In an acid-base neutralization reaction 43.74 mL of 0.500 M potassium hydroxide reacts with 50.00 mL of sulfuric acid solution.What is the concentration of the H₂SO₄ solution? A)0.219 M B)0.437 M C)0.875 M D)1.14 M

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Quizplus · Accepted Answer

The neutralization reaction between potassium hydroxide (KOH) and sulfuric acid (H2SO4) can be represented as: 2KOH + H2SO4 → K2SO4 + 2H2O. This shows that 2 moles of KOH react with 1 mole of H2SO4. The moles of KOH used are calculated as (0.04374 L) * (0.500 mol/L) = 0.02187 mol. Since it takes 2 moles of KOH to neutralize 1 mole of H2SO4, the moles of H2SO4 are half that of KOH, which is 0.010935 mol. The concentration of H2SO4 is then calculated by dividing the moles of H2SO4 by the volume of the solution in liters, which is 0.010935 mol / 0.05000 L = 0.2187 M, rounded to three significant figures as 0.219 M.

In an Acid-Base Neutralization Reaction 43