What volume of 5.75 M phosphoric acid is required to neutralize 22.2 mL of 2.22 M potassium hydroxide?
H3PO4(aq) + 3 KOH(aq) --> 3H2O(l) +K3PO4(aq)
Chemistry Gas Laws What volume of 5.75 M phosphoric acid is required to neutralize 22.2 mL of 2.22 M potassium hydroxide?
2017-04-24 12:25 pm
回答 (1)
2017-04-24 3:33 pm
Number of milli-moles of KOH reacted = (2.22 mmol/mL) × (22.2 mL) = 49.28 mmol
H₃PO₄(aq) + 3KOH(aq) → 3H₂O(l) + K₃PO₄(aq)
Molar ratio H₃PO₄ : KOH = 1 : 3
Number of milli-moles of H₃PO₄ required = (49.28 mmol) × (1/3) = 16.43 mmol
Volume of H₃PO₄ required = (16.43 mmol) / (5.75 mmol/mL) = 2.86 mL
H₃PO₄(aq) + 3KOH(aq) → 3H₂O(l) + K₃PO₄(aq)
Molar ratio H₃PO₄ : KOH = 1 : 3
Number of milli-moles of H₃PO₄ required = (49.28 mmol) × (1/3) = 16.43 mmol
Volume of H₃PO₄ required = (16.43 mmol) / (5.75 mmol/mL) = 2.86 mL
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