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Can someone explain to me the Kc in CHEp22?

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Well, I guess 24 hrs has just passed...
Can someone explain to me why Kc in Che P22 is not 0.44 but 2.25 or something?
I'm really comfusing right now... :(
 
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Logically bro i got .44 too , . The amount of NaOH taken was .0045 , from there .005 reacted with HCl , leaving .04 for the ethanoic acid to get neutralised. as 1mol Naoh neutralises 1mol ethanoic acid so .04 would neutralize .04 mol of ethanoic acid WHICH WAS LEFT AND IT WAS NOT TURNED TO ESTER , here we did the mistake. feeling very bad for it :(
 
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turboboy said:
Logically bro i got .44 too , . The amount of NaOH taken was .0045 , from there .005 reacted with HCl , leaving .04 for the ethanoic acid to get neutralised. as 1mol Naoh neutralises 1mol ethanoic acid so .04 would neutralize .04 mol of ethanoic acid WHICH WAS LEFT AND IT WAS NOT TURNED TO ESTER , here we did the mistake. feeling very bad for it :(

listen in the question they told us that a sample was taken when equilibrium was established so it means that after reacting 0.04 moles remained there so then as total were 0.1 moles and and 0.04 remained at eqm so moles consumed = 0.1 -0.04 = 0.06 the moles of products present at eqm. then just apply in the expression of kc and u get the answer.
 
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Balaj bro , i exactly did that in my exams , but a friend of mine explaned me this , the sample was titrated AFTER IT REACHED EQUILIBRIUM , that means IN EQUILIBRIUM .04 ethanoic was present and .06 reacted to form the ester.
 
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then whats the prob? i mean kc expression is products divided by reactants and we get (0.6)(0.6)/(0.4)(0.4)
 
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oh sorry i didnt read your reply well and just explained sorry about that. BUT tell me the expression for KC as one of my friend was saying water was to be omitted in it as water is pure liquid and others are in aqueous forms causing us to apply heterogenous eqn rules. if anyone can clear i will be delighted.
 
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mine was .6square/.4square , but it should be the opposite .4sq/.6sq , and not sure about ommiting water.
 
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