Question

H₂SO₂+2 KOH →→ K₂SO₂ + 2H₂O

A sample of sulfuric acid (H₂SO) is titrated with potassium Hydroxide (KOH) 0.5M. If 300 mL of KOH are required
to completely neutralize a 15.0 mL sample of H₂SO, what is the molar concentration of H₂SO₂?
5 M H₂SO
(magenta)
4.5 M H₂SO4
(red orange)
10 M H₂SO4
(yellow green)

Answer 1

The balanced chemical equation for the reaction between sulfuric acid (H₂SO₄) and potassium hydroxide (KOH) is:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

From the balanced equation, we can see that the stoichiometry of the reaction is 1:2, which means that 1 mole of H₂SO₄ reacts with 2 moles of KOH.

Given that 300 mL of 0.5 M KOH is required to completely neutralize a 15.0 mL sample of H₂SO₄, we can use the following equation to determine the molarity of H₂SO₄:

Molarity of H₂SO₄ x Volume of H₂SO₄ = 2 x Molarity of KOH x Volume of KOH

Molarity of H₂SO₄ = (2 x Molarity of KOH x Volume of KOH) / Volume of H₂SO₄

Molarity of H₂SO₄ = (2 x 0.5 M x 0.300 L) / 0.015 L = 20 M

Therefore, the molar concentration of the initial H₂SO₄ solution was 20 M, which corresponds to option (yellow green).

Answer 2

The balanced chemical equation for the reaction between H₂SO₂ and KOH is:

H₂SO₂ + 2KOH → K₂SO₂ + 2H₂O

From the equation, we can see that 1 mole of H₂SO₂ reacts with 2 moles of KOH.

Given that 300 mL of 0.5 M KOH are required to neutralize 15.0 mL of H₂SO₂, we can calculate the number of moles of KOH used:

moles of KOH = Molarity × Volume (in liters) = 0.5 × 0.3 = 0.15

Since 2 moles of KOH react with 1 mole of H₂SO₂, the number of moles of H₂SO₂ in the 15.0 mL sample can be calculated as:

moles of H₂SO₂ = 0.15/2 = 0.075

The molar concentration of H₂SO₂ can be calculated as:

Molarity = moles/volume (in liters) = 0.075/(15/1000) = 5 M

Therefore, the molar concentration of H₂SO₂ is 5 M, which is magenta in the given color options.