Question

One litre of a solution contains 18.9 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?

• A. 3:8
• B. 8:3
• C. 15:4
• D. 4:15

Answer 1

Answer: (D)

4:15

Answer 2

1. Calculate molar masses: $HNO_3$ (63 g/mol), NaOH (40 g/mol).
2. Determine moles per litre for each solution: $HNO_3$ (18.9 g / 63 g/mol = 0.3 mol/L), NaOH (3.2 g / 40 g/mol = 0.08 mol/L).
3. Calculate normality: $HNO_3$ is monoprotic ($n=1$), so $N_1 = 0.3$ N. NaOH is monobasic ($n=1$), so $N_2 = 0.08$ N.
4. Apply neutralization condition $N_1V_1 = N_2V_2$.
5. Solve for the volume ratio $V_1:V_2$.

Calculations Molar mass of $HNO_3 = 1 (\text{H}) + 14 (\text{N}) + 3 \times 16 (\text{O}) = 63$ g/mol. Molar mass of NaOH $= 23 (\text{Na}) + 16 (\text{O}) + 1 (\text{H}) = 40$ g/mol.

Concentration of $HNO_3$ solution = $\frac{18.9 \text{ g}}{1 \text{ L}} = 0.3$ mol/L. Concentration of NaOH solution = $\frac{3.2 \text{ g}}{1 \text{ L}} = 0.08$ mol/L.

Since $HNO_3$ is a monoprotic acid ($n=1$), its normality ($N_1$) is: $N_1 = \text{Molarity} \times n = 0.3 \text{ mol/L} \times 1 = 0.3$ N.

Since NaOH is a monobasic base ($n=1$), its normality ($N_2$) is: $N_2 = \text{Molarity} \times n = 0.08 \text{ mol/L} \times 1 = 0.08$ N.

For a neutral solution, the milliequivalents of acid must equal the milliequivalents of base: $N_1 V_1 = N_2 V_2$

Substituting the normality values: $0.3 \times V_1 = 0.08 \times V_2$

To find the volume ratio $V_1 : V_2$, we rearrange the equation: $\frac{V_1}{V_2} = \frac{0.08}{0.3}$ $\frac{V_1}{V_2} = \frac{8}{30} = \frac{4}{15}$ The volume ratio in which the $HNO_3$ solution and the NaOH solution must be mixed is $4:15$.