Final answer:
To neutralize 95.5 ml of a 0.336 M KOH solution, 11.30 milliliters of a 1.42 M H₂SO₄ solution are required according to stoichiometric calculations based on the balanced equation 2 KOH + H₂SO₄ → K₂SO₄ + 2 H₂O.
Step-by-step explanation:
The question relates to a stoichiometry problem involving a neutralization reaction between sulfuric acid (H₂SO₄) and potassium hydroxide (KOH). According to the balanced chemical equation 2 KOH(aq) + H₂SO₄(aq) → K₂SO₄(aq) + 2 H₂O(l), it takes two moles of KOH to neutralize one mole of H₂SO₄.
Firstly, calculate the number of moles of KOH needed to neutralize the H₂SO₄:
n(KOH) = 95.5 ml × 0.336 mol/L = 0.032082 mol
As the ratio of KOH to H₂SO₄ is 2:1, we need half the number of moles of H₂SO₄ to react with the given amount of KOH, which is 0.016041 mol of H₂SO₄.
Now, we can calculate the volume of the 1.42 M H₂SO₄ solution needed:
V(H₂SO₄) = 0.016041 mol / 1.42 mol/L = 0.01130352 L,
which is equivalent to 11.30 ml after converting liters to milliliters (1 L = 1000 mL).
Therefore, 11.30 milliliters of a 1.42 M H₂SO₄ solution are needed to neutralize 95.5 ml of a 0.336 M KOH solution.