An aqueous solution of barium hydroxide is standardized by titration with a 0.128 M solution of hydroiodic acid.
If 29.1 mL of base are required to neutralize 14.9 mL of the acid, what is the molarity of the barium hydroxide solution?
An aqueous solution of barium hydroxide is standardized by titration with a 0.128 M solution of hydroiodic acid.?
2016-10-24 4:30 am
回答 (2)
2016-10-24 9:23 am
2HI + Ba(OH)₂ → BaI₂ + 2H₂O
OR: Mole ratio HI : Ba(OH)₂ = 2 : 1
No. of milli-moles of HI = (0.128 mmol/mL) × (14.9 mL) = 1.9072 mmol
No. of milli-miles of Ba(OH)₂ = (1.9072 mmol) × (1/2) = 0.9536 mmol
Molarity of Ba(OH)₂ = (0.9536 mmol) / (29.1 mL) = 0.0328 M
OR: Mole ratio HI : Ba(OH)₂ = 2 : 1
No. of milli-moles of HI = (0.128 mmol/mL) × (14.9 mL) = 1.9072 mmol
No. of milli-miles of Ba(OH)₂ = (1.9072 mmol) × (1/2) = 0.9536 mmol
Molarity of Ba(OH)₂ = (0.9536 mmol) / (29.1 mL) = 0.0328 M
2016-10-24 4:32 am
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