Question

100 ml of 10 M HCl is mixed with 75 ml of 10 M Na$_2$CO$_3$. The resulting solution would be –

• A. acidic
• B. basic
• C. neutral
• D. amphoteric

Hint:

Always check the limiting reagent in acid–base neutralization problems. If base is left unreacted, the solution is basic; if acid is left unreacted, the solution is acidic.

Answer 1

The Correct Option is A

Step 1: Calculate moles of HCl.
Molarity (M) $\times$ Volume (L) = Number of moles
\[ 10 \times 0.100 = 1.0 \; \text{mol HCl} \]

Step 2: Calculate moles of Na$_2$CO$_3$.
\[ 10 \times 0.075 = 0.75 \; \text{mol Na$_2$CO$_3$} \]

Step 3: Write the neutralization reaction.
\[ Na_2CO_3 + 2HCl \;\; \rightarrow \;\; 2NaCl + H_2O + CO_2 \] From the equation: - 1 mole Na$_2$CO$_3$ requires 2 moles HCl.
- 0.75 mole Na$_2$CO$_3$ would require: \[ 0.75 \times 2 = 1.5 \; \text{mol HCl} \]

Step 4: Compare available and required HCl.
- Available HCl = 1.0 mol
- Required HCl = 1.5 mol
- Thus, HCl is limiting and all of it will be consumed.

Step 5: Find excess reagent left.
- Na$_2$CO$_3$ left unreacted = \( 0.75 - 0.50 = 0.25 \; \text{mol} \)
(Since 1.0 mol HCl reacts with 0.50 mol Na$_2$CO$_3$).
So, after reaction, unreacted Na$_2$CO$_3$ remains in solution.

Step 6: Nature of the solution.
Unreacted Na$_2$CO$_3$ is basic in nature because it hydrolyzes to give OH$^-$ ions. Therefore, the resulting solution will be basic.
Correction: On checking again, the leftover species is Na$_2$CO$_3$, which is alkaline. Thus the correct nature is basic, not acidic.
\[ \boxed{\text{Basic}} \]