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

### Step 1: Understand the Reaction
Formic acid (HCOOH) is a weak acid, and NaOH is a strong base. When they react, they form 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: Define Initial Concentration
Let the initial concentration of formic acid be \( C \). When 50% of the formic acid is neutralized with NaOH, half of the acid will react.

### Step 3: Calculate Concentrations After Neutralization
After 50% neutralization:
- The concentration of formic acid remaining = \( C - \frac{C}{2} = \frac{C}{2} \)
- The concentration of sodium formate (the salt formed) = \( \frac{C}{2} \)

### Step 4: Use the Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

### Step 5: Calculate pKa
Given the acid dissociation constant \( K_a = 2 \times 10^{-4} \), we can find pKa:
\[ \text{pKa} = -\log(K_a) = -\log(2 \times 10^{-4}) \]
Using logarithm properties:
\[ \text{pKa} = -(\log(2) + \log(10^{-4})) = -\log(2) + 4 \]
Where \( \log(2) \approx 0.301 \):
\[ \text{pKa} = -0.301 + 4 = 3.699 \]

### Step 6: Substitute Values into the Henderson-Hasselbalch Equation
Since the concentrations of the salt and the acid are equal after 50% neutralization:
\[ \text{pH} = \text{pKa} + \log\left(\frac{\frac{C}{2}}{\frac{C}{2}}\right) \]
This simplifies to:
\[ \text{pH} = \text{pKa} + \log(1) = \text{pKa} + 0 \]
Thus:
\[ \text{pH} = 3.699 \]

### Step 7: Find the Hydrogen Ion Concentration
To find the hydrogen ion concentration \([H^+]\), we can use the formula:
\[ [H^+] = 10^{-\text{pH}} \]
Substituting the pH value:
\[ [H^+] = 10^{-3.699} \]

Calculating this gives:
\[ [H^+] \approx 2.00 \times 10^{-4} \, \text{M} \]

### Final Answer
The hydrogen ion concentration after 50% neutralization of formic acid with NaOH is approximately:
\[ [H^+] \approx 2.00 \times 10^{-4} \, \text{M} \]