Question

50.0 mL of 0.10 M HCl(aq) is titrated with 0.20 M NaOH (aq). What is the volume of NaOH added at the equivalence point?

50.0 mL
75.0 mL
25.0 mL

Answer 1

At the equivalence point, 25.0 mL of 0.20 M NaOH is required to neutralize 50.0 mL of 0.10 M HCl, as they react in a 1:1 molar ratio.

Step-by-step explanation:

To determine the volume of NaOH needed at the equivalence point when titrating with HCl, one should utilize the concept of molarity and volume from the reaction stoichiometry. In this scenario, HCl and NaOH neutralize each other in a 1:1 molar ratio, according to the balanced equation:

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Given the initial conditions of 50.0 mL of 0.10 M HCl, we have:

• Moles of HCl = Molarity of HCl × Volume of HCl in liters = 0.10 moles/L × 0.0500 L = 0.0050 moles.

Since the titration is at the equivalence point, the molar amount of NaOH needed equals the molar amount of HCl. With a molarity of 0.20 M for NaOH, the volume of NaOH can be calculated using:

• Moles of NaOH = Molarity of NaOH × Volume of NaOH in liters
• 0.0050 moles = 0.20 mol/L × Volume of NaOH in liters
• Volume of NaOH = 0.0050 moles / 0.20 mol/L = 0.0250 L (or 25.0 mL)

Therefore, at the equivalence point, 25.0 mL of 0.20 M NaOH is needed to neutralize 50.0 mL of 0.10 M HCl.