`Ph` of `3xx10^(-3)M` solution of `H_(3)X` will be Assuming `alpha_(1)=1,alpha_(2)=1//3,alpha_(3)=` negligible

Options

A. `2.40`
B. `3.0`
C. `3.4771`
D. `4.0`

Explanation

To find the pH of a \(3 \times 10^{-3} M\) solution of \(H_3X\) given the dissociation constants and degrees of dissociation, we can follow these steps:

### Step 1: Understand the dissociation of the acid
The acid \(H_3X\) is a tribasic acid, which means it can dissociate in three steps:
1. \(H_3X \rightleftharpoons H^+ + H_2X^-\) (first dissociation)
2. \(H_2X^- \rightleftharpoons H^+ + HX^{2-}\) (second dissociation)
3. \(HX^{2-} \rightleftharpoons H^+ + X^{3-}\) (third dissociation)

### Step 2: Apply the given degrees of dissociation
- For the first dissociation, \(\alpha_1 = 1\) (complete dissociation).
- For the second dissociation, \(\alpha_2 = \frac{1}{3}\).
- For the third dissociation, \(\alpha_3\) is negligible.

### Step 3: Calculate the concentration of \(H^+\) ions from the first dissociation
From the first dissociation:
- Initial concentration of \(H_3X\) = \(C = 3 \times 10^{-3} M\)
- Concentration of \(H^+\) produced = \(C \cdot \alpha_1 = 3 \times 10^{-3} \times 1 = 3 \times 10^{-3} M\)

### Step 4: Calculate the concentration of \(H^+\) ions from the second dissociation
From the second dissociation:
- Initial concentration of \(H_2X^-\) = \(C \cdot \alpha_1 = 3 \times 10^{-3} M\)
- Concentration of \(H^+\) produced = \(C \cdot \alpha_1 \cdot \alpha_2 = 3 \times 10^{-3} \times \frac{1}{3} = 1 \times 10^{-3} M\)

### Step 5: Total concentration of \(H^+\) ions
Total concentration of \(H^+\) ions:
\[ [H^+] = [H^+]_{first} + [H^+]_{second} = 3 \times 10^{-3} + 1 \times 10^{-3} = 4 \times 10^{-3} M \]

### Step 6: Calculate the pH
Using the formula for pH:
\[ \text{pH} = -\log[H^+] \]
Substituting the value:
\[ \text{pH} = -\log(4 \times 10^{-3}) = -(\log 4 + \log 10^{-3}) = -\log 4 + 3 \]
Using \(\log 4 \approx 0.60\):
\[ \text{pH} = 3 - 0.60 = 2.40 \]

### Final Answer
The pH of the \(3 \times 10^{-3} M\) solution of \(H_3X\) is **2.40**.

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