Question

What volume (in ml) of 0.2 $M$ H$_2$SO$_4$ solution should be mixed with the 40 ml of 0.1 $M$ NaOH solution such that the resulting solution has the concentration of H$_2$SO$_4$ as $\frac{6}{55}M$?

• A. 70
• B. 45
• C. 30
• D. 58

Answer 1

Answer: (A)

70

Answer 2

Here's a step-by-step explanation:

1. Calculate initial moles of NaOH: $0.1 M \times 40 ml = 4 mmol$.
2. Let the volume of H$_2$SO$_4$ be $V$ ml. Initial moles of H$_2$SO$_4$: $0.2 M \times V ml = 0.2V mmol$.
3. The reaction is H$_2$SO$_4$ + 2NaOH $\rightarrow$ Na$_2$SO$_4$ + 2H$_2$O.
4. Since the final solution contains H$_2$SO$_4$, it is in excess, and NaOH is limiting.
5. 4 mmol of NaOH reacts with $4/2 = 2 mmol$ of H$_2$SO$_4$.
6. Moles of H$_2$SO$_4$ remaining = Initial moles - Reacted moles = $(0.2V - 2) mmol$.
7. Total volume of the solution = $(V + 40) ml$.
8. Final concentration of H$_2$SO$_4 = \frac{\text{Moles remaining}}{\text{Total volume}} = \frac{0.2V - 2}{V + 40}$.
9. Given final concentration is $\frac{6}{55}M$. So, $\frac{0.2V - 2}{V + 40} = \frac{6}{55}$.
10. Solve for V: $55(0.2V - 2) = 6(V + 40) \Rightarrow 11V - 110 = 6V + 240 \Rightarrow 5V = 350 \Rightarrow V = 70$.
11. Check: If $V=70$, initial H$_2$SO$_4 = 14 mmol$. Initial NaOH = 4 mmol. NaOH is limiting as $14/1 > 4/2$. This confirms the assumption.

Therefore, the answer is 70 ml.