Final answer:
To neutralize 50.0 mL of 0.150 M NaOH, 62.5 mL of 0.120 M HClO₃ is required.
Step-by-step explanation:
The volume of 0.120 M HClO₃ required to neutralize 50.0 mL of 0.150 M NaOH can be found through a stoichiometric calculation. Since HClO₃ and NaOH react in a 1:1 molar ratio (as both are strong acid and base, respectively), we use the equation:
molarity HClO₃ × volume HClO₃ = molarity NaOH × volume NaOH
Substituting the given values:
0.120 M × volume HClO₃ = 0.150 M × 50.0 mL
Now we solve for volume HClO₃:
volume HClO₃ = (0.150 M × 50.0 mL) / 0.120 M
volume HClO₃ = 62.5 mL
Therefore, 62.5 mL of 0.120 M HClO₃ is required to neutralize 50.0 mL of 0.150 M NaOH.