A 12.00 mL sample of sulfuric acid from an automobile battery requires 34.62 mL of 2.42 M sodium hydroxide solution for complete neutralization. What is the molarity of the sulfuric acid?
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O
molarity NaOH x liters NaOH = mols NaOH.
mols H2SO4 = 1/2 mols NaOH
molarity H2SO4 x L H2SO4 = mols H2SO4
solve for molarity H2SO4.
(2.42 mol NaOH / 1000 mL0 (34.62 mL) (1 mol H2SO4 / 2 mol NaOH) / (12.00 mL / 1000mL) = 3.49 M
right.
Yes, you are correct. To calculate the molarity of the sulfuric acid, you can use the stoichiometry of the neutralization reaction and the relationship between moles, volume, and molarity.
First, determine the moles of sodium hydroxide (NaOH) used in the reaction. To do this, multiply the molarity of NaOH (2.42 M) by the volume of NaOH used (34.62 mL):
(2.42 mol NaOH / 1000 mL) * 34.62 mL = 0.08317 mol NaOH
Since the balanced equation shows that it takes 2 moles of NaOH to react with 1 mole of sulfuric acid (H2SO4), divide the moles of NaOH by 2 to find the moles of H2SO4:
0.08317 mol NaOH / 2 = 0.04158 mol H2SO4
Next, calculate the molarity of the sulfuric acid by dividing the moles of H2SO4 by the volume of the sulfuric acid sample in liters:
(0.04158 mol H2SO4) / (12.00 mL / 1000 mL) = 3.49 M
So, the molarity of the sulfuric acid is 3.49 M.