What was the molarity of the KOH solution if 20.2 mL of 1.50 M H2SO4 was needed? The equation is
2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)
Express your answer with the appropriate units.
A volume of 80.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4).?
2019-02-19 8:41 pm
回答 (1)
2019-02-19 8:58 pm
2KOH(aq) + H₂SO₄(aq) → K₂SO₄(aq) + 2H₂O(l)
Mole ratio KOH : H₂SO₄ = 2 : 1
No. of moles of H₂SO₄ reacted = (1.50 mmol/mL) × (20.2 mmol) = 30.3 mmol
No. of moles of KOH reacted = (30.3 mmol) × 2 = 60.6 mmol
Molarity of KOH = (60.6 mmol) / (80.0 mL) = 0.758 M (3 sig. fig.)
Mole ratio KOH : H₂SO₄ = 2 : 1
No. of moles of H₂SO₄ reacted = (1.50 mmol/mL) × (20.2 mmol) = 30.3 mmol
No. of moles of KOH reacted = (30.3 mmol) × 2 = 60.6 mmol
Molarity of KOH = (60.6 mmol) / (80.0 mL) = 0.758 M (3 sig. fig.)
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