Question

10 mL of \(2\,\text{M}\) NaOH solution is added to 20 mL of \(1\,\text{M}\) HCl solution kept in a beaker. Now, 10 mL of this mixture is poured into a volumetric flask of 100 mL containing 2 moles of HCl and the volume is made up to the mark with distilled water. The solution in this flask is:

• A. \(0.2\,\text{M}\) NaCl solution
• B. \(20\,\text{M}\) HCl solution
• C. \(10\,\text{M}\) HCl solution
• D. Neutral solution

Hint:

Always track reactions stepwise:
First complete neutralization reactions.
Then consider dilution and additional reagents.
Salts like NaCl do not affect acid concentration.

Answer 1

The Correct Option is B

Concept:

Acid–base neutralization depends on the number of moles, not volume.
After neutralization, remaining species decide the nature of the solution.
Dilution does not change the number of moles, only concentration.
Step 1: Calculate moles of NaOH and HCl mixed initially. \[ \text{Moles of NaOH} = 2 \times \frac{10}{1000} = 0.02\,\text{mol} \] \[ \text{Moles of HCl} = 1 \times \frac{20}{1000} = 0.02\,\text{mol} \]
Step 2: Neutralization in the beaker. Since moles of NaOH = moles of HCl, complete neutralization occurs. \[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \] Moles of NaCl formed \(= 0.02\,\text{mol}\) Total volume of mixture \(= 30\,\text{mL}\)
Step 3: Amount of NaCl in 10 mL of this mixture. \[ \text{Moles of NaCl in 10 mL} = \frac{10}{30} \times 0.02 = 0.00667\,\text{mol} \]
Step 4: Contents of volumetric flask. The flask already contains \(2\,\text{mol}\) of HCl. NaCl does not react with HCl, so final moles of HCl remain: \[ 2\,\text{mol} \]
Step 5: Final concentration after dilution to 100 mL. \[ \text{Molarity of HCl} = \frac{2}{0.1} = 20\,\text{M} \]