Which of the following will be present in the soluiton formed when 50mL of 0.1 M HCl is mixed with 50mL of 0.1 M NaOH ?

Options

A. 4.5 m mole of `H^(+)`
B. 0.05 m mol of `OH^(-)`
C. 0.05 M NaCl
D. `10^(-7)` M of `H^(+)` ion

Answer

To solve the problem of what will be present in the solution formed when 50 mL of 0.1 M HCl is mixed with 50 mL of 0.1 M NaOH, we can follow these steps:

### Step 1: Calculate the number of moles of HCl and NaOH
- **HCl**:
\[ \text{Number of moles of HCl} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{M} \times 0.050 \, \text{L} = 0.005 \, \text{moles} \, (5 \, \text{mmol}) \]

- **NaOH**:
\[ \text{Number of moles of NaOH} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{M} \times 0.050 \, \text{L} = 0.005 \, \text{moles} \, (5 \, \text{mmol}) \]

### Step 2: Determine the reaction between HCl and NaOH
- The reaction between HCl (a strong acid) and NaOH (a strong base) can be represented as:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

### Step 3: Calculate the products formed
- Since both HCl and NaOH completely dissociate and react in a 1:1 ratio, we will have:
- 5 mmol of HCl reacts with 5 mmol of NaOH.
- This will produce 5 mmol of NaCl (salt) and 5 mmol of water.

### Step 4: Analyze the resulting solution
- After the reaction, we are left with:
- **NaCl** (sodium chloride) in solution.
- **Water** (H₂O).

### Step 5: Determine the concentration of NaCl in the solution
- The total volume of the solution after mixing is:
\[ 50 \, \text{mL} + 50 \, \text{mL} = 100 \, \text{mL} = 0.1 \, \text{L} \]
- The concentration of NaCl formed is:
\[ \text{Concentration of NaCl} = \frac{\text{Number of moles of NaCl}}{\text{Total volume}} = \frac{0.005 \, \text{moles}}{0.1 \, \text{L}} = 0.05 \, \text{M} \]

### Conclusion
The solution formed when 50 mL of 0.1 M HCl is mixed with 50 mL of 0.1 M NaOH will contain:
- Sodium chloride (NaCl)
- Water (H₂O)