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10 May, 20:46

A volume of 80.0 mL of aqueous potassium hydroxide (KOH) is titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 12.7 mL of 1.50 M H2SO4 was needed

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  1. 10 May, 21:06
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    0.478 M

    Explanation:

    Let's consider the neutralization reaction between KOH and H₂SO₄.

    2 KOH + H₂SO₄ → K₂SO₄ + 2 H₂O

    12.7 mL of 1.50 M H₂SO₄ react. The reacting moles of H₂SO₄ are:

    0.0127 L * 1.50 mol/L = 0.0191 mol

    The molar ratio of KOH to H₂SO₄ is 2:1. The reacting moles of KOH are 2 * 0.0191 mol = 0.0382 mol

    0.0382 moles of KOH are in 80.0 mL. The molarity of KOH is:

    M = 0.0382 mol/0.0800 L = 0.478 M
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