If an unknown sample contains 39.04% sulfuric acid by mass, then a 0.9368 g of that sample would require mL of 0.2389 M NaOH for neutralization. A) 31.22
B) 39.98
C) 7.80
D) 79.96
E) 15.61

Please give an answer and an explanation, thanks!

A) 31.22

Explanation:

The reaction of sulfuric acid with NaOH is:

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2H₂O

To solve this problem we need to determine the moles of acid that will react, and, using the chemical equation we can determine the moles of NaOH and the volume that a 0.2389M NaOH solution would require to neutralize it.

Moles H₂SO₄ (Molar mass: 98.08g/mol):

0.9368g * 39.04% = 0.3657g H₂SO₄ * (1mol / 98.08g) =

3.7289x10⁻³moles H₂SO₄

And moles of NaOH that you require to neutralize the acid are:

3.7289x10⁻³moles H₂SO₄ * (2 moles NaOH / 1 mole H₂SO₄) =

7.4578x10⁻³ moles NaOH

Using a 0.2389M NaOH solution:

7.4578x10⁻³ moles NaOH * (1L / 0.2389mol) = 0.03122L = 31.22mL

Right answer is:

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