Svinsern Posted January 18, 2017 Report Share Posted January 18, 2017 Unsure if this is the right place to ask this, but I have a result related problem for my IA. My topic is calculating KC for the hydrolysis of ethyl ethanoate, with titration as the chosen method. The concentration of ethyl ethanoate and water varied between different trial, but as the equilibrium law states only temerature should affect pH. However, when I calculated Kc for the different trials I ended up with five quite different values where the general trend as that Kc decreased as concentration of ethyl ethanoate decreased. I have no idea why this happened....is there any chemistry gurus out there who might know what can have happened? Reply Link to post Share on other sites More sharing options...
kw0573 Posted January 18, 2017 Report Share Posted January 18, 2017 How are you sure that equilibrium has been reached and maintained? What exactly are your measuring in the titration? Reply Link to post Share on other sites More sharing options...
Svinsern Posted January 19, 2017 Author Report Share Posted January 19, 2017 The solutions were left alone for two weeks and should reach equlibrium already after 48 hours. I was measuring the total amount of acid in the titration using NaOH. Reply Link to post Share on other sites More sharing options...
kw0573 Posted January 19, 2017 Report Share Posted January 19, 2017 I want to clarify that I will try to ask questions to help you identify potential flaws (that I probably am not aware of) yourself. Is the equilibrium constant correctly calculated, accounting for all species in the equilibrium? If the equilibrium is between water and ethyl acetate, why is the amount of acid of interest? Would -OH in any way, react with ethyl acetate, or in some other means alter the equilibrium? Reply Link to post Share on other sites More sharing options...
Svinsern Posted January 19, 2017 Author Report Share Posted January 19, 2017 The equilibrium constant has been calculated correctly, this I have confirmed with my teacher. Acid (HCl) is used as a catalyst for the reaction as the reaction is extremely on its own (hydrolysis of ethyl ethanoate). Therefore a fixed volume of HCl is added to every trial. This means that when performing the titration at equilibrium I get the total acid which consists of HCl and ethanoic acid. TO find ethanoic acid, which is one of the equilibrium values I had to calculate and subtract the total amount of HCl from total acid. Regarding OH-, I am not sure if that could have had any effect. Reply Link to post Share on other sites More sharing options...
kw0573 Posted January 19, 2017 Report Share Posted January 19, 2017 22 minutes ago, Svinsern said: The equilibrium constant has been calculated correctly, this I have confirmed with my teacher. Acid (HCl) is used as a catalyst for the reaction as the reaction is extremely on its own (hydrolysis of ethyl ethanoate). Therefore a fixed volume of HCl is added to every trial. This means that when performing the titration at equilibrium I get the total acid which consists of HCl and ethanoic acid. TO find ethanoic acid, which is one of the equilibrium values I had to calculate and subtract the total amount of HCl from total acid. Regarding OH-, I am not sure if that could have had any effect. Though you have correctly identified that HCl is a catalysis, you should read on reactivity of carbonyl/carboxyl compounds to determine whether the introduction of -OH may have impacted the reaction (during the time you performed the titration). Consider the two products, ethanoic acid and ethanol. NaOH deprotonates ethanoic acid, I agree, but you should also calculate if it can significantly deprotonate ethanol? What would be the equilibrium constant be between ethanol and NaOH? My final remark is regarding your observations of the final solution. Is there a single layer of liquid or multiple layers? In organic chemistry there is the "liquid-liquid extraction" technique, commonly known as "work-up", where you can separate different layers of liquids, which may be useful for you if you did observe 2 layers. Reply Link to post Share on other sites More sharing options...
Svinsern Posted January 19, 2017 Author Report Share Posted January 19, 2017 Thanks for the help. The amount of ethanol cannot be calculated easily from the experiment I performed, but is instead deduced through the mole ratio with ethanoic acid (1:1). I didnt observe two layers in the quilibrium solution as far as I can remember. I will discuss your suggestions with my teacher to see if I get anywhere. Again, thanks for the help Reply Link to post Share on other sites More sharing options...
kw0573 Posted January 19, 2017 Report Share Posted January 19, 2017 (edited) 15 minutes ago, Svinsern said: Thanks for the help. The amount of ethanol cannot be calculated easily from the experiment I performed, but is instead deduced through the mole ratio with ethanoic acid (1:1). I didnt observe two layers in the quilibrium solution as far as I can remember. I will discuss your suggestions with my teacher to see if I get anywhere. Again, thanks for the help You can use mol ratio if you assume that NaOH only deprotonates ethanoic acid and HCl. But you can also use the pKa of ethanol to calculate the equilibrium constant between ethanol and NaOH so see how accurate that assumption is. For example if you were to discover that the ethanol - NaOH equilibrium is 3, then that would mean the assumption is invalid and the acid titrated minus HCl added accounts for both ethanoic acid, as well as some part of ethanol. If you look up the solubility of ethyl ethanoate in water, you would see that it is not very soluble. Clearly in an equilibrium, you would expect ethyl ethanoate to be present in a own layer. Edited January 19, 2017 by kw0573 Reply Link to post Share on other sites More sharing options...
Svinsern Posted January 19, 2017 Author Report Share Posted January 19, 2017 Hmm I see. The thing tht surprised me the most was that the trend was nearly linear. As I was decreasing the ethyl ethanoate concentration, Kc increased linearly with it. Reply Link to post Share on other sites More sharing options...
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