A student uses 25.00ml of a 0.0525 mol/L calcium hydroxide solution to titrate 0.111 mol/L hydrochloric acid solution. What volume of acid is needed for complete neutralization?
I got the answer as 11.8 ml (no sig digs)
The answer key says its 23.6 ml
can someone tell me what to do?
More Chemistry!?
2017-01-24 11:33 am
回答 (1)
2017-01-24 11:43 am
Method 1 :
Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O
OR: Mole ratio Ca(OH)₂ : HCl = 1 : 2
No. of milli-moles of Ca(OH)₂ reacted = (0.0525 mmol/ml) × (25.00 ml) = 1.31 mmol
No. of milli-moles of HCl needed = (1.31 mmol) × 2 = 2.62 mmol
Volume of HCl needed = (2.62 mmol) / (0.111 mmol/ml) = 23.6 ml (3 sig. digs.)
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Method 2 :
Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O
OR: Mole ratio Ca(OH)₂ : HCl = 1 : 2
No. of moles of Ca(OH)₂ reacted = (0.0525 mol/L) × (25.00/1000 L) = 0.00131 mol
No. of milli-moles of HCl needed = (0.00131 mol) × 2 = 0.00262 mol
Volume of HCl needed = (0.00262 mol) / (0.111 mol/L) = 0.0236 L = 23.6 ml (3 sig. digs.)
Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O
OR: Mole ratio Ca(OH)₂ : HCl = 1 : 2
No. of milli-moles of Ca(OH)₂ reacted = (0.0525 mmol/ml) × (25.00 ml) = 1.31 mmol
No. of milli-moles of HCl needed = (1.31 mmol) × 2 = 2.62 mmol
Volume of HCl needed = (2.62 mmol) / (0.111 mmol/ml) = 23.6 ml (3 sig. digs.)
====
Method 2 :
Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O
OR: Mole ratio Ca(OH)₂ : HCl = 1 : 2
No. of moles of Ca(OH)₂ reacted = (0.0525 mol/L) × (25.00/1000 L) = 0.00131 mol
No. of milli-moles of HCl needed = (0.00131 mol) × 2 = 0.00262 mol
Volume of HCl needed = (0.00262 mol) / (0.111 mol/L) = 0.0236 L = 23.6 ml (3 sig. digs.)
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