50% neutralisation of a solution of formic acid `(k_(a)=2xx10^(-4))` with NaOH would result in a solution having a hydrogen ion concentration of :

Options

A. `2xx10^(-4)`
B. 3.7
C. 2.7
D. 1.85

Explanation

To solve the problem of determining the hydrogen ion concentration after 50% neutralization of formic acid with NaOH, we can follow these steps:

### Step 1: Understand the Neutralization Process
When formic acid (HCOOH) is neutralized by sodium hydroxide (NaOH), it forms sodium formate (HCOONa) and water. The reaction can be represented as:
\[ \text{HCOOH} + \text{NaOH} \rightarrow \text{HCOONa} + \text{H}_2\text{O} \]

### Step 2: Identify the Concept of 50% Neutralization
50% neutralization means that half of the formic acid remains unreacted, and half has been converted to its conjugate base (sodium formate). If we start with an initial concentration \( C \) of formic acid, after 50% neutralization:
- Concentration of unreacted formic acid (HCOOH) = \( \frac{C}{2} \)
- Concentration of sodium formate (HCOONa) formed = \( \frac{C}{2} \)

### Step 3: Use the Henderson-Hasselbalch Equation
The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

In this case:
- \( [\text{Salt}] = \frac{C}{2} \)
- \( [\text{Acid}] = \frac{C}{2} \)

### Step 4: Substitute Values into the Equation
Substituting into the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{\frac{C}{2}}{\frac{C}{2}}\right) \]
Since the ratio of salt to acid is 1:
\[ \log(1) = 0 \]
Thus:
\[ \text{pH} = \text{pK}_a \]

### Step 5: Calculate pK_a from K_a
Given \( K_a = 2 \times 10^{-4} \), we can find \( pK_a \):
\[ \text{pK}_a = -\log(K_a) = -\log(2 \times 10^{-4}) \]
Using logarithmic properties:
\[ \text{pK}_a = -\log(2) + 4 \]
Approximating \( \log(2) \approx 0.301 \):
\[ \text{pK}_a \approx 4 - 0.301 = 3.699 \]

### Step 6: Find the Hydrogen Ion Concentration
Since \( \text{pH} = \text{pK}_a \), we have:
\[ \text{pH} \approx 3.699 \]
To find the hydrogen ion concentration:
\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-3.699} \]

Calculating this gives:
\[ [\text{H}^+] \approx 2 \times 10^{-4} \]

### Final Answer
The hydrogen ion concentration after 50% neutralization of formic acid with NaOH is approximately:
\[ [\text{H}^+] = 2 \times 10^{-4} \, \text{M} \]