Answer to Question #103760 in Chemistry for Nashaly

Question

What is the molarity of a 86.16 mL sulfuric acid solution (H2SO4) which is just neutralized by 178.10 mL of 0.19 M sodium hydroxide (NaOH)?

Answer 1

The reaction of neutralization is described as following:

2NaOH + H₂SO₄ = Na₂SO4 + 2H₂O

As 178.10 mL of 0.19 M sodium hydroxide was used:

n(NaOH) = c(NaOH) × V(NaOH),

where n - number of moles, c - molar concentration, V - volume.

From here:

n(NaOH) = 0.19 M × 178.10 mL = 0.19 M × 0.17810 L = 0.034 mol.

As a result, 0.034 mol of soidum hydroxide was used. According to the reaction, the number of moles of neutralized sulfuric acid equals:

n(H₂SO₄) = 0.034 mol / 2 = 0.017 mol

Finally, the molarity the molarity of a 86.16 mL sulfuric acid solution containing 0.017 mol is:

c(H₂SO₄) = n(H₂SO₄) / V(H₂SO₄) = 0.017 mol / 86.16 mL = 0.017 mol / 0.08616 L = 0.197 L = 197 mL

Answer: 197 mL