Final answer:
To neutralize 32.7 ml of 0.767 M HCl, 59.58 ml of 0.421 M NaOH is needed, calculated by using the molarity and volume of HCl to find the moles and applying stoichiometry.
Step-by-step explanation:
Titration Calculation
To calculate the volume of 0.421 M NaOH needed to neutralize 32.7 ml of 0.767 M HCl in a titration, we first need to understand the reaction taking place:
HCl + NaOH → NaCl + H₂O
This is a neutralization reaction where 1 mole of HCl reacts with 1 mole of NaOH. Using the molarity and volume of HCl, we can find the moles of HCl:
# mol HCl = volume of HCl (L) × molarity of HCl (M) = 0.0327 L × 0.767 M = 0.0250829 mol HCl
Since the stoichiometry of the reaction is 1:1, we have the same number of moles of NaOH reacting with HCl. Next, we calculate the required volume of NaOH:
# mol NaOH = # mol HCl = 0.0250829 mol
Volume NaOH (L) = # mol NaOH ÷ molarity of NaOH (M) = 0.0250829 mol ÷ 0.421 M ≈ 0.05958 L = 59.58 ml
Therefore, 59.58 ml of 0.421 M NaOH is needed to neutralize 32.7 ml of 0.767 M HCl.