One litre of a solution contains 15.12 g of `HNO_(3)` and one litre of another solution contains 3.2 g of NaOH. In what volume ratio must these solutions be mixed to obtain a neutral solution ?

Options

A. `1/3`
B. `2/3`
C. `8/3`
D. `3/8`

Answer

To find the volume ratio in which the solutions of HNO3 and NaOH must be mixed to obtain a neutral solution, we can follow these steps:

### Step 1: Calculate the number of moles of HNO3
Given:
- Mass of HNO3 = 15.12 g
- Molar mass of HNO3 = 63 g/mol

Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
\[ \text{Number of moles of HNO3} = \frac{15.12 \text{ g}}{63 \text{ g/mol}} = 0.24 \text{ moles} \]

### Step 2: Calculate the number of moles of NaOH
Given:
- Mass of NaOH = 3.2 g
- Molar mass of NaOH = 40 g/mol

Using the same formula:
\[ \text{Number of moles of NaOH} = \frac{3.2 \text{ g}}{40 \text{ g/mol}} = 0.08 \text{ moles} \]

### Step 3: Determine the dissociation of HNO3 and NaOH
- HNO3 dissociates to give 0.24 moles of \(H^+\) ions.
- NaOH dissociates to give 0.08 moles of \(OH^-\) ions.

### Step 4: Set up the neutralization reaction
For neutralization:
\[ H^+ + OH^- \rightarrow H_2O \]
To achieve neutrality, the number of moles of \(H^+\) must equal the number of moles of \(OH^-\).

### Step 5: Find the volume ratio for neutralization
Let \(V_1\) be the volume of HNO3 solution and \(V_2\) be the volume of NaOH solution. The moles of \(H^+\) and \(OH^-\) can be expressed as:
\[ \text{Moles of } H^+ = 0.24 \times \frac{V_1}{1 \text{ L}} = 0.24 V_1 \]
\[ \text{Moles of } OH^- = 0.08 \times \frac{V_2}{1 \text{ L}} = 0.08 V_2 \]

For neutralization:
\[ 0.24 V_1 = 0.08 V_2 \]

### Step 6: Solve for the volume ratio
Rearranging the equation gives:
\[ \frac{V_1}{V_2} = \frac{0.08}{0.24} = \frac{1}{3} \]
Thus, the volume ratio of HNO3 to NaOH is:
\[ \text{Volume ratio} = 1 : 3 \]

### Final Answer
The solutions must be mixed in a volume ratio of 1:3 to obtain a neutral solution.

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