Question

20 mL of 0.1 M HCl is added to 30 mL of 0.1 M NaOH. To this solution, extra 50 mL of water was added. What is the molarity of the final solution formed?

• A. 0.1 M
• B. 0.01 M
• C. 0.5 M
• D. 0.05 M

Hint:

In neutralization problems, always: 1. Calculate initial moles of acid and base. 2. Identify the limiting reagent. 3. Account for the total volume after mixing and dilution.

Answer 1

The Correct Option is B

To determine the molarity of the final solution, we follow these steps: Calculate moles of HCl and NaOH: - Moles of HCl = $ 20 \text{ mL} \times 0.1 \text{ M} = 2 \text{ mmol} $ - Moles of NaOH = $ 30 \text{ mL} \times 0.1 \text{ M} = 3 \text{ mmol} $ Neutralization reaction: \[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \] - HCl is the limiting reagent (2 mmol reacts completely). - Remaining NaOH = $ 3 \text{ mmol} - 2 \text{ mmol} = 1 \text{ mmol} $. Total volume of final solution: \[ 20 \text{ mL} + 30 \text{ mL} + 50 \text{ mL} = 100 \text{ mL} \] Molarity of remaining NaOH: \[ \text{Molarity} = \frac{1 \text{ mmol}}{100 \text{ mL}} = 0.01 \text{ M} \] Thus, the molarity of the final solution is (2).

Answer 2

Step 1: Calculate the moles of HCl and NaOH before mixing.
- Moles of HCl = Molarity × Volume = \(0.1 \, \text{M} \times 0.020 \, \text{L} = 0.002 \, \text{mol}\).
- Moles of NaOH = \(0.1 \, \text{M} \times 0.030 \, \text{L} = 0.003 \, \text{mol}\).

Step 2: Determine the reaction between HCl and NaOH.
HCl and NaOH react in a 1:1 ratio:
\[ \text{HCl} + \text{NaOH} \to \text{NaCl} + \text{H}_2\text{O}. \]
- Since moles of NaOH (0.003 mol) > moles of HCl (0.002 mol), all HCl will react.
- Remaining moles of NaOH = \(0.003 - 0.002 = 0.001 \, \text{mol}\).

Step 3: Calculate the total volume after mixing and dilution.
Total volume = \(20 \, \text{mL} + 30 \, \text{mL} + 50 \, \text{mL} = 100 \, \text{mL} = 0.1 \, \text{L}\).

Step 4: Calculate the molarity of the final solution.
Since only NaOH remains unreacted:
\[ \text{Molarity} = \frac{\text{moles of NaOH}}{\text{total volume}} = \frac{0.001}{0.1} = 0.01 \, \text{M}. \]

Step 5: Conclusion.
The molarity of the final solution is 0.01 M.