In an acid-base neutralization reaction,38.74 mL of 0.500 mol L^-1 potassium hydroxide reacts with 50.00 mL of sulfuric acid solution.What is the concentration of the H₂SO₄ solution? A)0.194 mol L^-1 B)0.387 mol L^-1 C)0.775 mol L^-1 D)1.29 mol L^-1

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Quizplus · Accepted Answer

The balanced chemical equation for the reaction between potassium hydroxide (KOH) and sulfuric acid (H2SO4) is 2KOH + H2SO4 → K2SO4 + 2H2O. This shows that 2 moles of KOH react with 1 mole of H2SO4. Given that 38.74 mL of 0.500 M KOH is used, the moles of KOH are (38.74 mL) * (0.500 mol/L) = 0.01937 mol. Since the ratio of KOH to H2SO4 is 2:1, the moles of H2SO4 needed are half of that of KOH, which is 0.01937 mol / 2 = 0.009685 mol. The concentration of H2SO4 is then the moles of H2SO4 divided by the volume of the H2SO4 solution in liters, which is 0.009685 mol / 0.05000 L = 0.194 mol/L.

In an Acid-Base Neutralization Reaction,38