Buffers question...help!
The question is from the June 2011 paper, OCR exam board.
"The 'magic tang' in many sweets is obtained by use of acidic buffers. A sweet manufacturer carried out tasting tests with consumers and identified the acid taste that gives the 'magic tang' to a sweet.
The manufacturer was convinced that the 'magic tang' would give the company a competitive edge and he asked the company's chemists to identify the chemicals needed to generate the required taste. The chemists' findings would be a key factor in the success of the sweets.
The team of chemists identified that a pH of 3.55 was required and they worked to develop a buffer at this pH. The chemists decided to use one of the acids in Table 4.1 (see attachments) and a salt of the acid to prepare this buffer.
"
I don't know how I'm supposed to check for pH values if they haven't given me ANY value for concentration of the acids.
"The 'magic tang' in many sweets is obtained by use of acidic buffers. A sweet manufacturer carried out tasting tests with consumers and identified the acid taste that gives the 'magic tang' to a sweet.
The manufacturer was convinced that the 'magic tang' would give the company a competitive edge and he asked the company's chemists to identify the chemicals needed to generate the required taste. The chemists' findings would be a key factor in the success of the sweets.
The team of chemists identified that a pH of 3.55 was required and they worked to develop a buffer at this pH. The chemists decided to use one of the acids in Table 4.1 (see attachments) and a salt of the acid to prepare this buffer.
•
Deduce the chemicals required by the chemists to prepare this buffer.
•
Calculate the relative concentrations of the acid and its salt needed by the chemist to make this buffer.
•
Comment on the validity of the prediction that the pH of the sweet would give the sweets the 'magic tang'.
I don't know how I'm supposed to check for pH values if they haven't given me ANY value for concentration of the acids.
*bump*
http://www.a-levelchemistry.co.uk/OCR%20Chemistry%20A/Unit%205/F325_MS_June11.pdf
Scroll down to page 10
I didn't get how to do this either, it was a stupidly worded question and I'm quite bad at remembering and rearranging the equations anyway.. There's some sort of link between pKa values and pH, I think pH values are only a bit different to pKa but I've forgotten why, but anyway the acid you use is lactic acid and then its' salt which is sodium lactate.
Scroll down to page 10
I didn't get how to do this either, it was a stupidly worded question and I'm quite bad at remembering and rearranging the equations anyway.. There's some sort of link between pKa values and pH, I think pH values are only a bit different to pKa but I've forgotten why, but anyway the acid you use is lactic acid and then its' salt which is sodium lactate.
Original post by subzero0137
*bump*
pH = 3.55
[H+] = ?
Ka = ?
pKa = ?
Original post by chopsticks
http://www.a-levelchemistry.co.uk/OCR%20Chemistry%20A/Unit%205/F325_MS_June11.pdf
Scroll down to page 10 I didn't get how to do this either, it was a stupidly worded question and I'm quite bad at remembering and rearranging the equations anyway.. There's some sort of link between pKa values and pH, I think pH values are only a bit different to pKa but I've forgotten why, but anyway the acid you use is lactic acid and then its' salt which is sodium lactate.
Scroll down to page 10
I didn't get how to do this either, it was a stupidly worded question and I'm quite bad at remembering and rearranging the equations anyway.. There's some sort of link between pKa values and pH, I think pH values are only a bit different to pKa but I've forgotten why, but anyway the acid you use is lactic acid and then its' salt which is sodium lactate.Remember that pH = pKa at Half-neutralisation point
Original post by thegodofgod
Remember that pH = pKa at Half-neutralisation point
Oh yeah that's it! I knew it was something like that
thanks bro!Funny enough my teacher showed us this question.
The right thing to do is to use lactic acid. His reasons were because it's pKa is closest value to the pH.
I just accepted that.
I tried to figure it out the PH mathematically but i couldn't get it.
I did the following using the infomation from Lactic acid.
10^-3.86 which is (10^pKa)
Ans = Ka
Ka = [H+][H+]
SqrtKA
-logAns
Any help/tips/corrections greatly welcomed
The right thing to do is to use lactic acid. His reasons were because it's pKa is closest value to the pH.
I just accepted that.
I tried to figure it out the PH mathematically but i couldn't get it.
I did the following using the infomation from Lactic acid.
10^-3.86 which is (10^pKa)
Ans = Ka
Ka = [H+][H+]
SqrtKA
-logAns
Any help/tips/corrections greatly welcomed
Original post by thegodofgod
Remember that pH = pKa at Half-neutralisation point
Half-neutralisation? I'm not sure if that's in the specification. I've got the OCR textbook, and I've checked the relevant chapters. It doesn't mention anything about pH equaling pKa in any situation.
Is it to do with the titration curves?
Original post by Woodworth
Funny enough my teacher showed us this question.
The right thing to do is to use lactic acid. His reasons were because it's pKa is closest value to the pH.
The right thing to do is to use lactic acid. His reasons were because it's pKa is closest value to the pH.
Yeah, this is what the first part is looking for; the pH of a buffer solution operates at roughly pKa ± 1. So the closer the pH to the pKa the better.
I tried to figure it out the PH mathematically but i couldn't get it.
I did the following using the infomation from Lactic acid.
10^-3.86 which is (10^pKa)
Ans = Ka
Ka = [H+][H+]
SqrtKA
-logAns
Any help/tips/corrections greatly welcomed
I did the following using the infomation from Lactic acid.
10^-3.86 which is (10^pKa)
Ans = Ka
Ka = [H+][H+]
SqrtKA
-logAns
Any help/tips/corrections greatly welcomed
There's an equation called the Henderson–Hasselbalch equation that's useful here:
pH = pKa + log ([A-]/[HA])
Plug the numbers in and find the [A-]/[HA] ratio (salt to acid)
Original post by subzero0137
Half-neutralisation? I'm not sure if that's in the specification. I've got the OCR textbook, and I've checked the relevant chapters. It doesn't mention anything about pH equaling pKa in any situation.
Is it to do with the titration curves?
Is it to do with the titration curves?
Sorry, it's on my spec (AQA), not sure about others.
Original post by thegodofgod
Sorry, it's on my spec (AQA), not sure about others.
i think it's called equivalence point on my syllabus (ocr chemsitry A)
Original post by chopsticks
Original post by chopsticks
i think it's called equivalence point on my syllabus (ocr chemsitry A)
It's also called this on the Edexcel specification.
Original post by chopsticks
i think it's called equivalence point on my syllabus (ocr chemsitry A)
So titration curves and buffer solutions are linked?
Original post by subzero0137
So titration curves and buffer solutions are linked?
yep- they're both part of the same topic if you're doing the ocr syllabus as me - it really helps reading through the textbook and doing some exam questions because it links it all in together.
buffer solutions minimise ph changes, and titration curves (i'm assuming you mean the ones where it's a neutralisation) show neutralisation, during which ph changes.
Original post by chopsticks
i think it's called equivalence point on my syllabus (ocr chemsitry A)
Original post by NutterFrutter
It's also called this on the Edexcel specification.
You know the point when titrating strong acids/bases and there's that really steep point - like when it goes up by 7 or 8 pH's when just adding more base?
Half-way between the top and bottom of that steep line is the equivalence point, and it's also called the half-neutralisation point
Original post by thegodofgod
Original post by thegodofgod
You know the point when titrating strong acids/bases and there's that really steep point - like when it goes up by 7 or 8 pH's when just adding more base?
Half-way between the top and bottom of that steep line is the equivalence point, and it's also called the half-neutralisation point
Half-way between the top and bottom of that steep line is the equivalence point, and it's also called the half-neutralisation point
The equivalence point is when two reactants are mixed in exactly the same proportions indicated by the reaction equation, it's also known as the neutral point but this isn't always the case.
That's the Edexcel definition.
Hi, can anyone plz guide what contents we should add if question is ' preparation of buffers of different ph and molarity'
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