chemical equilibrium cal.

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2011-12-23 9:27 am

Determining the equilibrium constant for an esterification reaction
CH3COOH(l)+CH3CH(OH)CH3(l)<----->CH3COOCH(CH3)2(l)+H2O(l)
Step1:Mix 0.300 mole of ethanoic acid and 0.300 mole of propan-2-ol in a pear-shaped flask.
Step 2:Withdraw 1.00 cm3 of this mixture and add to a conical flask containing 25 cm3 of distilled water. Titrate the contents of the conical flask against 0.250 mol dm-3 sodium hydroxide solution. 30.0 cm3 of NaOH is required.

Calculation:
CH3COOH(l)+NaOH(aq)<----->CH3COONa(aq)+H2O(l)
no.of moles of NaOH=no. of moles of CH3COOH
0.250 mol dm-3 x (30.0/1000) dm3 = [CH3COOH(l)] x (1.00 /1000) dm3
[CH3COOH(l)] = 7.50 mol dm-3
therefore, concentration of CH3COOH(l) in the original mixture is 7.50 mol dm-3

我唔明紅色個度,點解係用1.00 cm3 而唔係用(1.00+25) = 26 cm3(total volume of the solution wo...)?

回答 (1)

知識就是力量

2011-12-23 8:51 pm

✔ 最佳答案

The actual no. of moles of CH3COOH is the concentration * volume used, we use the 1 cm^3 but not the 26 cm^3. If the volume of water is taken into the account, the no. of moles calculated will be larger than actual one.

For example, suppose the concentration of CH3COOH in the original solution is 7.5 M,
no. of moles of CH3COOH = 7.5 * 1/1000 = 7.5 * 10^-3 mol (1 cm^3 of the mixture used)
no. of moles of CH3COOH = 7.5 * 26/1000 = 0.195 mol (26 cm^3 of mixture used, but this is false because water is added to dilute the solution and it cannot increase the no. of moles of CH3COOH in the solution mixture. It just lowers the mixture concentration.)

參考: Knowledge is power.

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