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Mr.GreyCells

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A 25.00 mL sample of 0.100 mol L-1 acetic acid solution needed 22.20 mL of a NaOH solution for complete neutralization. What volume of the same NaOH solution is needed to neutralize 25.00 mL of 0.100 mol L-1 sulfuric acid solution?
A. 11.10 mL
B. 22.20 mL
C. 33.30 mL
D. 44.40 mL
Thanks
 

medaspirant

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A 25.00 mL sample of 0.100 mol L-1 acetic acid solution needed 22.20 mL of a NaOH solution for complete neutralization. What volume of the same NaOH solution is needed to neutralize 25.00 mL of 0.100 mo L-1 sulfuric acid solution?
A. 11.10 mL
B. 2.20 mL
C. 33.30 mL
D. 44.40 mL
Thanks
Sup I did the q, don't know if I did it correctly if someone can verify that would be gr8, read the green first then blue
 

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Mr.GreyCells

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Thanks for your reply but I can't view the attachment.
Do you know where I could get the solutions of Abbotsleigh trial 2021 paper from which this question was taken?
 

medaspirant

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Thanks for your reply but I can't view the attachment.
Do you know where I could get the solutions of Abbotsleigh trial 2021 paper from which this question was taken?
sos dont know bout that but u should be able to see the attachment now sos for b4 forgot to upload lmao
 

Mr.GreyCells

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Wait why did you take the number of moles of H+ ions released by CH3COOH, the same as the number of moles of ch3cooh? Isn't it a weak acid therefore dissociates partially to give CH3COO- and H+ and CH3COOH still remains in the solution?
Sup I did the q, don't know if I did it correctly if someone can verify that would be gr8, read the green first then blue
 

medaspirant

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Wait why did you take the number of moles of H+ ions released by CH3COOH, the same as the number of moles of ch3cooh? Isn't it a weak acid therefore dissociates partially to give CH3COO- and H+ and CH3COOH still remains in the solution?
disclaimer i dont know if this correct but yh this is my logic:
for neutralisation reactions, the strength of the acid is irrelevant as the base is stronger than water. all ionisable protons will be irreversibly removed by the base.
 

Mr.GreyCells

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Oh I thought neutralization was all about strength of acids and bases.
disclaimer i dont know if this correct but yh this is my logic:
for neutralisation reactions, the strength of the acid is irrelevant as the base is stronger than water. all ionisable protons will be irreversibly removed by the base.
 

jazz519

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Strength of an acid / base is referring to the ionisation reaction where it dissociates in a solvent like water. It is not the same kind of principle when it is reacting with a base. If either of the acid or the base is strong the reaction will go to completion.

It is true that we need to know the strength of acid/base for determining things like the pH of the equivalence point and conductivity graphs, however for the purpose of deciding amount of a substance reacted it jut comes back to the basics you learnt in year 11.

Therefore only molar ratios for the reactions need to be compared and you can find the volume.

Also, there is no need to actually calculate the values like the above user did so.

The volumes and concentrations of the acid solutions are the same. The reactions take place as follows:
CH3COOH(aq) + NaOH(aq) --> NaCH3COO(aq) + H2O(l)
H2SO4(aq) + 2NaOH(aq) --> Na2SO4(aq) + 2H2O(l)

Twice as much NaOH is needed i.e. twice the volume, so 44.40 mL would be used.



If the question changed the acid to acetic acid vs hydrochloric acid as an extra example as the reactions would be 1:1 in the ratios, the volume used would be the same.
 

Mr.GreyCells

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I had a question:
Wouldn't it require more of naoh to neutralise h2so4 than it took to neutralise ch3cooh? like way more than double?
Don't know if my question makes sense lol.
Strength of an acid / base is referring to the ionisation reaction where it dissociates in a solvent like water. It is not the same kind of principle when it is reacting with a base. If either of the acid or the base is strong the reaction will go to completion.

It is true that we need to know the strength of acid/base for determining things like the pH of the equivalence point and conductivity graphs, however for the purpose of deciding amount of a substance reacted it jut comes back to the basics you learnt in year 11.

Therefore only molar ratios for the reactions need to be compared and you can find the volume.

Also, there is no need to actually calculate the values like the above user did so.

The volumes and concentrations of the acid solutions are the same. The reactions take place as follows:
CH3COOH(aq) + NaOH(aq) --> NaCH3COO(aq) + H2O(l)
H2SO4(aq) + 2NaOH(aq) --> Na2SO4(aq) + 2H2O(l)

Twice as much NaOH is needed i.e. twice the volume, so 44.40 mL would be used.



If the question changed the acid to acetic acid vs hydrochloric acid as an extra example as the reactions would be 1:1 in the ratios, the volume used would be the same.
 

jazz519

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I had a question:
Wouldn't it require more of naoh to neutralise h2so4 than it took to neutralise ch3cooh? like way more than double?
Don't know if my question makes sense lol.
No it's exactly double. The molar ratios are 2:1 as opposed to 1:1. The volumes obtained in a titration are linked to molar ratios and by the maths it can't be any different.

Can you explain why you're kind of thinking that then I could explain the misunderstanding in that knowledge
 

Mr.GreyCells

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This particular question I now understand after you explained it.
However, I find the number of moles of substance before dissociation and the moles after dissociation thing confusing.
When you make a solution of a substance for eg an acid, wouldn't the water be in dissociated form too thereby providing more h+ ions and oh-ions which would kind mess up the number of h+ ions?
Wouldn't we have to take the volume of water into consideration when we react this solution with a base?
 

jazz519

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This particular question I now understand after you explained it.
However, I find the number of moles of substance before dissociation and the moles after dissociation thing confusing.
When you make a solution of a substance for eg an acid, wouldn't the water be in dissociated form too thereby providing more h+ ions and oh-ions which would kind mess up the number of h+ ions?
Wouldn't we have to take the volume of water into consideration when we react this solution with a base?
The amount of H+ ions from the water and OH- ions is negligible. It would not really have an effect even if there was one. pH = 7 has a [H+] = 10^-7 mol/L and [OH-] = 10^-7 mol/L. Think about what that means for an acid concentration then if you had 0.1 mol/L acid solution it's going to barely change it because the concentration is 1 million times greater
 

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