Question

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 :

• A. 0.2 M NaCl solution
• B. 20 M HCl solution
• C. 10 M HCl solution
• D. Neutral solution

Hint:

First, determine the composition of the mixture after the first reaction. Then, consider what is added to the volumetric flask and calculate the final concentration of HCl in the flask, taking into account the total volume.

Answer 1

The Correct Option is B

First, let's analyze the reaction between NaOH and HCl in the beaker: Moles of NaOH = Molarity × Volume (in L) = 2 M × (10 / 1000) L = 0.02 moles Moles of HCl = Molarity × Volume (in L) = 1 M × (20 / 1000) L = 0.02 moles The reaction is: \[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \] From the stoichiometry, 1 mole of NaOH reacts with 1 mole of HCl. 
Since we have 0.02 moles of each, they will completely neutralize each other. 
The resulting solution in the beaker will contain NaCl and water, and will be neutral. 
Now, 10 mL of this neutral solution (containing NaCl and water) is poured into a 100 mL volumetric flask containing 2 moles of HCl. The volume is then made up to 100 mL with distilled water. 
The amount of HCl already present in the flask is 2 moles. The addition of 10 mL of the neutral solution from the beaker does not add any significant amount of HCl. 
The total volume of the solution in the flask is 100 mL = 0.1 L. 
The molarity of HCl in the flask is: \[ \text{Molarity of HCl} = \frac{\text{Moles of HCl}}{\text{Volume of solution in L}} = \frac{2 \, \text{moles}}{0.1 \, \text{L}} = 20 \, \text{M} \] 
The solution in the flask is 20 M HCl solution.