`H_(3)PO_(4)+H_(2) Leftrightarrow H_(3)O^(+)+H_(2)PO_(4)^(-),pK_(1)=2.15`
`H_3PO_(4)^(-)+H_(2)O Leftrightarrow H_(3)O^(+)+HPO_(4)^(2-), pK_(2)=7.20`
Hence pH of 0.01 M `NaH_(2)PO_(4)` is

Options

A. 9.35
B. 4.675
C. 2.675
D. `7.350`

Answer

To find the pH of a 0.01 M solution of NaH₂PO₄, we can use the given pK values of the relevant acid-base equilibria. Here's the step-by-step solution:

### Step 1: Identify the relevant equilibria
The dissociation of phosphoric acid (H₃PO₄) can be represented in two steps:
1. H₃PO₄ ⇌ H₃O⁺ + H₂PO₄⁻ (pK₁ = 2.15)
2. H₂PO₄⁻ + H₂O ⇌ H₃O⁺ + HPO₄²⁻ (pK₂ = 7.20)

### Step 2: Understand the species in solution
NaH₂PO₄ is the sodium salt of the dihydrogen phosphate ion (H₂PO₄⁻). In solution, it will dissociate completely to give Na⁺ and H₂PO₄⁻ ions.

### Step 3: Use the Henderson-Hasselbalch equation
For a buffer solution containing a weak acid (H₂PO₄⁻) and its conjugate base (HPO₄²⁻), we can use the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]

In this case, we can use the average of pK₁ and pK₂ to find the pH of the solution:
\[ \text{pH} = \frac{\text{pK}_1 + \text{pK}_2}{2} \]

### Step 4: Calculate the average pK
Substituting the values of pK₁ and pK₂:
\[ \text{pH} = \frac{2.15 + 7.20}{2} \]
\[ \text{pH} = \frac{9.35}{2} \]
\[ \text{pH} = 4.675 \]

### Step 5: Conclusion
The pH of a 0.01 M solution of NaH₂PO₄ is approximately 4.675.