Chemistry: Post your doubts here!

[nehaoscar]

Why can't it be HZ + OH- ==> H2O + Z-
And pls explain the second part of the ms ... addition of base
Adding OH- ... where did the H+ come from ... there's no H+ in the equation :/

 

[Konstantino Nikolas]

View attachment 59025
View attachment 59026
Why can't it be HZ + OH- ==> H2O + Z-
And pls explain the second part of the ms ... addition of base
Adding OH- ... where did the H+ come from ... there's no H+ in the equation :/

The weak acid in a buffer solution is in an equilibrium which is : HZ <==> H+ + Z-
I guess since it's a solution, they are adding H2O.

Then, the added OH- ions combine with H+ to form water. This reduces the concentration of H+. The position of equilibrium shifts to the right and more of the acid ionises to form H+ and Z- .

 

[qwertypoiu]

View attachment 59025
View attachment 59026
Why can't it be HZ + OH- ==> H2O + Z-
And pls explain the second part of the ms ... addition of base
Adding OH- ... where did the H+ come from ... there's no H+ in the equation :/

What you must understand is that there is actually no such thing as H+ existing by itself. H+ is a proton, and it would be too unstable to stay like that. As soon as it finds an H2O molecule, it binds to it to form the H3O+ ion. So really, whenever we talk about H+ in a solution in chemistry, we mean the H3O+. I suggest you think of H+ and H3O+ as interchangeable terms for the same thing.

A weak acid dissociates in a solution by reacting with water. This is what actually happens:
HZ + H2O <-> Z- + H3O+

But sometimes we just ignore the H2O and just say:
HZ <-> H+ + Z-
You can think of it as subtracting H2O from both sides of the equation, but it isn't what happens really.

The weak acid being in a reversible reaction above when placed in water is very useful for us. It can act as a BUFFER.

When acid is added (we say H+ is added), equilibrium shifts to left side based on Le Chatelier's Principle, so change in pH is resisted.
When alkali is added, OH- reacts with "H+" to make H2O molecules; H+ ion concentration is reduced when base is added to buffer.
Le Chatelier Principle means that equilibrium will shift to right to restore this "H+" and so change in pH is resisted again.

Hope that's clear.

 

[Konstantino Nikolas]

What you must understand is that there is actually no such thing as H+ existing by itself. H+ is a proton, and it would be too unstable to stay like that. As soon as it finds an H2O molecule, it binds to it to form the H3O+ ion. So really, whenever we talk about H+ in a solution in chemistry, we mean the H3O+. I suggest you think of H+ and H3O+ as interchangeable terms for the same thing.

A weak acid dissociates in a solution by reacting with water. This is what actually happens:
HZ + H2O <-> Z- + H3O+

But sometimes we just ignore the H2O and just say:
HZ <-> H+ + Z-
You can think of it as subtracting H2O from both sides of the equation, but it isn't what happens really.

The weak acid being in a reversible reaction above when placed in water is very useful for us. It can act as a BUFFER.

When acid is added (we say H+ is added), equilibrium shifts to left side based on Le Chatelier's Principle, so change in pH is resisted.
When alkali is added, OH- reacts with "H+" to make H2O molecules; H+ ion concentration is reduced when base is added to buffer.
Le Chatelier Principle means that equilibrium will shift to right to restore this "H+" and so change in pH is resisted again.

Hope that's clear.

Thanks for the detailed explanation.
So when base is added, the OH- ions, in reality, react with the H3O+ ions?

 

[nehaoscar]

What you must understand is that there is actually no such thing as H+ existing by itself. H+ is a proton, and it would be too unstable to stay like that. As soon as it finds an H2O molecule, it binds to it to form the H3O+ ion. So really, whenever we talk about H+ in a solution in chemistry, we mean the H3O+. I suggest you think of H+ and H3O+ as interchangeable terms for the same thing.

A weak acid dissociates in a solution by reacting with water. This is what actually happens:
HZ + H2O <-> Z- + H3O+

But sometimes we just ignore the H2O and just say:
HZ <-> H+ + Z-
You can think of it as subtracting H2O from both sides of the equation, but it isn't what happens really.

The weak acid being in a reversible reaction above when placed in water is very useful for us. It can act as a BUFFER.

When acid is added (we say H+ is added), equilibrium shifts to left side based on Le Chatelier's Principle, so change in pH is resisted.
When alkali is added, OH- reacts with "H+" to make H2O molecules; H+ ion concentration is reduced when base is added to buffer.
Le Chatelier Principle means that equilibrium will shift to right to restore this "H+" and so change in pH is resisted again.

Hope that's clear.

Yes thankyou soooo much!

 

[nehaoscar]

So the answer is J
But why not G?? Rotating the left side of G by 180 degrees will give F right?? :/

 

[Eugene99]

What does the reaction of ammonium chloride and Iron(II) sulfate produce? (can it produce ammonium iron(II) sulfate???)

 

[qwertypoiu]

Thanks for the detailed explanation.
So when base is added, the OH- ions, in reality, react with the H3O+ ions?

Do you mean instead of H+? Yes!

What actually happens:

H3O+ + OH- <-> 2H2O

So you see H3O+ donates the proton to OH-, so that TWO water molecules are formed.

We normally simplify this by "subtracting" H2O from both sides:

H+ + OH- <-> H2O

 

[qwertypoiu]

View attachment 59070
So the answer is J
But why not G?? Rotating the left side of G by 180 degrees will give F right?? :/

Rotating the left side of G by 180° actually will not give you F. For this you must try to imagine this compound in 3D.

For G we have:

1. CO2H that COMES OUT of page
2. NH2 that GOES INTO the page.
3. H that is along same plane as page

Now, if you were to rotate this 180°, you'd have a CO2H group that GOES INTO the page, and NH2 that COMES OUT of page. Try think about it. You'll see.

Is this what F is? No. F has CO2H that still comes out of page, so it is not identical to G

 

[nehaoscar]

Rotating the left side of G by 180° actually will not give you F. For this you must try to imagine this compound in 3D.

For G we have:

1. CO2H that COMES OUT of page
2. NH2 that GOES INTO the page.
3. H that is along same plane as page

Now, if you were to rotate this 180°, you'd have a CO2H group that GOES INTO the page, and NH2 that COMES OUT of page. Try think about it. You'll see.

Is this what F is? No. F has CO2H that still comes out of page, so it is not identical to G

Ohh! I see ... but so when you rotate it 180 degrees wouldn't it be
1. COOH is along the same plane as page
2. NH2 goes into page
3. H comes out of page
like so:

Instead of "Now, if you were to rotate this 180°, you'd have a CO2H group that GOES INTO the page, and NH2 that COMES OUT of page."

 

[qwertypoiu]

Ohh! I see ... but so when you rotate it 180 degrees wouldn't it be
1. COOH is along the same plane as page
2. NH2 goes into page
3. H comes out of page
like so:
View attachment 59077
Instead of "Now, if you were to rotate this 180°, you'd have a CO2H group that GOES INTO the page, and NH2 that COMES OUT of page."

I have tried thinking about it, but I still think the CO2H group would go into page after 180 rotation.
I don't see how it could be out of page initially, and then become same plane as paper. That would require non-180° rotation.
I may be misunderstanding something though. Let's see what others have to say.

 

[nehaoscar]

I have tried thinking about it, but I still think the CO2H group would go into page after 180 rotation.
I don't see how it could be out of page initially, and then become same plane as paper. That would require non-180° rotation.
I may be misunderstanding something though. Let's see what others have to say.

Haha alright
Just to confirm ... the stick means same plane as paper, the dotted line means into the page and the bold line means out of the page right?

 

[qwertypoiu]

Haha alright
Just to confirm ... the stick means same plane as paper, the dotted line means into the page and the bold line means out of the page right?

Yep!

 

[Konstantino Nikolas]

Do you mean instead of H+? Yes!

What actually happens:

H3O+ + OH- <-> 2H2O

So you see H3O+ donates the proton to OH-, so that TWO water molecules are formed.

We normally simplify this by "subtracting" H2O from both sides:

H+ + OH- <-> H2O

Oh oh ok ... That was news to me. Thanks

 

[Catherine_1]

View attachment 59070
So the answer is J
But why not G?? Rotating the left side of G by 180 degrees will give F right?? :/

See since this is in three dimensions and not two when you rotate G the COOH which is coming out of the plane of paper would now go into the plane of paper and NH2 would come out,however rotating J from the right side the OH bond will go in and H will come out which is what we want.

 

[nehaoscar]

See since this is in three dimensions and not two when you rotate G the COOH which is coming out of the plane of paper would now go into the plane of paper and NH2 would come out,however rotating J from the right side the OH bond will go in and H will come out which is what we want.

Thanks! Okay so I get why it's J but then this small doubt

So rotating G would look like that so can you explain how when "you rotate G the COOH which is coming out of the plane of paper would now go into the plane of paper"
Because according to the diagram on rotation it would be along the same plane of the paper...

 

[Lola_sweet]

does anyone have good notes or videos for Hess's law?
and can someone please tell me how do we know if an experiment done in Paper 3 exothermic or endothermic?

 

[Rizwan Javed]

and can someone please tell me how do we know if an experiment done in Paper 3 exothermic or endothermic?

If the temperature of the solution which you're experimenting with rises, then the experiment is exomethermic otherwise if it drops then it'll be endothermic.

 

[Lola_sweet]

If the temperature of the solution which you're experimenting with rises, then the experiment is exomethermic otherwise if it drops then it'll be endothermic.

thank you so much

 

[The Sarcastic Retard]

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s15_qp_43.pdf
Q4(c). I need exact detailed solution. Anyone?
in c(ii) where is concentration of Ag given?