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August 05, 2021, 05:02:40 am

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#### SynX

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« Reply #3765 on: August 06, 2019, 03:45:08 pm »
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Use the Henderson-hasselbalch equation .  It is given like this :

pH = pKA + log base 10 of (conjugate base concentration/ conjugate acid concentration).
Since we have the pH, manipulate the equation to find the concentration of acetic acid and ion.
Correct me if I am wrong.
Wow, that’s great! I just have a hard time memorizing “Henderson-hasselbalch equation” all I know about this is the pH=-log(aH+), by doing some research I found out that pKa=-log(Ka), which is really similar to the pH equation, as well as pOH equation. But the definition of Ka is “where "Ka" is the “equilibrium constant for the ionization of the acid”. Seems the pH and pOH value of a substance adds up to a constant around 14. Just learning these at school. But the concept of pKa is confusing!
Plus, I just figured out the function of the log button by playing the calculator a few weeks ago...
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#### horse9996

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« Reply #3766 on: August 12, 2019, 06:18:53 pm »
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Wow, that’s great! I just have a hard time memorizing “Henderson-hasselbalch equation” all I know about this is the pH=-log(aH+), by doing some research I found out that pKa=-log(Ka), which is really similar to the pH equation, as well as pOH equation. But the definition of Ka is “where "Ka" is the “equilibrium constant for the ionization of the acid”. Seems the pH and pOH value of a substance adds up to a constant around 14. Just learning these at school. But the concept of pKa is confusing!
Plus, I just figured out the function of the log button by playing the calculator a few weeks ago...

Ka is literally just any old equilibrium constant, the a just means that its for an acid. pKa is the same as pH, except with Ka rather than [H+]. This is to make values easier to work with and more comparable. Also, pKa + pKb = 14 (pKb is the same thing but the dissociation of a weak base rather than acid - denoted by b rather than a). To remember the Henderson-Hasselbalch equation, remember 'aha' - since the log is [A-]/[HA] = [base]/[acid]
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#### louisaaa01

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« Reply #3767 on: September 04, 2019, 02:56:58 pm »
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For 17 - I've just assumed that the acetic acid completely dissociates. I know that isn't the case, but they haven't given any Ka values - are we expected to just know Ka for acetic acid? (But even if it only partially dissociates, shouldn't the mass required be greater than what I've calculated?)

Thank you.
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#### horse9996

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« Reply #3768 on: September 08, 2019, 03:27:16 pm »
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For 17 - I've just assumed that the acetic acid completely dissociates. I know that isn't the case, but they haven't given any Ka values - are we expected to just know Ka for acetic acid? (But even if it only partially dissociates, shouldn't the mass required be greater than what I've calculated?)

Thank you.

For 17 you need the Ka value - I would assume it should be given to you? Check the data sheet. Form a RICE table. The equilibrium concentration of H+ ions can be determined from the pH and then let the initial concentration of acetic acid be x. By using the equilibrium constant expression and the Ka value, you should be able to calculate x and therefore the mass of acetic acid needed.

For 18 it is as follows:
- Not A - cellulose is a polymer which requires glucose as its monomer so this doesn't make sense as coming from ethylene
- Not C - glucose can't be formed from ethylene in 1 step as far as I know
- Not D - ethanol can't undergo polymerisation and doesn't form any polymers
- B - styrene can be formed from ethylene (one of the hydrogens is replaced with a benzene) and this undergoes addition polymerisation to form a polymer (polystyrene)
Hence the answer must be B.

I hope that makes sense
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#### Katie-E

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« Reply #3769 on: September 13, 2019, 08:51:41 pm »
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Hey all,

Does anyone know of any tutors in the Bankstown/ 2200 postcode area who are available for some year 12 HSC Chemistry tutoring?

#### DamnDhruv

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« Reply #3770 on: October 11, 2019, 07:55:31 pm »
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hello I have the atar notes topic notes ook for chemistry wherein i found out that they wrote potassium chloride is not soluble. Now is this  a genuine mistake or am just missing something?
here is the image for reference:

#### Erutepa

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« Reply #3771 on: October 11, 2019, 08:14:16 pm »
+2
hello I have the atar notes topic notes ook for chemistry wherein i found out that they wrote potassium chloride is not soluble. Now is this  a genuine mistake or am just missing something?
here is the image for reference:
Potassium chloride is definitely soluble. Unfortunately, you will come across the odd mistake/typo in pretty much all resources from textbooks to study notes like these, so just be wary.
« Last Edit: October 11, 2019, 08:33:38 pm by Erutepa »
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#### DamnDhruv

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« Reply #3772 on: October 11, 2019, 11:52:25 pm »
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Wow, that’s great! I just have a hard time memorizing “Henderson-hasselbalch equation” all I know about this is the pH=-log(aH+), by doing some research I found out that pKa=-log(Ka), which is really similar to the pH equation, as well as pOH equation. But the definition of Ka is “where "Ka" is the “equilibrium constant for the ionization of the acid”. Seems the pH and pOH value of a substance adds up to a constant around 14. Just learning these at school. But the concept of pKa is confusing!
Plus, I just figured out the function of the log button by playing the calculator a few weeks ago...

its on the fomula sheet i think

#### RACHEL1111

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« Reply #3773 on: October 29, 2019, 09:35:44 am »
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Hi, Could anyone help me with this questions, question 8 part b.
when the answer says ' sextet' , I understand that it refers there is carbon next door attaching 5 proton. But as far as I can see, there isn't any such carbon.
Also, I am a bit confused about N+1 rule because the next door carbon could refer to either left or right? So, which side is it?
« Last Edit: October 29, 2019, 09:40:36 am by RACHEL1111 »

#### louisaaa01

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« Reply #3774 on: October 29, 2019, 09:50:17 am »
+1
Hi, Could anyone help me with this questions, question 8 part b.
when the answer says ' sextet' , I understand that it refers there is carbon next door attaching 5 proton. But as far as I can see, there isn't any such carbon.
Also, I am a bit confused about N+1 rule because the next door carbon could refer to either left or right? So, which side is it?

Hi!

Since carbon forms a maximum of four bonds (as it is tetravalent), a single carbon actually can't bond with five protons. The N+1 rule refers to the carbons on both sides of the carbon you're thinking of.

For instance, if we consider the diagram on the left, the second carbon from the left is a CH2 group, bonded to a CH3 group on the left and a CH2 group on the right. An arrangement like this will produce a sextet in the H-NMR spectrum - there is a carbon atom attached to two carbon atoms with five protons in total that are directly attached, so using the n+1 rule, there is a sextet.

The other sextet from the diagram on the left, I think, comes from the carbon attached to the hydroxyl group. Note that it is also attached to a CH3 and a CH2 group (five identical hydrogens on either side), so a sextet is present.

When we do the N+1 rule, it's also worth noting that a hydroxyl group will show a singlet (it is not attached to an atom directly attached to any other hydrogens).

Also, when using the N+1 rule on adjacent hydrogens, if an -OH group is present on the adjacent carbon, you don't add this. For instance, if we consider the third carbon on the left diagram, we see that it is attached to a CH2 group, and a carbon attached to a H and an OH. In this case, we look only at the hydrogens directly attached to adjacent carbons - so the third carbon in this diagram is attached to a carbon with 2 hydrogens directly attached and another carbon with only one hydrogen attached (3 in total). This produces a quartet by the N+1 rule.

Similar logic is applied to the structural formula on the right.

I hope this helps! If you have any more questions, don't hesitate to ask.
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#### RACHEL1111

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« Reply #3775 on: October 29, 2019, 02:49:03 pm »
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Hi!

Since carbon forms a maximum of four bonds (as it is tetravalent), a single carbon actually can't bond with five protons. The N+1 rule refers to the carbons on both sides of the carbon you're thinking of.

For instance, if we consider the diagram on the left, the second carbon from the left is a CH2 group, bonded to a CH3 group on the left and a CH2 group on the right. An arrangement like this will produce a sextet in the H-NMR spectrum - there is a carbon atom attached to two carbon atoms with five protons in total that are directly attached, so using the n+1 rule, there is a sextet.

The other sextet from the diagram on the left, I think, comes from the carbon attached to the hydroxyl group. Note that it is also attached to a CH3 and a CH2 group (five identical hydrogens on either side), so a sextet is present.

When we do the N+1 rule, it's also worth noting that a hydroxyl group will show a singlet (it is not attached to an atom directly attached to any other hydrogens).

Also, when using the N+1 rule on adjacent hydrogens, if an -OH group is present on the adjacent carbon, you don't add this. For instance, if we consider the third carbon on the left diagram, we see that it is attached to a CH2 group, and a carbon attached to a H and an OH. In this case, we look only at the hydrogens directly attached to adjacent carbons - so the third carbon in this diagram is attached to a carbon with 2 hydrogens directly attached and another carbon with only one hydrogen attached (3 in total). This produces a quartet by the N+1 rule.

Similar logic is applied to the structural formula on the right.

I hope this helps! If you have any more questions, don't hesitate to ask.

That's very helpful! Thank you very much!

#### RACHEL1111

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« Reply #3776 on: October 29, 2019, 03:02:19 pm »
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Hey, Could anyone help me with this question with chemical structure?
I lose track of the second part of solution.
1. For the highly shifted (around 4.2ppm) peak which represents those two functional group: is because that they are unshielded that they have high ppm?
2. And I am completely confused from the sentence: the second group will experience a higher peak... For example: what does height of the peak mean? what does alkyl portion mean? After all, please help me with the way to eliminate option 2.

#### classof2019

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« Reply #3777 on: October 29, 2019, 04:17:38 pm »
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A 25.0mL sample of acetic acid has a pH of 1.30. It is mixed with 20.0mL of 0.130mol/L calcium hydroxide. What is the pH of this mixture?
A. 2.4
B. 2.9
C. 11.1
D. 11.6

I'm consistently getting a pH of about 12.9, which isn't even one of the options! So I'm not sure whether I'm making a mistake or the question itself is flawed. Any assistance is greatly appreciated. Cheers.

#### InnererSchweinehund

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« Reply #3778 on: October 30, 2019, 08:29:25 am »
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A 25.0mL sample of acetic acid has a pH of 1.30. It is mixed with 20.0mL of 0.130mol/L calcium hydroxide. What is the pH of this mixture?
A. 2.4
B. 2.9
C. 11.1
D. 11.6

I'm consistently getting a pH of about 12.9, which isn't even one of the options! So I'm not sure whether I'm making a mistake or the question itself is flawed. Any assistance is greatly appreciated. Cheers.

Hi there!

I did the question (working below) and got the answer of pH = 11.1

I'm not 100% sure if this is correct so if someone could check it and let me (and classof2019) know, that would be great!

Thanks!

(ps. Sorry if the image quality is bad!)

#### RACHEL1111

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« Reply #3779 on: October 30, 2019, 08:47:50 am »
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Hi there!

I did the question (working below) and got the answer of pH = 11.1

I'm not 100% sure if this is correct so if someone could check it and let me (and classof2019) know, that would be great!

Thanks!

(ps. Sorry if the image quality is bad!)

Hi, I might be wrong. There are some point in your solution that I don't understand.
Since the ratio between calcium hydroxide and acetate acid is 1:2, 1.25*10-3 mols of CH3COOH should react with 6.25*10-4 mols of Ca(OH)2. So,Ca(OH)2 left would be 1.975*10-3 mols.
And I think we should do 1.975*10-3 / volume which is 4.5*10-2 to get concentration of OH- first and substitute it into -log formula.
But I still did not get the answer though...