A sample of 25.0 cm 3 of aqueous sodium hydroxide of concentration 0.5 mol/dm 3 was
titrated against aqueous hydrochloric acid of unknown concentration. It required
16.7cm 3 of hydrochloric acid solution for neutralisation.
How many moles of sodium hydroxide were measured out?
How many moles of sodium hydroxide had reacted at the end point?
What was the concentration of the aqueous hydrochloric acid solution?
TITRATION QUESTION?
2019-03-06 7:02 pm
回答 (1)
2019-03-06 7:34 pm
A sample of 25.0 cm³ of aqueous sodium hydroxide of concentration 0.5 mol/dm³ was titrated against aqueous hydrochloric acid of unknown concentration. It required 16.7 cm³ of hydrochloric acid solution for neutralisation. How many moles of sodium hydroxide were measured out?
Moles of NaOH = (0.5 mol/dm³) × (25.0/1000 dm³) = 0.0125 mol
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How many moles of sodium hydroxide had reacted at the end point?
Moles of NaOH = 0.0125 mol
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What was the concentration of the aqueous hydrochloric acid solution?
NaOH + HCl → NaCl + H₂O
Mole ratio NaOH : HCl = 1 : 1
Moles of HCl = Moles of NaOH = 0.0125 mol
Concentration of HCl = (0.0125 mol) / (16.7/1000 dm³) = 0.749 mol/dm³
Moles of NaOH = (0.5 mol/dm³) × (25.0/1000 dm³) = 0.0125 mol
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How many moles of sodium hydroxide had reacted at the end point?
Moles of NaOH = 0.0125 mol
====
What was the concentration of the aqueous hydrochloric acid solution?
NaOH + HCl → NaCl + H₂O
Mole ratio NaOH : HCl = 1 : 1
Moles of HCl = Moles of NaOH = 0.0125 mol
Concentration of HCl = (0.0125 mol) / (16.7/1000 dm³) = 0.749 mol/dm³
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