Final answer:
The balanced equation for the reaction of sulfuric acid with sodium hydroxide is H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l). To find the molarity of sulfuric acid in rainwater, we use the given titration data, which, after calculations, reveals the molarity to be approximately 0.00345 M.
Step-by-step explanation:
In the case of titrating acid rain that contains sulfuric acid, the balanced chemical equation for the reaction with sodium hydroxide (NaOH) would be:
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
Since sulfuric acid is a diprotic acid, it releases two protons (H+ ions), the titration requires two hydroxide ions from NaOH for each sulfate ion to form water and sodium sulfate.
Using the provided titration data, we can calculate the molarity of the sulfuric acid in the rainwater. A 20.0 mL sample of acid rain requires 1.7 mL of 0.0811 M NaOH.
- Moles of NaOH = Volume (L) × Molarity = 0.0017 L × 0.0811 M = 1.3787 × 10-4 moles
- According to the equation, 1 mole of sulfuric acid reacts with 2 moles of NaOH. Therefore, moles of H2SO4 = 1.3787 × 10-4 moles / 2 = 6.8935 × 10-5 moles
- The molarity of H2SO4 = Moles / Volume (L) = 6.8935 × 10-5 moles / 0.020 L = 0.00344675 M
Therefore, the molarity of the rainwater is approximately 0.00345 M, indicating a relatively low concentration of sulfuric acid.