Question

Two acids A and B are titrated separately. 25 mL of 0.5 M Na$_2$CO$_3$ solution requires 10 mL of A and 40 mL of B for complete neutralisation. The volume (in L) of A and B required to produce 1 L of 1 N acid solution respectively are

• A. 0.2, 0.8
• B. 0.8, 0.2
• C. 0.3, 0.7
• D. 0.7, 0.3

Hint:

Titration: Milliequivalents of acid = Milliequivalents of base.

Answer 1

The Correct Option is A

For complete neutralization, the milliequivalents of acid must equal the milliequivalents of base.
Milliequivalents of Na$_2$CO$_3$ = $25 \, \text{mL} \times 0.5 \, \text{M} \times 2$ (since Na$_2$CO$_3$ has 2 replaceable H$^+$) = $25$ milliequivalents.
For acid A: Milliequivalents of A = $10 \, \text{mL} \times N_A$, where $N_A$ is the normality of A. Since milliequivalents are equal, $10 N_A = 25$, so $N_A = 2.5 \, \text{N}$. To make 1 L of 1 N solution, we need $V_A$ liters of A such that $V_A \times 2.5 \, \text{N} = 1 \, \text{L} \times 1 \, \text{N}$. Thus, $V_A = 0.4 \, \text{L}$ or 400 mL. (The options seem inconsistent with this calculation. Perhaps there's a misunderstanding about what "produce 1L of 1N acid" means or a typo in the question/options). This seems to be an issue with the provided options, as they don't match the calculated result. It might be beneficial to double check the question and options to ensure accuracy.
For acid B: Milliequivalents of B = $40 \, \text{mL} \times N_B$. So, $40 N_B = 25$, and $N_B = 0.625 \, \text{N}$. To make 1 L of 1 N solution, we need $V_B$ liters of B such that $V_B \times 0.625 = 1 \times 1$, thus $V_B = 1.6 \, \text{L}$ or 1600 mL. (Again, the options provided don't match the calculation).