Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Sush_eera on June 21, 2016, 01:34:46 PM
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Can some one help me with this question from part (a) - (c) ? I dont even know how to start it off or what to do. I know we have to do back titration and stuff and work out the moles backwards but i can't figure out how exactly to do it.
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have you thought about what reaction is taking place in step 2? What are the reactants and what might the products be?
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have you thought about what reaction is taking place in step 2? What are the reactants and what might the products be?
For reaction 2 i got the equation to be CH3COONa+H2SO4---> CH3COOH+Na2SO4
For reaction one I am still not sure whether CH3COONa will dissociate into CH3COO- and Na+ or into CH3COO- and OH-
If my equation for reaction 2 is correct then I guess when the products are reacted with NaOH i will get the equation to be
CH3COOH + NaOH ----> CH3COONa + H2O
i think i am okay with the equations
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Dissolution of sodium acetate gives you ions in solution. Dissolution is accompanied by hydrolysis of salt but this is unimportant for step 2.
For step 3 - remember about an excess of sulfuric acid (reaction is needed)
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Dissolution of sodium acetate gives you ions in solution. Dissolution is accompanied by hydrolysis of salt but this is unimportant for step 2.
For step 3 - remember about an excess of sulfuric acid (reaction is needed)
But dont we need it in step 1?
Is the reaction esterification with H2SO4? I am lost
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have you thought about what reaction is taking place in step 2? What are the reactants and what might the products be?
For reaction 2 i got the equation to be CH3COONa+H2SO4---> CH3COOH+Na2SO4
For reaction one I am still not sure whether CH3COONa will dissociate into CH3COO- and Na+ or into CH3COO- and OH-
If my equation for reaction 2 is correct then I guess when the products are reacted with NaOH i will get the equation to be
CH3COOH + NaOH ----> CH3COONa + H2O
i think i am okay with the equations
This sounds reasonable. So with the data you have, which species can you work out what number of moles you have?
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have you thought about what reaction is taking place in step 2? What are the reactants and what might the products be?
For reaction 2 i got the equation to be CH3COONa+H2SO4---> CH3COOH+Na2SO4
For reaction one I am still not sure whether CH3COONa will dissociate into CH3COO- and Na+ or into CH3COO- and OH-
If my equation for reaction 2 is correct then I guess when the products are reacted with NaOH i will get the equation to be
CH3COOH + NaOH ----> CH3COONa + H2O
i think i am okay with the equations
This sounds reasonable. So with the data you have, which species can you work out what number of moles you have?
I this i can work out the no of moles of H2SO4 in step 2.
Which will be
No. of moles= Conc.×vol./ 1000
No. of moles of H2SO4 would be= 0.1×50/1000= 0.005 M
comparing the mole ration i would have 0.005 moles of CH3COONa since it is a one to one mole ration
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Good, that gives you the number of moles of sulfuric acid. Why do you think the mole ratio is 1:1?
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Good, that gives you the number of moles of sulfuric acid. Why do you think the mole ratio is 1:1?
because the the co-efficient is 1 infront of sulphuric acid and ch3coona
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Looking at the equation again, you've got the right reactants and products but actually the equation isn't balanced. Compare the number of hydrogen / sodium atoms on each side.
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Looking at the equation again, you've got the right reactants and products but actually the equation isn't balanced. Compare the number of hydrogen / sodium atoms on each side.
CH3COONa+1/2H2SO4---> CH3COOH+1/2 Na2SO4
multiply by 2
2CH3COONa+H2SO4---> 2CH3COOH+Na2SO4
i now got the mole ration to be 2:1
That means the no. of moles of CH3COONa would be 0.005 *2 =0.01 moles of CH3cooNa
is it? but how can i find the no. of moles of ch3coona used to find the excess moles of NaOH??
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Good work. What can you figure out from the data you have for reaction 3?
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Good work. What can you figure out from the data you have for reaction 3?
No.of moles of NaOH uses? but i not too sure if the titre value is the volume of NaOh used or volume of the product formed in reaction 2? ??? :'(
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When it says the reaction mixture is titrated against NaOH, they mean whatever is in the reaction vessel (beaker / conical flask or whatever) is placed under the burette so the titre relates to the NaOH. Once you work out the number of moles of NaOH, what can you use that to find out?
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When it says the reaction mixture is titrated against NaOH, they mean whatever is in the reaction vessel (beaker / conical flask or whatever) is placed under the burette so the titre relates to the NaOH. Once you work out the number of moles of NaOH, what can you use that to find out?
OH i seee.. i can work out the no. of moles of ch3cooh
so the no. of moles of NaOH wouldbe 0.00267 moles of naOH
according the equation CH3COOH + NaOH ----> CH3COONa + H2O
since its a 1:1 ratio i can find the no. of CH3COOH used which would be 0.00267 moles as well
since they said H2SO4 was in excess then CH3COONa has to be the limiting reagent in part 2
also since the product which is CH3COOH is dependent on the limiting reating the no. of moles of CH3COONa used in reaction 2 is 0.00267 moles.
To find the no. of moles of excess H2SO4 it would be 0.005- 0.00267= 0.00233 moles of H2SO4 i excess .. is it?
for part (ii) to find the no. of moles of H2SO4 consumed it would be 0.00267 moles
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Look again at the first question and see if you can find any information that conflicts with your suggestion
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Look again at the first question and see if you can find any information that conflicts with your suggestion
Is there something wrong in what I said? Cus i couldnt figure out where i went wrong
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You need to go backwards. You use the information from step 3 to figure out how much excess sulfuric acid reacted with the sodium hydroxide. From that you work out how much ethanoic acid reacted with the sulfuric acid in step 2 and from that you work out what proportion of the 10g of powder in step 1 consisted of sodium ethanoate. My impression is that you were thinking that all of the 10g in step 1 was sodium ethanoate.
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You need to go backwards. You use the information from step 3 to figure out how much excess sulfuric acid reacted with the sodium hydroxide. From that you work out how much ethanoic acid reacted with the sulfuric acid in step 2 and from that you work out what proportion of the 10g of powder in step 1 consisted of sodium ethanoate. My impression is that you were thinking that all of the 10g in step 1 was sodium ethanoate.
But why would H2SO4 react with NaOH.. i though it would only react with ethanoic acid from the reaction i used before
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But why would H2SO4 react with NaOH.
Because that's the way these substances behave. Whenever you have strong acid and strong base you can be sure they will react. When there is a mixture strongest bases/acids react first.
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But why would H2SO4 react with NaOH.
Because that's the way these substances behave. Whenever you have strong acid and strong base you can be sure they will react. When there is a mixture strongest bases/acids react first.
This is interesting. I was starting to think that my suggestions were flawed. We use the sulfuric acid to turn the sodium ethanoate into ethanoic acid, then we titrate whatever sulfuric acid is left over against the sodium hydroxide, which tells us how much sulfuric acid was left over and hence how much sodium ethanoate there was to start with. But suddenly I realised that if the whole reaction mixture is titrated against the sodium hydroxide, it will neutralise first the sulfuric acid and then the ethanoic acid. In effect, it will neutralise all the protons that were originally associated with the sulfuric acid. This feels like a problem.
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IMHO this is a poor question - ethanoic acid is way too strong for this kind of approach (at least in water solution). While sometimes choice of an indicator makes it possible to separately titrate two acids in one sample, I am not convinced it is possible here.
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I'm glad you agree. Presumably whoever wrote it is expecting us to pretend that the ethanoic acid does not interfere with the titration of excess sulfuric acid against sodium hydroxide, as unrealistic as that might be.