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17.4 mL of unknown H2SO4 solution was titrated by 0.20 M of NaOH. 41.51 mL of that NaOH solution was consumed to exactly completely neutralize the acid. What is the original molarity (in M) of the unknown H2SO4 solution?

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Answer:

0.24 M

Step-by-step explanation:

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

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation,

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

The mole ratio of the base, NaOH (n₆) = 2

Finally, we shall determine the molarity of the acid. This can be obtained as follow:

Volume of acid, H₂SO₄ (Vₐ) = 17.4 mL

Molarity of base, NaOH (M₆) = 0.20 M

Volume of base, NaOH (V₆) = 41.51 mL

Molarity of acid, H₂SO₄ (Mₐ) =?

MₐVₐ / M₆V₆ = nₐ / n₆

Mₐ × 17.4 / 0.2 × 41.51 = 1 / 2

Mₐ × 17.4 / 8.302 = 1 / 2

Cross multiply

Mₐ × 17.4 × 2 = 8.302

Mₐ × 34.8 = 8.302

Divide both side by 34.8

Mₐ = 8.302 / 34.8

Mₐ = 0.24 M

Therefore, the molarity of the acid is 0.24 M

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