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Chemistry Question 54 – JEE-MAIN 2025

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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 :

First, determine the nature and amount of species present after the initial acid-base reaction.

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Ninja Strategy•Magnitude Check

Recognize that 2 moles of HCl in 100 mL will result in an extremely high HCl concentration, immediately eliminating options suggesting neutrality or a dominant NaCl presence.

Step 1: Analyze the initial reaction in the beaker✦ Active

Calculate the moles of NaOH and HCl in the initial mixture:

Moles of NaOH=0.010 L×2 M=0.02 mol Moles of HCl=0.020 L×1 M=0.02 mol

Since the moles are equal, a complete neutralization occurs, forming 0.02 mol of NaCl. The total volume of this mixture is 10 mL+20 mL=30 mL. The concentration of NaCl in this mixture is 0.02 mol0.030 L=23 M.

Step 2: Calculate moles transferred to the volumetric flask○ Expand

When 10 mL of the initial mixture is transferred to the volumetric flask, the moles of NaCl transferred are:

Moles of NaCl transferred=23 M×0.010 L=0.023 mol

The volumetric flask already contains 2 moles of HCl.

Step 3: Determine the final composition and concentration in the volumetric flask○ Expand

The final solution in the 100 mL volumetric flask contains 2 moles of HCl and 0.023 moles of NaCl. The total volume is 100 mL=0.100 L. The concentration of HCl in the flask is:

Concentration of HCl=2 mol0.100 L=20 M

The concentration of NaCl is 0.02/3 mol0.100 L=0.020.3 M≈0.067 M. The solution is predominantly an HCl solution with a concentration of 20 M.

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