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…

Step 1 — Neutralisation in the beaker

Moles of $\ce{NaOH} = 2 \times 0.010 = 0.02$ mol; Moles of $\ce{HCl} = 1 \times 0.020 = 0.02$ mol

Both completely neutralise: $\ce{NaOH + HCl -> NaCl + H2O}$

Result: 0.02 mol $\ce{NaCl}$ in 30 mL — neutral solution.

Step 2 — Transfer to flask

10 mL of the neutral $\ce{NaCl}$ solution is poured into the flask. This contains no excess acid or base.

Step 3 — Contents of 100 mL flask

The flask already contains 2 mol $\ce{HCl}$. The added neutral solution does not consume any $\ce{HCl}$.

Total $\ce{HCl}$ = 2 mol in 100 mL = 0.1 L

Step 4 — Molarity

$M = \tfrac{2}{0.1} = 20$ M $\ce{HCl}$

Key Points to Remember:

• Equal moles of strong acid and base completely neutralise each other
• The neutral $\ce{NaCl}$ solution does not react with $\ce{HCl}$ in the flask
• Molarity = moles $\div$ volume (L)