10 mL of 2 M NaOH solution is added to 20 mL of 1 M HCl solution kept in a beaker. Now, 10 mL of this…

10 mL of 2 M NaOH solution is added to 20 mL of 1 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 made the volume upto the mark with distilled water. The solution in this flask is :
  1. 0.2 M NaCl solution
  2. 20 M HCl solution
  3. 10 M HCl solution
  4. Neutral solution

Solution

When $10 \mathrm{ml}, 2 \mathrm{M} \mathrm{NaOH}$ solution is added to 20 ml of 1 M HCl solution :
$\mathrm{NaOH}+\mathrm{HCl} \rightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}$
Initial : MV $=2 \times 0.1 \quad \mathrm{MV}=1 \times 0.2$
$=0.2 \text { mole } \quad=0.2 \mathrm{~mole}$
Final $0 \quad 0$
$\therefore$ Resulting solution becomes neutral.
Now when 10 mol of above solution is poured into a flask containing 2 mole HCl and made solution 100 ml will distilled water.
Molarity of $\mathrm{HCl}=\frac{2}{100} \times 1000=20$

Asked in: JEE Main 2025 (03 Apr Shift 2)

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