Answer to Question #137516 in Chemistry for Jan

Question

Sulfuric acid is needed in in a lead storage battery. A 25.00 mL sample of “battery acid” taken from the functioning battery requires 53.7 mL of 4.552 M NaOH for complete neutralization. What is the molarity of the battery acid?

Answer 1

Molarity of the battery acid = C_M(H₂SO₄) = unknown

Volume of battery acid = V(H₂SO₄) = 25.00 mL

Molarity of sodium hydroxide = C_M(NaOH) = 4.552 M

Volume of sodium hydroxide = V(NaOH) = 53.70 mL

Solution:

The reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH) is:

H₂SO₄ + 2NaOH = Na₂SO₄ + 2H₂O

According to the equation: n(H₂SO₄) = n(NaOH)/2

Therefore,

2 × C_M(H₂SO₄) × V(H₂SO₄) = C_M(NaOH) × V(NaOH)

2 × C_M(H₂SO₄) × (25.00 mL) = (4.552 M) × (53.70 mL)

C_M(H₂SO₄) = (4.552 M × 53.70 mL) / (2 × 25.00 mL) = 4.8888 M = 4.889 M

Molarity of the battery acid = C_M(H₂SO₄) = 4.889 M.

Answer: The molarity of the battery acid is 4.889 M.