The reaction of H₂SO₄ with NaOH is represented by the equation
H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)
What volume of 0.752 M H₂SO₄ is required to neutralize 33.7 mL of 0.483 M NaOH?

Question

Quizplus · Accepted Answer

First, calculate the moles of NaOH using its concentration and volume: $0.483 \, \text{M} \times 0.0337 \, \text{L} = 0.01627 \, \text{moles}$. Since the reaction requires 2 moles of NaOH for every mole of $H_2SO_4$, the moles of $H_2SO_4$ needed are half of the NaOH moles: $0.01627 \, \text{moles} / 2 = 0.008135 \, \text{moles}$. To find the volume of $H_2SO_4$ required, use its concentration: $0.008135 \, \text{moles} / 0.752 \, \text{M} = 0.0108 \, \text{L}$ or $10.8 \, \text{mL}$.

The Reaction of H2SO4 with NaOH Is Represented by the Equation

The Reaction of H₂SO₄ with NaOH Is Represented by the Equation