Equal volumes of `0.50` M of HCl, `0.25` M of NaOH and `0.75` M of NaCl are mixed. The molarity of the NaCl solution :

Options

A. `(3)/(4)M`
B. `(1)/(3)M`
C. `(1)/(2)M`
D. `2.00` M

Answer

To find the molarity of the NaCl solution after mixing equal volumes of 0.50 M HCl, 0.25 M NaOH, and 0.75 M NaCl, we can follow these steps:

### Step 1: Define the Volume of Each Solution
Let’s assume we take a volume \( V \) (in liters) of each solution. Since we are mixing equal volumes, we can denote the volume of each solution as \( V \).

### Step 2: Calculate Moles of Each Component
1. **Moles of HCl**:
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} = 0.50 \, \text{M} \times V = 0.50V \, \text{moles} \]

2. **Moles of NaOH**:
\[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume} = 0.25 \, \text{M} \times V = 0.25V \, \text{moles} \]

3. **Moles of NaCl**:
\[ \text{Moles of NaCl} = \text{Molarity} \times \text{Volume} = 0.75 \, \text{M} \times V = 0.75V \, \text{moles} \]

### Step 3: Determine the Reaction Between HCl and NaOH
HCl and NaOH will react in a 1:1 molar ratio:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

- The limiting reagent will determine how many moles of NaCl are formed. Since we have:
- 0.50V moles of HCl
- 0.25V moles of NaOH

HCl is in excess, and NaOH is the limiting reagent. Therefore, all of the NaOH will react.

### Step 4: Calculate Moles of NaCl Formed
Since 0.25V moles of NaOH will react with 0.25V moles of HCl, the moles of NaCl produced will also be:
\[ \text{Moles of NaCl produced} = 0.25V \, \text{moles} \]

### Step 5: Calculate Total Moles of NaCl After Mixing
Initially, we had 0.75V moles of NaCl. After the reaction, we add the moles produced:
\[ \text{Total moles of NaCl} = 0.75V + 0.25V = 1.00V \, \text{moles} \]

### Step 6: Calculate the Total Volume of the Mixture
The total volume of the mixture after combining the three solutions is:
\[ \text{Total Volume} = V + V + V = 3V \]

### Step 7: Calculate the Molarity of the NaCl Solution
The molarity of the NaCl solution can now be calculated using the total moles of NaCl and the total volume:
\[ \text{Molarity of NaCl} = \frac{\text{Total moles of NaCl}}{\text{Total Volume}} = \frac{1.00V}{3V} = \frac{1.00}{3} \, \text{M} \approx 0.33 \, \text{M} \]

### Final Answer
The molarity of the NaCl solution after mixing is approximately **0.33 M**.

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