125 ml of 8% w/v NaOH solution (sp. gravity 1) is added to 125 ml of 10% w/v HCl solution. The nature of resultant solution would be ____

Options

A. Acidic
B. Basic
C. Neutral
D. Can not be predicted

Answer

To determine the nature of the resultant solution when 125 ml of 8% w/v NaOH solution is mixed with 125 ml of 10% w/v HCl solution, we can follow these steps:

### Step 1: Calculate the mass of NaOH in the solution
Given that the NaOH solution is 8% w/v, this means there are 8 grams of NaOH in 100 ml of solution.

For 125 ml of NaOH solution:
\[ \text{Mass of NaOH} = \left(\frac{8 \text{ g}}{100 \text{ ml}}\right) \times 125 \text{ ml} = 10 \text{ g} \]

### Step 2: Calculate the number of moles of NaOH
The molar mass of NaOH is 40 g/mol. To find the number of moles of NaOH:
\[ \text{Moles of NaOH} = \frac{\text{Mass of NaOH}}{\text{Molar mass of NaOH}} = \frac{10 \text{ g}}{40 \text{ g/mol}} = 0.25 \text{ moles} \]

### Step 3: Calculate the mass of HCl in the solution
For the HCl solution, which is 10% w/v, this means there are 10 grams of HCl in 100 ml of solution.

For 125 ml of HCl solution:
\[ \text{Mass of HCl} = \left(\frac{10 \text{ g}}{100 \text{ ml}}\right) \times 125 \text{ ml} = 12.5 \text{ g} \]

### Step 4: Calculate the number of moles of HCl
The molar mass of HCl is 36.5 g/mol. To find the number of moles of HCl:
\[ \text{Moles of HCl} = \frac{\text{Mass of HCl}}{\text{Molar mass of HCl}} = \frac{12.5 \text{ g}}{36.5 \text{ g/mol}} \approx 0.342 \text{ moles} \]

### Step 5: Compare the moles of HCl and NaOH
NaOH dissociates to give OH⁻ ions, and HCl dissociates to give H⁺ ions. The reaction can be represented as:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

From the calculations:
- Moles of OH⁻ from NaOH = 0.25 moles
- Moles of H⁺ from HCl = 0.342 moles

### Step 6: Determine the nature of the resultant solution
Since the moles of H⁺ (0.342 moles) are greater than the moles of OH⁻ (0.25 moles), the solution will be acidic in nature.

### Final Answer:
The nature of the resultant solution is **acidic**.

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