acids and bases HELP
A buffer solution is formed when 2.00 g of sodium hydroxide are added to 1.00 dm3 of
a 0.220 mol dm–3 solution of ethanoic acid.
Calculate the pH at 298 K of this buffer solution.
when you do the calculation : At the start: 0.22 moles ethanoic and 0.05 moles NaOH.Therefore, 0.05 moles of ethanoic acid will be neutralised.Therefore, this means that the number of moles of ethanoic left for the buffer is 0.22 - 0.05 = 0.17 moles....and 0.05 moles of the salt is formed as it is a 1:1 mole ratio.The concentrations are the same as the numbers of moles in this case because it's a 1 dm3 solution.
[HA] = 0.17
[A-] = 0.05
..then calculate pH using the Ka to get 4.24 ?
(I didnt write this but my question is
why is the 0.17 divided by 0.05?
also how do you get 0.05 here :"and 0.05 moles of the salt is formed as it is a 1:1 mole ratio"
a 0.220 mol dm–3 solution of ethanoic acid.
Calculate the pH at 298 K of this buffer solution.
when you do the calculation : At the start: 0.22 moles ethanoic and 0.05 moles NaOH.Therefore, 0.05 moles of ethanoic acid will be neutralised.Therefore, this means that the number of moles of ethanoic left for the buffer is 0.22 - 0.05 = 0.17 moles....and 0.05 moles of the salt is formed as it is a 1:1 mole ratio.The concentrations are the same as the numbers of moles in this case because it's a 1 dm3 solution.
[HA] = 0.17
[A-] = 0.05
..then calculate pH using the Ka to get 4.24 ?
(I didnt write this but my question is
why is the 0.17 divided by 0.05?
also how do you get 0.05 here :"and 0.05 moles of the salt is formed as it is a 1:1 mole ratio"
Hi! I saw this and have been trying to work it out.
Where did you get this question, may I ask? And also the explanation.
Do you know that 4.24 is definitely the correct answer? Also which exam board are you doing? Though I don’t think this would affect ability to do this question but just in case.
This is a picture of my workings explained, which will hopefully help with your last question on your post.
What do you mean by “why is the 0.17 divided by 0.05? As I can’t see in your post where this has been done. Hopefully we can try and work out the answer together haha.
Where did you get this question, may I ask? And also the explanation.
Do you know that 4.24 is definitely the correct answer? Also which exam board are you doing? Though I don’t think this would affect ability to do this question but just in case.
This is a picture of my workings explained, which will hopefully help with your last question on your post.
What do you mean by “why is the 0.17 divided by 0.05? As I can’t see in your post where this has been done. Hopefully we can try and work out the answer together haha.
Original post by Kitty_01
Hi! I saw this and have been trying to work it out.
Where did you get this question, may I ask? And also the explanation.
Do you know that 4.24 is definitely the correct answer? Also which exam board are you doing? Though I don’t think this would affect ability to do this question but just in case.
This is a picture of my workings explained, which will hopefully help with your last question on your post.
What do you mean by “why is the 0.17 divided by 0.05? As I can’t see in your post where this has been done. Hopefully we can try and work out the answer together haha.
Where did you get this question, may I ask? And also the explanation.
Do you know that 4.24 is definitely the correct answer? Also which exam board are you doing? Though I don’t think this would affect ability to do this question but just in case.
This is a picture of my workings explained, which will hopefully help with your last question on your post.
What do you mean by “why is the 0.17 divided by 0.05? As I can’t see in your post where this has been done. Hopefully we can try and work out the answer together haha.
Hi, I do AQA a-level chemistry, i found it on physics and maths tutor a while ago- i think it's OCR. yeah 4.24 is the answer on the mark scheme along with the same working out you did but then it divides 0.17 by 0.05 and i don't know why. it says it where its written [HA] and then [A-] underneath with 0.17 and 0.05 next to each- not very clear tho. this if from the Ka formula i think but i'm not sure where the [A-] came from.
i did [H]= square root of Ka multiplied by HA but the markscheme doesnt include square root and also divides by A- which is 0.05. i was wondering why they divide by 0.05
Original post by Kitty_01
Hi! I saw this and have been trying to work it out.
Where did you get this question, may I ask? And also the explanation.
Do you know that 4.24 is definitely the correct answer? Also which exam board are you doing? Though I don’t think this would affect ability to do this question but just in case.
This is a picture of my workings explained, which will hopefully help with your last question on your post.
What do you mean by “why is the 0.17 divided by 0.05? As I can’t see in your post where this has been done. Hopefully we can try and work out the answer together haha.
Where did you get this question, may I ask? And also the explanation.
Do you know that 4.24 is definitely the correct answer? Also which exam board are you doing? Though I don’t think this would affect ability to do this question but just in case.
This is a picture of my workings explained, which will hopefully help with your last question on your post.
What do you mean by “why is the 0.17 divided by 0.05? As I can’t see in your post where this has been done. Hopefully we can try and work out the answer together haha.
Ka is 1.74x10*-5 and also answer is 4.23
Original post by sorhgv
A buffer solution is formed when 2.00 g of sodium hydroxide are added to 1.00 dm3 of
a 0.220 mol dm–3 solution of ethanoic acid.
Calculate the pH at 298 K of this buffer solution.
when you do the calculation : At the start: 0.22 moles ethanoic and 0.05 moles NaOH.Therefore, 0.05 moles of ethanoic acid will be neutralised.Therefore, this means that the number of moles of ethanoic left for the buffer is 0.22 - 0.05 = 0.17 moles....and 0.05 moles of the salt is formed as it is a 1:1 mole ratio.The concentrations are the same as the numbers of moles in this case because it's a 1 dm3 solution.
[HA] = 0.17
[A-] = 0.05
..then calculate pH using the Ka to get 4.24 ?
(I didnt write this but my question is
why is the 0.17 divided by 0.05?
also how do you get 0.05 here :"and 0.05 moles of the salt is formed as it is a 1:1 mole ratio"
a 0.220 mol dm–3 solution of ethanoic acid.
Calculate the pH at 298 K of this buffer solution.
when you do the calculation : At the start: 0.22 moles ethanoic and 0.05 moles NaOH.Therefore, 0.05 moles of ethanoic acid will be neutralised.Therefore, this means that the number of moles of ethanoic left for the buffer is 0.22 - 0.05 = 0.17 moles....and 0.05 moles of the salt is formed as it is a 1:1 mole ratio.The concentrations are the same as the numbers of moles in this case because it's a 1 dm3 solution.
[HA] = 0.17
[A-] = 0.05
..then calculate pH using the Ka to get 4.24 ?
(I didnt write this but my question is
why is the 0.17 divided by 0.05?
also how do you get 0.05 here :"and 0.05 moles of the salt is formed as it is a 1:1 mole ratio"
You are using the acid dissociation equation:
ka = [H+][A-]/[HA]
which rearranges to:
[H+] = ka[HA]/[A-]
now substitute your mol values into the equation. For a buffer the dissolved components are always in the same volume therefore you can do the calculation using the mol quantities after neutralisation.
Original post by charco
You are using the acid dissociation equation:
ka = [H+][A-]/[HA]
which rearranges to:
[H+] = ka[HA]/[A-]
now substitute your mol values into the equation. For a buffer the dissolved components are always in the same volume therefore you can do the calculation using the mol quantities after neutralisation.
ka = [H+][A-]/[HA]
which rearranges to:
[H+] = ka[HA]/[A-]
now substitute your mol values into the equation. For a buffer the dissolved components are always in the same volume therefore you can do the calculation using the mol quantities after neutralisation.
thank you so much, i cant believe i missed that. could you also tell me why some questions avoid square rooting while others square root the calculation?
Original post by sorhgv
Ka is 1.74x10*-5 and also answer is 4.23
How did you get Ka? When I sub values into the Ka expression I don't get that
Original post by Kitty_01
How did you get Ka? When I sub values into the Ka expression I don't get that
Ka is given to you so you dont have to
Original post by sorhgv
thank you so much, i cant believe i missed that. could you also tell me why some questions avoid square rooting while others square root the calculation?
They are two different types of calculation.
When you are determining the pH of a weak acid you know that the concentration of hydrogen ions and the concentrations of the anions is the same:
HA <==> H+ + A-
So you can substitute the anion concentration for another hydrogen
ka = [H+][A-]/[HA] becomes ka = [H+]2/[HA]
But in a buffer the concentration of hydrogen ions does NOT equal the concentration of the conjugate base ions (usually), so you have to substitute in all of the values.
If, by coincidence they are equal then ka = [H+] and pH = pKa
Check out my work on buffers in this video:
Original post by sorhgv
Ka is given to you so you dont have to
Ohh brill.
Original post by charco
They are two different types of calculation.
When you are determining the pH of a weak acid you know that the concentration of hydrogen ions and the concentrations of the anions is the same:
HA <==> H+ + A-
So you can substitute the anion concentration for another hydrogen
ka = [H+][A-]/[HA] becomes ka = [H+]2/[HA]
But in a buffer the concentration of hydrogen ions does NOT equal the concentration of the conjugate base ions (usually), so you have to substitute in all of the values.
If, by coincidence they are equal then ka = [H+] and pH = pKa
Check out my work on buffers in this video:
When you are determining the pH of a weak acid you know that the concentration of hydrogen ions and the concentrations of the anions is the same:
HA <==> H+ + A-
So you can substitute the anion concentration for another hydrogen
ka = [H+][A-]/[HA] becomes ka = [H+]2/[HA]
But in a buffer the concentration of hydrogen ions does NOT equal the concentration of the conjugate base ions (usually), so you have to substitute in all of the values.
If, by coincidence they are equal then ka = [H+] and pH = pKa
Check out my work on buffers in this video:
thank you! your help was really useful. i understand it now
Quick Reply
Related discussions
- Learning Transition metal complex ions.
- Chemistry question help pls
- how to identify strong acids and alkalis???
- Chem question help pls
- A level carboxylic acids
- Acidic alkaline basic metals and non metals
- Help with Acid-Bases
- A level chemistry
- Chemistry question help please
- Cheimstry
- Help with IB Chemistry IA
- How hard is ACID-Base equilbira in a level chemistry ?
- What is logarithms (chemistry help)
- Chemistry
- Chemistry MCQ help please
- Chemistry A-Level: Mechanism Help Please x
- Chemistry F325- End Point/ Equivalence Point
- A level chemistry help - pt 2
- what is formed when ammonium sulfate reacts with HCl acid?
- transition metals (a level chemistry)(help me)
Latest
Trending
