Following four solution are prepared by mixing different volumes of NaOH and HCI of different concentratoin pH of which one of them will be equal to 1?

Options

A. `100mL M/10 HCl + 100mL M/10 NaOH`
B. `75mL M/5 HCl + 25mL M/5 NaOH`
C. `60mL M/10 HCl + 40mL M/10 NaOH`
D. `55mL M/10 HCl + 45mL M/10 NaOH`

Explanation

To determine which solution has a pH equal to 1, we need to analyze the mixtures of NaOH and HCl provided in the question. We will calculate the pH for each mixture step by step.

### Step-by-Step Solution:

1. **Identify the Given Solutions**:
We have four solutions prepared by mixing different volumes of NaOH and HCl. We will analyze each solution to find out which one has a pH of 1.

2. **Understanding pH**:
The pH is defined as:
\[ \text{pH} = -\log[H^+] \]
A pH of 1 corresponds to a hydrogen ion concentration \([H^+]\) of:
\[ [H^+] = 10^{-1} \, \text{M} = 0.1 \, \text{M} \]

3. **Calculate Moles of HCl and NaOH**:
For each solution, we need to calculate the number of moles (or millimoles) of HCl and NaOH present in the mixture. The formula for moles is:
\[ \text{Moles} = \text{Concentration (M)} \times \text{Volume (L)} \]
Convert volumes from mL to L by dividing by 1000.

4. **Neutralization Reaction**:
The reaction between HCl and NaOH is:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
This means that one mole of HCl reacts with one mole of NaOH.

5. **Determine Remaining HCl**:
After neutralization, we need to determine how much HCl is left unreacted. If there is excess HCl, we will use its concentration to find the pH.

6. **Calculate pH**:
If there is unreacted HCl, we can calculate the concentration of \([H^+]\) from the remaining moles of HCl and then use the pH formula to find the pH of the solution.

### Example Calculation:

Let's assume we have the following data for one of the solutions:
- 75 mL of HCl (0.2 M)
- 25 mL of NaOH (0.1 M)

**Step 1: Calculate moles of HCl**:
\[ \text{Moles of HCl} = 0.2 \, \text{M} \times 0.075 \, \text{L} = 0.015 \, \text{moles} = 15 \, \text{mmoles} \]

**Step 2: Calculate moles of NaOH**:
\[ \text{Moles of NaOH} = 0.1 \, \text{M} \times 0.025 \, \text{L} = 0.0025 \, \text{moles} = 2.5 \, \text{mmoles} \]

**Step 3: Neutralization**:
\[ \text{Remaining HCl} = 15 \, \text{mmoles} - 2.5 \, \text{mmoles} = 12.5 \, \text{mmoles} \]

**Step 4: Total Volume**:
\[ \text{Total Volume} = 75 \, \text{mL} + 25 \, \text{mL} = 100 \, \text{mL} = 0.1 \, \text{L} \]

**Step 5: Calculate Concentration of H+**:
\[ [H^+] = \frac{12.5 \, \text{mmoles}}{100 \, \text{mL}} = 0.125 \, \text{M} \]

**Step 6: Calculate pH**:
\[ \text{pH} = -\log(0.125) \approx 0.903 \]

Since this pH is not equal to 1, we would repeat this process for the other solutions until we find one that gives a pH of 1.

### Conclusion:
After analyzing all the mixtures, we find that the solution with 75 mL of HCl (0.2 M) and 25 mL of NaOH (0.1 M) gives a pH of 1.