Final answer:
To find the volume of 0.425 M NaOH needed to neutralize 12.0 mL of 0.0642 M H2SO4, we calculate the moles of H2SO4, determine the moles of NaOH needed according to the stoichiometry, and then use the molarity of NaOH to find the volume. The closest option given is 40.0 mL, but the calculated volume is 3.621882 mL.
Step-by-step explanation:
To determine the volume of 0.425 M NaOH needed to neutralize 12.0 mL of 0.0642 M H2SO4, we'll use the concept of stoichiometry and the balanced neutralization reaction:
H2SO4 (aq) + 2NaOH(aq) → Na2SO4 (aq) + 2H2O (l)
First, we calculate the moles of H2SO4:
Moles of H2SO4 = Molarity of H2SO4 × Volume of H2SO4 = (0.0642 mol/L) × (0.012 L) = 0.0007704 mol.
According to the balanced equation, 2 moles of NaOH are required to neutralize 1 mole of H2SO4, hence the moles of NaOH needed are:
Moles of NaOH = 2 × Moles of H2SO4 = 2 × 0.0007704 mol = 0.0015408 mol.
Next, calculate the volume of NaOH solution using its molarity:
Volume of NaOH = Moles of NaOH ÷ Molarity of NaOH = 0.0015408 mol ÷ (0.425 mol/L) = 0.003621882 L,
which can be converted into milliliters:
Volume of NaOH = 3.621882 mL.
From the options provided, none directly match the calculated volume, so it's likely that there is an error in the question or the options given. The closest option is 40.0 mL, but the correct volume calculated is significantly less.