Answer :
Final answer:
To find the molarity of the H₂SO₄ solution, calculate the moles of KOH used, then determine the moles of H₂SO₄. Since the reaction ratio is 1:2 for H₂SO₄ to KOH, the moles of H₂SO₄ is half of that of KOH. Divide this by the volume of H₂SO₄ in liters to find the molarity, which is approximately 1.25 M. b) 1.25 M
Explanation:
To calculate the molarity of the H₂SO₄ solution, we need to use the stoichiometry of the neutralization reaction between H₂SO₄ and KOH. The balanced chemical equation for the reaction is:
H₂SO₄(aq) + 2KOH(aq) → K₂SO₄(aq) + 2H₂O(l)
From the equation, we see that one mole of H₂SO₄ reacts with two moles of KOH. Therefore, the molar amount of H₂SO₄ is half the molar amount of KOH used in the reaction.
First, we calculate the amount (in moles) of KOH:
moles KOH = Molarity of KOH × Volume of KOH (in liters)
moles KOH = 1.55 M × 0.0305 L = 0.047275 moles KOH
Since it takes two moles of KOH to neutralize one mole of H₂SO₄, the moles of H₂SO₄ is half of this:
moles H₂SO₄ = 0.047275 moles KOH / 2 = 0.0236375 moles H₂SO₄
Now, we can find the molarity of H₂SO₄ by dividing the moles of H₂SO₄ by the volume of H₂SO₄ solution (in liters).
Molarity of H₂SO₄ = moles H₂SO₄ / Volume of H₂SO₄ (in liters)
Molarity of H₂SO₄ = 0.0236375 moles H₂SO₄ / 0.0186 L
The molarity of the H₂SO₄ solution is therefore 1.2707 M, which can be rounded to 1.27 M when considering significant figures. The closest answer choice to 1.27 M is (b) 1.25 M.
