A 25 ml aliquot of vinegar was diluted to 250mL. Titration of 50 mL aliquots of the diluted solution required an average of 34.88 ml of 0.096M NaOH.
(a) Express the acidity of the vinegar as percent (w/v) acetic acid.
Thanks~~~
Titration + dilution calculation~HELP
2008-06-15 5:42 pm
回答 (2)
2008-06-15 7:08 pm
✔ 最佳答案
Consider the titration between 50 mL diluted vinegar and NaOH.CH3COOH + NaOH → CH3COONa + H2O
Mole ratio CH3COOH : NaOH = 1 : 1
No. of moles of NaOH = MV = 0.096 x (34.88/1000) = 0.003348 mol
No. of moles of CH3COOH in 50 mL of diluted vinegar = 0.003348 mol
No. of moles of CH3COOH in 250 mL of diluted vinegar
= 0.003348 x (250/50)
= 0.01674 mol
No. of moles of CH3COOH in 25 mL of original vinegar = 0.01674 mol
Molar mass of CH3COOH = 12x2 + 1x4 + 16x2 = 60 g mol-1
Mass of CH3COOH in 25 mL of original vinegar
= mol x (molar mass)
= 0.01674 x 60
= 1.004 g
Acidity of the original vinegar
= (1.004/25) x 100%
= 4.016%
2008-06-15 11:09:44 補充：
The 「acidity of the original vinegar」should be the「concentration of the original vinegar」instead.
2008-06-15 6:00 pm
H+ + NaOH --> Na+ + H2O (1)
NO. OF MOLE OF NaOH: 34.88 /1000 X 0.096=0.00334848
BY (1) No.of mole of vinegar = no .of mole of NaOH =0.00334848
No. of mole = Volume X acidity
Acidity : 0.00334848/ (50/1000) =0.0669696
THE ORIGIN (NOT YEY DILUTE) VINEGAE ACIDIRY IS"
0.0669696/ (25/250) = 0.669696
=0.67 M
NO. OF MOLE OF NaOH: 34.88 /1000 X 0.096=0.00334848
BY (1) No.of mole of vinegar = no .of mole of NaOH =0.00334848
No. of mole = Volume X acidity
Acidity : 0.00334848/ (50/1000) =0.0669696
THE ORIGIN (NOT YEY DILUTE) VINEGAE ACIDIRY IS"
0.0669696/ (25/250) = 0.669696
=0.67 M
收錄日期: 2021-04-23 23:08:41
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