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: 1. Always identify the limiting reagent first 2. Total volume includes all added liquids (reactants + solvents) 3. Molarity = $\frac{\text{moles of solute}}{\text{total volume in liters}}$

Answer 1

The Correct Option is B

Calculate initial moles: $\text{Moles of HCl} = 20\,\text{mL}$$ \times 0.1\,\text{M} = 2\,\text{mmol} $
$\text{Moles of NaOH} = 30\,\text{mL} \times 0.1\,\text{M} = 3\,\text{mmol}$
Neutralization reaction: \[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \] - HCl is limiting (2 mmol reacts completely)
- Excess NaOH = $3\,\text{mmol} - 2\,\text{mmol} = 1\,\text{mmol}$ Total volume after dilution: \[ 20\,\text{mL (HCl)} + 30\,\text{mL (NaOH)} + 50\,\text{mL (water)} = 100\,\text{mL} \] Final molarity calculation: \[ \text{Molarity} = \frac{1\,\text{mmol}}{100\,\text{mL}} = 0.01\,\text{M} \] Thus, the correct answer 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.