Henderson's equation is `pH = pK_(a) + log.(["salt"])/(["acid"])`. If the acid gets half neutralized the value of pH will be : `[pK_(a) = 4.30]`

Options

A. 4.3
B. 2.15
C. 8.6
D. 7

Explanation

To solve the problem, we will use Henderson's equation:

\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{salt}]}{[\text{acid}]}\right) \]

Given:
- \(\text{pK}_a = 4.30\)
- The acid is half neutralized.

### Step-by-Step Solution:

1. **Understanding Half Neutralization**:
- When an acid is half neutralized, the concentration of the salt (conjugate base) becomes equal to half of the original concentration of the acid.
- If we denote the initial concentration of the acid as \([HA]\) and the concentration of the salt as \([A^-]\), after half neutralization, we have:
\[ [A^-] = \frac{1}{2} [HA] \]

2. **Substituting into Henderson's Equation**:
- When the acid is half neutralized, the concentration of the salt becomes equal to the concentration of the acid:
\[ [A^-] = \frac{1}{2} [HA] \implies \frac{[\text{salt}]}{[\text{acid}]} = \frac{\frac{1}{2} [HA]}{[HA]} = \frac{1}{2} \]

3. **Calculating the Logarithm**:
- Now, substitute this ratio into the Henderson equation:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{1}{2}\right) \]
- We know that \(\log\left(\frac{1}{2}\right) = -\log(2)\). The value of \(\log(2) \approx 0.301\):
\[ \text{pH} = 4.30 - 0.301 \]

4. **Final Calculation**:
- Now, perform the subtraction:
\[ \text{pH} = 4.30 - 0.301 = 4.00 \]

### Final Answer:
The value of pH when the acid gets half neutralized is approximately **4.00**.