The pH of a solution obtained by mixing 50 ml of 0.4 M HCl with 50 ml of 0.2 M NaOH is

Options

A. `-log 2`
B. `-log 2 xx 10^(-1)`
C. `7.0`
D. `2.0`

Answer

To find the pH of the solution obtained by mixing 50 ml of 0.4 M HCl with 50 ml of 0.2 M NaOH, we can follow these steps:

### Step 1: Calculate the moles of HCl and NaOH

1. **Moles of HCl:**
\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 0.4 \, \text{M} \times 0.050 \, \text{L} = 0.020 \, \text{moles} \]

2. **Moles of NaOH:**
\[ \text{Moles of NaOH} = \text{Concentration} \times \text{Volume} = 0.2 \, \text{M} \times 0.050 \, \text{L} = 0.010 \, \text{moles} \]

### Step 2: Determine the limiting reactant and the remaining moles

- HCl and NaOH react in a 1:1 ratio:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

- Since we have 0.020 moles of HCl and 0.010 moles of NaOH, NaOH is the limiting reactant.

- **Moles of HCl remaining:**
\[ \text{Remaining HCl} = 0.020 - 0.010 = 0.010 \, \text{moles} \]

- **Moles of NaOH remaining:**
\[ \text{Remaining NaOH} = 0.010 - 0.010 = 0 \, \text{moles} \]

### Step 3: Calculate the total volume of the solution

- Total volume after mixing:
\[ \text{Total Volume} = 50 \, \text{ml} + 50 \, \text{ml} = 100 \, \text{ml} = 0.100 \, \text{L} \]

### Step 4: Calculate the concentration of remaining HCl

- **Concentration of HCl:**
\[ \text{Concentration of HCl} = \frac{\text{Moles of HCl remaining}}{\text{Total Volume}} = \frac{0.010 \, \text{moles}}{0.100 \, \text{L}} = 0.1 \, \text{M} \]

### Step 5: Calculate the pH of the solution

- Since HCl is a strong acid, we can directly use its concentration to find the pH:
\[ \text{pH} = -\log[\text{H}^+] = -\log[0.1] = 1 \]

### Final Answer:
The pH of the solution obtained by mixing 50 ml of 0.4 M HCl with 50 ml of 0.2 M NaOH is **1**.

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