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What is the molarity of the H₂SO₄ solution, 18.6 mL of which neutralizes 30.5 mL of 1.55 M KOH solution?
(a) 1.0 M
(b) 1.5 M
(c) 2.0 M
(d) 2.5 M

Answer :

Final answer:

The molarity of the H₂SO₄ solution can be found by calculating the moles of KOH used, adjusting for the stoichiometry of the reaction, and dividing by the volume of H₂SO₄ solution. The molarity is approximately 1.27 M, and the closest given option is b) 1.5 M

Explanation:

The student is asking about the molarity of an H₂SO₄ solution that neutralizes a given volume of a 1.55 M KOH solution. In a titration, the molarity (M) of an acid can be calculated using the volume of the base (Vb), the molarity of the base (Mb), and the volume of the acid (Va), taking into account the stoichiometry of the reaction.



First, we write the balanced chemical equation for the reaction:



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



According to the stoichiometry of the reaction, one mole of sulfuric acid reacts with two moles of potassium hydroxide. We calculate the number of moles of KOH used:



n(KOH) = M(KOH) × V(KOH) = 1.55 M × 0.0305 L = 0.047275 mol



Since the ratio of H₂SO₄ to KOH is 1:2, the moles of H₂SO₄ will be half of the moles of KOH:



n(H₂SO₄) = 0.047275 mol / 2 = 0.0236375 mol



Finally, we calculate the molarity of the H₂SO₄ solution:



M(H₂SO₄) = n(H₂SO₄) / V(H₂SO₄) = 0.0236375 mol / 0.0186 L ≈ 1.27 M



When we review the options provided in the question, none of them exactly matches our calculated value. It is possible that there was a rounding error in the question's options or a typo. However, given the options, the closest correct choice to our calculated molarity is option (b) 1.5 M.

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