Antacid Analysis... General Chem 2?
2020-09-24 12:11 am
A hydrochloric acid solution consisting of a volume of 98.5 mL with a concentration of 0.945M HCl is ascertained for a back titration. An antacid table added to the acidic solution. Then, a back titration is performed with a 0.78M NaOH solution The initial volume reading of the sodium hydroxide solution on the buret is 32.3mL. The titration comes to completion with a final reading of 112.4mL on the buret. What are the number moles of antacid?
回答 (1)
2020-09-24 12:29 am
Consider the back titration of the NaOH against the excess HCl:
NaOH + HCl → NaCl + H₂O
Mole ratio NaOH : (excess HCl) = 1 : 1
No. of moles of NaOH added = (0.78 mol/L) × [(112.4 - 32.3)/1000 L] = 0.0625 mol
No. of moles of excess HCl = 0.0624 mol
Consider the titration of HCl against the antacid:
Assume that HCl reacts with the antacid in mole ratio 1 : 1.
Total no. of moles of HCl added = (0.945 mol/L) × (98.5/1000 L) = 0.0931 mol
No. of moles of HCl reacted with the antacid = (0.0931 - 0.0625) mol = 0.0306 mol
No. of moles of the antacid = 0.0306 mol
NaOH + HCl → NaCl + H₂O
Mole ratio NaOH : (excess HCl) = 1 : 1
No. of moles of NaOH added = (0.78 mol/L) × [(112.4 - 32.3)/1000 L] = 0.0625 mol
No. of moles of excess HCl = 0.0624 mol
Consider the titration of HCl against the antacid:
Assume that HCl reacts with the antacid in mole ratio 1 : 1.
Total no. of moles of HCl added = (0.945 mol/L) × (98.5/1000 L) = 0.0931 mol
No. of moles of HCl reacted with the antacid = (0.0931 - 0.0625) mol = 0.0306 mol
No. of moles of the antacid = 0.0306 mol
收錄日期: 2021-04-24 08:02:11
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