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

To solve the question, let's analyze the steps involved in the whole process:

Determine the reaction between NaOH and HCl in the initial mixture:

• We have 10 mL of 2 M NaOH. Therefore, moles of NaOH = \(0.01 \, \text{L} \times 2 \, \text{mol/L} = 0.02 \, \text{mol}\).
• We have 20 mL of 1 M HCl. Therefore, moles of HCl = \(0.02 \, \text{L} \times 1 \, \text{mol/L} = 0.02 \, \text{mol}\).
• NaOH and HCl react in a 1:1 ratio: \(\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}\).
• Both reactants have the same number of moles (0.02 mol each), so they will completely neutralize each other, resulting in a neutral solution of NaCl with no excess HCl or NaOH.

Consider what happens when 10 mL of this neutral solution is added to a volumetric flask containing HCl:

• The mixture is poured into a 100 mL volumetric flask containing 2 moles of HCl.
• Since the initial solution was neutral (containing only NaCl and water), adding it does not change the HCl concentration already in the flask.

Calculate the final concentration of HCl in the volumetric flask:

• Total volume in the flask is 100 mL or 0.1 L.
• Moles of HCl initially in the flask = 2 mol.
• Concentration of HCl in the flask = \(\frac{\text{Number of moles}}{\text{Volume in L}} = \frac{2 \, \text{mol}}{0.1 \, \text{L}} = 20 \, \text{M}\).

From this calculation, we see that the solution in the flask becomes a 20 M HCl solution.

Thus, the correct answer is: 20 M HCl solution.

Answer 2

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.