1))` of `H_(2)CO_(3) = 4.3 xx 10^(-7)`
What is the `pH` of neutral rain water ?

Options

A. `5.64`
B. `7.00`
C. `5.85`
D. `7.40`

Explanation

To find the pH of neutral rainwater in equilibrium with carbon dioxide, we can follow these steps:

### Step 1: Calculate the mole fraction of CO₂ in water
Using Henry's Law, we can relate the partial pressure of CO₂ to its concentration in water. The formula is:

\[ P_{CO_2} = K_H \cdot C_{CO_2} \]

Where:
- \(P_{CO_2} = 1.25 \times 10^{-4} \, \text{atm}\)
- \(K_H = 1.25 \times 10^6 \, \text{torr}\)

First, convert the pressure from atm to torr:
\[ 1 \, \text{atm} = 760 \, \text{torr} \implies P_{CO_2} = 1.25 \times 10^{-4} \, \text{atm} \times 760 \, \text{torr/atm} = 9.5 \times 10^{-2} \, \text{torr} \]

Now, we can rearrange Henry's Law to find the concentration of CO₂ in water:
\[ C_{CO_2} = \frac{P_{CO_2}}{K_H} \]

Substituting the values:
\[ C_{CO_2} = \frac{9.5 \times 10^{-2} \, \text{torr}}{1.25 \times 10^6 \, \text{torr}} \approx 7.6 \times 10^{-8} \, \text{mol/L} \]

### Step 2: Calculate the concentration of carbonic acid (H₂CO₃)
Since 1 mole of CO₂ forms 1 mole of H₂CO₃, the concentration of H₂CO₃ is equal to the concentration of CO₂:
\[ C_{H_2CO_3} = 7.6 \times 10^{-8} \, \text{mol/L} \]

### Step 3: Calculate the concentration of H⁺ ions
The first dissociation constant \(K_{a1}\) for carbonic acid is given as:
\[ K_{a1} = 4.3 \times 10^{-7} \]

Using the formula for the dissociation of H₂CO₃:
\[ H_2CO_3 \rightleftharpoons H^+ + HCO_3^- \]

The expression for \(K_{a1}\) is:
\[ K_{a1} = \frac{[H^+][HCO_3^-]}{[H_2CO_3]} \]

Assuming that the concentration of H⁺ ions produced is \(x\):
\[ K_{a1} = \frac{x^2}{C_{H_2CO_3} - x} \approx \frac{x^2}{C_{H_2CO_3}} \quad \text{(since } x \text{ is small compared to } C_{H_2CO_3}\text{)} \]

Substituting the values:
\[ 4.3 \times 10^{-7} = \frac{x^2}{7.6 \times 10^{-8}} \]

Solving for \(x\):
\[ x^2 = 4.3 \times 10^{-7} \times 7.6 \times 10^{-8} \]
\[ x^2 = 3.268 \times 10^{-14} \]
\[ x = \sqrt{3.268 \times 10^{-14}} \approx 1.8 \times 10^{-7} \, \text{mol/L} \]

### Step 4: Calculate the pH
The pH is calculated using the formula:
\[ pH = -\log[H^+] \]
Substituting the value of \(H^+\):
\[ pH = -\log(1.8 \times 10^{-7}) \approx 6.74 \]

### Final Answer
The pH of neutral rainwater is approximately **6.74**.

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