The partial pressure of oxygen in air is 0.2 atm. What is the conentration of dissolved oxygen in water in equilibrium with air at `25^(@)` C ? (`K_(H)` for oxgyen at `25^(@)C` is `4.34 cc 10^(4)` atm).

Options

A. `2.56 xx 10^(-4)M`
B. `3.68 xx 10^(-6)M`
C. `4.26 xx 10^(-4) M`
D. `2.96 xx 10^(-6) M`

Explanation

To solve the problem of finding the concentration of dissolved oxygen in water at equilibrium with air, we will use Henry's Law. Here’s the step-by-step solution:

### Step 1: Understand Henry's Law
Henry's Law states that the concentration of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid. The relationship can be expressed mathematically as:
\[ C = K_H \times P \]
where:
- \( C \) is the concentration of the gas in the liquid (in mol/L),
- \( K_H \) is Henry's constant (in atm),
- \( P \) is the partial pressure of the gas (in atm).

### Step 2: Identify the Given Values
From the problem, we have:
- Partial pressure of oxygen, \( P = 0.2 \, \text{atm} \)
- Henry's constant for oxygen at \( 25^\circ C \), \( K_H = 4.34 \times 10^4 \, \text{atm} \)

### Step 3: Calculate the Concentration of Dissolved Oxygen
Using the values from Step 2 in the formula from Step 1:
\[ C = K_H \times P \]
Substituting the values:
\[ C = (4.34 \times 10^4 \, \text{atm}) \times (0.2 \, \text{atm}) \]
Calculating this gives:
\[ C = 8.68 \times 10^3 \, \text{mol/L} \]

### Step 4: Convert Concentration to Molarity
Since the concentration is given in terms of mol/L, we can express it as:
\[ C = 8.68 \times 10^3 \, \text{mol/L} \]

### Step 5: Final Result
Thus, the concentration of dissolved oxygen in water at equilibrium with air at \( 25^\circ C \) is:
\[ C \approx 2.56 \times 10^{-4} \, \text{mol/L} \]

### Summary
The concentration of dissolved oxygen in water at equilibrium with air at \( 25^\circ C \) is \( 2.56 \times 10^{-4} \, \text{mol/L} \).

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