Posted by yaya on Monday, May 7, 2012 at 4:39pm.
Subject – equilibria
Pure ethanoic acid (25.0cm3, CH3COOH), pure ethanol (35.0 cm3, C2H5OH) and pure water (20.0cm3, H2O) were mixed in a sealed flask at room temperature. The flask was placed in a heated water bath and maintained at 323K until equilibrium was established.
CH3COOH(aq) + C2H5OH(aq) ? CH3 COOC2H5(aq) + H2O(l)
When equilibrium was attained the molar concentration of ethanoic acid was determined by titration as follows. A sample of the equilibrium mixture was taken (10.0cm3) and titrated with a standard solution of aqueous potassium hydroxide (KOH; 1.00mol dm-3). The volume of aqueous KOH required for complete reaction with ethanoic acid in the equilibrium sample of the reaction mixture was 25.5cm3.
a) (i) Predict the value pH of the titration mixture at the equivalence point. Explain your answer.
(ii) What chemical indicator might you use for this titration?
(iii) Write a balanced equation for the reaction of aqueous potassium hydroxide (KOH(aq)) with aqueous ethanoic acid (CH3COOH(aq)).
(iv) Calculate the amount of unreacted aqueous ethanoic acid (CH3COOH(aq)) in the sample taken for titration with aqueous potassium hydroxide(KOH(aq)). Clearly show the steps in your calculation.
v) Hence calculate the amount of aqueous ethanoic acid CH3COOH(aq)) in the sealed flask at equilibrium. Clearly show the steps in your calculation and comment on any assumptions that you make.
The questions which follow will require the use of the following data. The questions relate to the experiment described at the beginning of questions
Substance ===============Density at room temperature (g cm-3)
Ethanoic acid========= 1.05
Ethanol ============ 0.790
Water ============= 1.00
Element======== Molar mass (g mol-1)
- chemistry - DrBob222, Monday, May 7, 2012 at 4:53pm
This is a long question. What do you not understand about it? Much of it is smokescreen. We can help you through the rough spots.
- chemistry - yaya, Monday, May 7, 2012 at 7:59pm
DrBob222! I am so happy that you took a look at this question, I was hoping when posting that YOU would help me out...honestly.
Well I know its a long question and to be honest I kinda need help with all of it...I have got the first part semi figured out but would still need to finish it...I would be thankful for any help.
- chemistry - yaya, Tuesday, May 8, 2012 at 5:15pm
I am really stuck and would need some help..is there anyone out there in cyberworld that can help me ...pleeeeeeaaaaaassssseeeee
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