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timrelawan06
02.08.2021 •
Chemistry
If it takes 72 mL of 9 M KOH to neutralize 415 mL of sulfuric
acid (H2SO4) solution, what is the concentration of the H2SO4
solution?
M
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Ответ:
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
H₂SO₄ + 2KOH —> K₂SO₄ + 2H₂O
From the balanced equation above,
Mole ratio of the acid, H₂SO₄ (nₐ) = 1
Mole ratio of the base, KOH (n₆) = 2
Finally, we shall determine the concentration of H₂SO₄. This can be obtained as follow:
Volume of base, KOH (V₆) = 72 mL
Concentration of base, KOH (C₆) = 9 M
Volume acid, H₂SO₄ (Vₐ) = 415 mL
Mole ratio of the acid, H₂SO₄ (nₐ) = 1
Mole ratio of the base, KOH (n₆) = 2
Concentration of acid, H₂SO₄ (Cₐ) =?CₐVₐ / C₆V₆ = nₐ/n₆
Cₐ × 415 / 9 × 72 = 1/2
Cₐ × 415 / 648 = 1/2
Cross multiply
Cₐ × 415 × 2 = 648
Cₐ × 830 = 648
Divide both side by 830
Cₐ = 648 / 830
Cₐ = 0.78 MTherefore, the concentration of H₂SO₄ is 0.78 M
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Ответ:
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Explanation: