Phosphorous acid, `H_3PO_3` is a diprotic acid with `K_a`. values `5 xx 10^(-2) mol L^(-1)` and `2 xx 10 mol L^(-1)` respectively for its two stages. Calculate the pH of a 0.01 M solution of `H_3PO_3` (aq).

Explanation

To calculate the pH of a 0.01 M solution of phosphorous acid, \( H_3PO_3 \), we will follow these steps:

### Step 1: Understand the dissociation of \( H_3PO_3 \)
Phosphorous acid is a diprotic acid, meaning it can donate two protons (H⁺ ions). The dissociation can be represented in two stages:
1. \( H_3PO_3 \rightleftharpoons H^+ + H_2PO_3^- \) (First dissociation)
2. \( H_2PO_3^- \rightleftharpoons H^+ + HPO_3^{2-} \) (Second dissociation)

### Step 2: Identify the \( K_a \) values
The given \( K_a \) values for the two dissociation stages are:
- \( K_{a1} = 5 \times 10^{-2} \, \text{mol L}^{-1} \)
- \( K_{a2} = 2 \times 10^{-7} \, \text{mol L}^{-1} \)

### Step 3: Set up the equilibrium expression for the first dissociation
For the first dissociation, we can write the equilibrium expression as follows:
\[ K_{a1} = \frac{[H^+][H_2PO_3^-]}{[H_3PO_3]} \]
Assuming the initial concentration of \( H_3PO_3 \) is 0.01 M, and letting \( x \) be the concentration of \( H^+ \) produced, we can express the concentrations at equilibrium:
- \( [H_3PO_3] = 0.01 - x \)
- \( [H^+] = x \)
- \( [H_2PO_3^-] = x \)

Thus, the equation becomes:
\[ K_{a1} = \frac{x \cdot x}{0.01 - x} = \frac{x^2}{0.01 - x} \]

### Step 4: Simplify the equation
Since \( K_{a1} \) is relatively large, we can assume \( x \) is small compared to 0.01, allowing us to simplify:
\[ K_{a1} \approx \frac{x^2}{0.01} \]
Substituting \( K_{a1} \):
\[ 5 \times 10^{-2} = \frac{x^2}{0.01} \]
\[ x^2 = 5 \times 10^{-2} \times 0.01 = 5 \times 10^{-4} \]

### Step 5: Solve for \( x \)
Taking the square root:
\[ x = \sqrt{5 \times 10^{-4}} = 2.236 \times 10^{-2} \, \text{mol L}^{-1} \]

### Step 6: Calculate the pH
The concentration of \( H^+ \) ions is equal to \( x \):
\[ [H^+] = 2.236 \times 10^{-2} \, \text{mol L}^{-1} \]
Now, we can calculate the pH:
\[ \text{pH} = -\log[H^+] = -\log(2.236 \times 10^{-2}) \]

### Step 7: Final calculation
Using a calculator:
\[ \text{pH} \approx 1.651 \]

### Final Answer
The pH of a 0.01 M solution of \( H_3PO_3 \) is approximately **1.651**.

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