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.