[Sush_eera]

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.

[thetada]

have you thought about what reaction is taking place in step 2? What are the reactants and what might the products be?

[Sush_eera]

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

[AWK]

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)

[Sush_eera]

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

[thetada]

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?

[Sush_eera]

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

[thetada]

Good, that gives you the number of moles of sulfuric acid. Why do you think the mole ratio is 1:1?

[Sush_eera]

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

[thetada]

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.

[Sush_eera]

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??

[thetada]

Good work. What can you figure out from the data you have for reaction 3?

[Sush_eera]

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?

[thetada]

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?

[Sush_eera]

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