To a 50 ml of `0.1` M HCl solution , 10 ml of `0.1` M NaOH is added and the resulting solution is diluted to 100 ml. What is change in pH of the HCl solution ?

Options

A. `4.398`
B. `0.398`
C. `0.1 M`
D. None of these

Explanation

To solve the problem step by step, we will follow the outlined procedure:

### Step 1: Calculate the initial moles of HCl
Given:
- Volume of HCl = 50 ml = 0.050 L
- Molarity of HCl = 0.1 M

Using the formula:
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume (L)} \]
\[ \text{Moles of HCl} = 0.1 \, \text{mol/L} \times 0.050 \, \text{L} = 0.005 \, \text{mol} \]

### Step 2: Calculate the moles of NaOH added
Given:
- Volume of NaOH = 10 ml = 0.010 L
- Molarity of NaOH = 0.1 M

Using the same formula:
\[ \text{Moles of NaOH} = 0.1 \, \text{mol/L} \times 0.010 \, \text{L} = 0.001 \, \text{mol} \]

### Step 3: Determine the reaction and calculate the remaining moles
The reaction between HCl and NaOH is:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

From the reaction, we see that 1 mole of HCl reacts with 1 mole of NaOH.

- Initial moles of HCl = 0.005 mol
- Initial moles of NaOH = 0.001 mol

After the reaction:
- Moles of HCl remaining = Initial moles of HCl - Moles of NaOH
\[ \text{Remaining moles of HCl} = 0.005 \, \text{mol} - 0.001 \, \text{mol} = 0.004 \, \text{mol} \]

### Step 4: Calculate the concentration of HCl after dilution
After the reaction, the total volume of the solution is diluted to 100 ml = 0.1 L.

Using the formula for concentration:
\[ \text{Concentration of HCl} = \frac{\text{Remaining moles of HCl}}{\text{Total volume (L)}} \]
\[ \text{Concentration of HCl} = \frac{0.004 \, \text{mol}}{0.1 \, \text{L}} = 0.04 \, \text{M} \]

### Step 5: Calculate the pH of the resulting solution
The concentration of H\(^+\) ions is equal to the concentration of HCl since HCl is a strong acid and dissociates completely.

Using the formula for pH:
\[ \text{pH} = -\log[\text{H}^+] \]
\[ \text{pH} = -\log(0.04) = -\log(4 \times 10^{-2}) = -\log(4) + 2 \]

Using the approximate value of \(-\log(4) \approx 0.6\):
\[ \text{pH} \approx 2 - 0.6 = 1.4 \]

### Step 6: Calculate the change in pH
Initial pH of HCl solution = 1

Change in pH:
\[ \text{Change in pH} = \text{Final pH} - \text{Initial pH} \]
\[ \text{Change in pH} = 1.4 - 1 = 0.4 \]

### Final Answer
The change in pH of the HCl solution is approximately **0.4**.

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