`H_(3)PO_(4)` is a weak triprotic acid, approximate pH 0.1 M `Na_2HPO_(4)`(aq.) is calculated by:

Options

A. `1/2[pK_(a1)+pK_(a2)]`
B. `1/2[pK_(a2)+pK_(a3)]`
C. `1/2[pK_(a1)+pK_(a3)]`
D. `pK_(a1)+pK_(a2)`

Answer

To calculate the approximate pH of a 0.1 M Na₂HPO₄ solution, we can follow these steps:

### Step 1: Understand the Nature of Na₂HPO₄
Na₂HPO₄ is the sodium salt of the weak triprotic acid H₃PO₄. In solution, it dissociates to form HPO₄²⁻ ions, which can act as a weak base.

**Hint:** Recognize that sodium salts of weak acids can affect the pH of a solution, and the anion (HPO₄²⁻) will determine the pH.

### Step 2: Identify Relevant Equilibria
The dissociation of H₃PO₄ can be represented as:
1. H₃PO₄ ⇌ H₂PO₄⁻ + H⁺ (Ka₁)
2. H₂PO₄⁻ ⇌ HPO₄²⁻ + H⁺ (Ka₂)
3. HPO₄²⁻ ⇌ PO₄³⁻ + H⁺ (Ka₃)

Since we are dealing with HPO₄²⁻, we need to focus on the equilibria involving HPO₄²⁻.

**Hint:** Identify which dissociation constants (Ka values) are relevant for the species present in the solution.

### Step 3: Use the Henderson-Hasselbalch Equation
For a salt like Na₂HPO₄, we can use the Henderson-Hasselbalch equation to find the pH:
\[ \text{pH} = \frac{1}{2} (\text{pKa}_2 + \text{pKa}_3) \]

This is because HPO₄²⁻ is the conjugate base of H₂PO₄⁻ and the acid dissociation constants Ka₂ and Ka₃ are involved.

**Hint:** Remember that the pKa values are related to the strength of the acid; the lower the pKa, the stronger the acid.

### Step 4: Find pKa Values
We need the pKa values for H₃PO₄:
- pKa₁ ≈ 2.15
- pKa₂ ≈ 7.21
- pKa₃ ≈ 12.67

**Hint:** Look up the pKa values for the acid in question, as they are essential for calculating the pH.

### Step 5: Calculate the pH
Now we can substitute the pKa values into the equation:
\[ \text{pH} = \frac{1}{2} (7.21 + 12.67) \]
\[ \text{pH} = \frac{1}{2} (19.88) \]
\[ \text{pH} = 9.94 \]

**Hint:** Ensure that you perform the arithmetic correctly when averaging the pKa values.

### Final Answer
The approximate pH of a 0.1 M Na₂HPO₄ solution is **9.94**.