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

To solve this problem, we need to evaluate the changes in the chemical species concentrations step-by-step as described in the scenario.

1. First, add 10 mL of a \(2\,\text{M}\) NaOH solution to 20 mL of a \(1\,\text{M}\) HCl solution:
   • Moles of NaOH = \(10\,\text{mL} \times \frac{2}{1000}\,\text{mol/mL} = 0.02\,\text{mol}\)
   • Moles of HCl = \(20\,\text{mL} \times \frac{1}{1000}\,\text{mol/mL} = 0.02\,\text{mol}\)
2. 10 mL of this neutral solution (essentially containing only NaCl and water) is poured into a volumetric flask that contains 2 moles of HCl and is then diluted to 100 mL:
3. The 10 mL added to the flask alters the overall concentration but does not change the moles of HCl already present as NaCl does not react with HCl. So, the final solution in the flask mostly consists of the HCl added separately.
4. Concentration of HCl in the flask after dilution:
   • Since we initially have 2 moles of HCl in a solution made up to 100 mL, the molarity (M) of HCl is calculated as follows:
   • Molarity \(M\) = \(\frac{\text{moles of HCl}}{\text{volume in L}}\) = \(\frac{2\,\text{mol}}{0.1\,\text{L}}\) = \(20\,\text{M}\)

The resulting solution in the flask is a \(20\,\text{M}\) HCl solution.