`20 cm^(3)` of `xM` solution of `HCl` is exactly neutralised by `40cm^(3)` of `0.05 M NaOH` solutions, the `pH` of `HCl` solution is

Options

A. `1.0`
B. `2`
C. `1.5`
D. `2.5`

Explanation

To find the pH of the HCl solution, we need to follow these steps:

### Step 1: Understand the neutralization reaction
HCl is a strong acid that dissociates completely in solution to give H⁺ ions and Cl⁻ ions. NaOH is a strong base that dissociates completely to give Na⁺ ions and OH⁻ ions. The neutralization reaction can be represented as:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

### Step 2: Write the equation for equivalents
The number of equivalents of acid (HCl) will equal the number of equivalents of base (NaOH) at the point of neutralization. The formula for equivalents is:
\[ \text{n}_1 \cdot \text{V}_1 = \text{n}_2 \cdot \text{V}_2 \]
Where:
- \( n_1 \) = normality of HCl
- \( V_1 \) = volume of HCl solution (in liters)
- \( n_2 \) = normality of NaOH
- \( V_2 \) = volume of NaOH solution (in liters)

### Step 3: Calculate the normality of NaOH
Given that the concentration of NaOH is 0.05 M and its n-factor is 1 (since it produces one OH⁻ ion), the normality (N) of NaOH is:
\[ n_2 = 0.05 \, \text{N} \]

### Step 4: Convert volumes from cm³ to liters
Convert the volumes from cm³ to liters:
- \( V_1 = 20 \, \text{cm}^3 = 20 \times 10^{-3} \, \text{L} \)
- \( V_2 = 40 \, \text{cm}^3 = 40 \times 10^{-3} \, \text{L} \)

### Step 5: Substitute values into the equation
Substituting the known values into the equation:
\[ n_1 \cdot (20 \times 10^{-3}) = (0.05) \cdot (40 \times 10^{-3}) \]

### Step 6: Solve for \( n_1 \)
Now we can solve for \( n_1 \):
\[ n_1 \cdot (20 \times 10^{-3}) = 0.05 \cdot (40 \times 10^{-3}) \]
\[ n_1 \cdot 20 = 0.05 \cdot 40 \]
\[ n_1 \cdot 20 = 2 \]
\[ n_1 = \frac{2}{20} = 0.1 \, \text{N} \]

### Step 7: Determine the concentration of H⁺ ions
Since HCl is a strong acid, the normality is equal to the molarity:
\[ \text{Molarity of HCl} = 0.1 \, \text{M} \]

### Step 8: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Substituting the concentration of H⁺ ions:
\[ \text{pH} = -\log(0.1) = 1 \]

### Final Answer
The pH of the HCl solution is **1**.

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