Question

What volume of a 0.00945-M solution of potassium hydroxide would be required to titrate 50.00 mL of a sample of acid rain with an H₂SO₄ concentration of 1.23 × 10^(-4) M. H₂SO₄(aq)+2KOH(aq)⟶K₂SO₄(aq)+2H₂O(l)

a) 15.75 mL
b) 23.62 mL
c) 31.50 mL
d) 39.37 mL

Answer 1

The calculated volume of 0.00945-M KOH solution required to titrate a 50.00 mL sample of 1.23 × 10^-4 M H2SO4 is 1.3 mL, which does not match the provided options.

Step-by-step explanation:

To calculate the volume of 0.00945-M potassium hydroxide solution needed to titrate a 50.00 mL sample of acid rain containing 1.23 × 10-4 M sulfuric acid (H2SO4), we use the stoichiometry of the balanced reaction:

H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l)

Since the molar ratio of H2SO4 to KOH is 1:2, we need double the amount of moles of KOH to react with the given moles of sulfuric acid:

Moles of H2SO4 = Molarity of H2SO4 × Volume of acid rain in liters

Moles of H2SO4 = (1.23 × 10-4 M) × (50.00 mL × (1 L / 1000 mL))
= 6.15 × 10-6 moles

Moles of KOH required = 2 × Moles of H2SO4
= 2 × (6.15 × 10-6 moles)
= 1.23 × 10-5 moles

Now, to find the volume of KOH needed:

Volume of KOH = Moles of KOH / Molarity of KOH

Volume of KOH = (1.23 × 10-5 moles) / (0.00945 M)
= 1.3 × 10-3 L
= 1.3 mL

This calculated volume does not match any of the provided answer choices, indicating a possible error in the question or choices.