Calculate the `pH` of resulting solution obtained by mixing `50 mL` of `0.6N HCl` and `50 ml ` of `0.3 N NaOH`

Options

A. `0.1`
B. `0.8`
C. `2.1`
D. 4

Explanation

To calculate the pH of the resulting solution obtained by mixing 50 mL of 0.6 N HCl and 50 mL of 0.3 N NaOH, follow these steps:

### Step 1: Calculate the moles of HCl and NaOH
1. **For HCl:**
- Normality (N) = 0.6 N
- Volume (V) = 50 mL = 0.050 L
- Moles of HCl = Normality × Volume = 0.6 N × 0.050 L = 0.03 moles

2. **For NaOH:**
- Normality (N) = 0.3 N
- Volume (V) = 50 mL = 0.050 L
- Moles of NaOH = Normality × Volume = 0.3 N × 0.050 L = 0.015 moles

### Step 2: Determine the reaction between HCl and NaOH
The reaction between HCl and NaOH is:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

### Step 3: Calculate the remaining moles after the reaction
- Initially, we have:
- Moles of HCl = 0.03 moles
- Moles of NaOH = 0.015 moles

- During the reaction:
- NaOH will completely react with HCl since it is the limiting reagent.
- Moles of HCl remaining = 0.03 moles - 0.015 moles = 0.015 moles
- Moles of NaOH remaining = 0.015 moles - 0.015 moles = 0 moles

### Step 4: Calculate the concentration of H⁺ ions
- The total volume of the solution after mixing = 50 mL + 50 mL = 100 mL = 0.1 L
- Concentration of H⁺ ions = Moles of HCl remaining / Total volume = 0.015 moles / 0.1 L = 0.15 M

### Step 5: Calculate the pH of the solution
- pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
- Substitute the concentration of H⁺ ions:
\[ \text{pH} = -\log(0.15) \]
- Using a calculator:
\[ \text{pH} \approx 0.8239 \]

### Final Answer
The pH of the resulting solution is approximately **0.82**.

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