Question

What volume of a 3.5 M HCl solution should be added to a 200 mL 2.4 M HCl solution and finally diluted to 500 mL so that the molarity of solution becomes 1.24 M?

• A. 65 mL
• B. 50 mL
• C. 40 mL
• D. 90 mL

Answer 1

Answer: (C)

40 mL

Answer 2

Let $V$ be the volume (in mL) of the 3.5 M HCl solution to be added.

The number of moles of HCl in the initial 200 mL of 2.4 M solution is: $n_1 = \text{Molarity}_1 \times \text{Volume}_1 = 2.4 \, \text{mol/L} \times 200 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 2.4 \times 0.2 \, \text{mol} = 0.48 \, \text{mol}$.

The number of moles of HCl in $V$ mL of 3.5 M solution is: $n_2 = \text{Molarity}_2 \times \text{Volume}_2 = 3.5 \, \text{mol/L} \times V \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 3.5 \times \frac{V}{1000} \, \text{mol} = 0.0035V \, \text{mol}$.

When these two solutions are mixed, the total number of moles of HCl is the sum of the moles from each solution: $n_{total} = n_1 + n_2 = 0.48 + 0.0035V \, \text{mol}$.

This mixture is then diluted to a final volume of 500 mL, and the final molarity is 1.24 M. The total number of moles of HCl in the final solution is: $n_{final} = \text{Molarity}_{final} \times \text{Volume}_{final} = 1.24 \, \text{mol/L} \times 500 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 1.24 \times 0.5 \, \text{mol} = 0.62 \, \text{mol}$.

Since the total number of moles of solute remains constant during dilution, the total moles from the initial solutions must equal the total moles in the final solution: $n_{total} = n_{final}$ $0.48 + 0.0035V = 0.62$

Now, we solve for $V$: $0.0035V = 0.62 - 0.48$ $0.0035V = 0.14$ $V = \frac{0.14}{0.0035}$

To simplify the division, multiply the numerator and denominator by 10000: $V = \frac{0.14 \times 10000}{0.0035 \times 10000} = \frac{1400}{35}$ $V = \frac{140 \times 10}{35} = \frac{(4 \times 35) \times 10}{35} = 4 \times 10 = 40$.

The volume $V$ is in mL.

So, 40 mL of the 3.5 M HCl solution should be added.