How many milliliters of 0.350 M H2SO4 are required to neutralize 60.0 mL of 0.300 M NaOH?

25.7 mL of H₂SO₄

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

The mole ratio of acid, H₂SO₄ (nₐ) = 1

The mole ratio of base, NaOH (n₆) = 2

Finally, we shall determine the volume of H₂SO₄ needed for the reaction. This can be obtained as shown below:

Molarity of acid, H₂SO₄ (Mₐ) = 0.350 M

Volume of base, NaOH (V₆) = 60 mL

Molarity of base, NaOH (M₆) = 0.300 M

Volume of acid, H₂SO₄ (Vₐ) = ?

MₐVₐ / M₆V₆ = nₐ/n₆

0.350 × Vₐ / 0.3 × 60 = 1/2

0.350 × Vₐ / 18 = 1/2

Cross multiply

0.350 × 2 × Vₐ = 18

0.7 × Vₐ = 18

Divide both side by 0.7

Vₐ = 18 / 0.7

Vₐ = 25.7 mL

Thus, 25.7 mL of H₂SO₄ is needed for the reaction.

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