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 :

Solution

Chemistry ma’am · 3 steps

The first mixing just makes neutral salt water. The 2 moles of HCl in the flask decide everything.

1. First mixture. NaOH: 10 mL × 2 M = 20 mmol. HCl: 20 mL × 1 M = 20 mmol. They neutralise each other exactly, so the beaker holds only 20 mmol of NaCl in 30 mL. It has no leftover acid or base.
2. The 10 mL portion. This portion carries 20/3 ≈ 6.7 mmol of NaCl, which is neutral. It adds no H⁺ to the flask, so it does not change the acid concentration.

   [HCl] = (2 mol)/(0.1 L) = 20 M

3. So. The flask holds 2 mol of HCl in 100 mL, which is a 20 M HCl solution. It also contains a little NaCl (about 0.067 M) as a spectator. The answer is (b).

The idea to keep. In multi-step mixing, track millimoles of acid and base at every stage. A salt from exact neutralisation is a spectator, and only the leftover or added H⁺ or OH⁻ sets the nature of the solution.

Answer from the NTA final answer key. Solution by PrepWiser.