# chemistry buffer question

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#1
A chemist needs Ph 4 of a buffer solution of HNO2 and NaNO2, Ka for HNO2 = 4x10^-4, conc for both are 0.25 moldm-3, what volumes would you use for both in a 250cm3 buffer?

Got this question on a past paper and just cant figure it out...

(answer is 200cm3 NaNO2, 50cm3 HNO2)
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1 year ago
#2
(Original post by blossomhibbert99)
A chemist needs Ph 4 of a buffer solution of HNO2 and NaNO2, Ka for HNO2 = 4x10^-4, conc for both are 0.25 moldm-3, what volumes would you use for both in a 250cm3 buffer?

Got this question on a past paper and just cant figure it out...

(answer is 200cm3 NaNO2, 50cm3 HNO2)
So you just need to work out the ratio of moles of each needed in the buffer.

ka = [H+][A-]/[HA]

ka/[H+] = [A-]/[HA]

this means salt/weak acid ratio

pH 4: [H+] = 1 x 10-4
ka = 4 x 10-4

hence ka/[H+] = 4 = [A-]/[HA] = [salt] /[weak acid]

so the salt concentration has to be 4 x the weak acid concentration. As the two solutions are the same concentration then you have to mix four volumes of salt with 1 volume of acid.
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#3
How do you get ka / [H+]? ahh exams tmr super stressed also cheers for your help 0
1 year ago
#4
(Original post by blossomhibbert99)
How do you get ka / [H+]? ahh exams tmr super stressed also cheers for your help Ka is given to you, and you know the pH you want is 4. So antilog(-4) gives you [H+]

Then just form the equation, subbing in values for Ka and [H+]:

Ka = [H+][salt] / [acid]

rearranging, as mentioned above, you get [salt] / [acid] = 4, so that's 1:4, total 5 parts

250 / 5 = 50.

So overall you want 50cm^3 acid, 200cm^3 salt
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#5
i get everything apart from the rearranging.. If you have the normal equation, then do ka x [HA], you get:

[H+] x [A-] = ka x [HA]

Then how do you get [A-] / [HA]??
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1 year ago
#6
What exam board is this??? Please not AQA...
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#7
What exam board is this??? Please not AQA...
nah OCR haha fml exam is tomorrow im so ****ed aghh
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1 year ago
#8
(Original post by blossomhibbert99)
i get everything apart from the rearranging.. If you have the normal equation, then do ka x [HA], you get:

[H+] x [A-] = ka x [HA]

Then how do you get [A-] / [HA]??
The expression for ka is defined as:

Ka = [H+][A-] / [HA]
Start from there, and you can divide both sides by [H+].

That leaves Ka / [H+] = [A-] / [HA]

If your expression does start as Ka x [HA] = [H+][A-], you just have to divide both sides by both [HA] and [H+] to find the needed expression
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4 months ago
#9
(Original post by blossomhibbert99)
A chemist needs Ph 4 of a buffer solution of HNO2 and NaNO2, Ka for HNO2 = 4x10^-4, conc for both are 0.25 moldm-3, what volumes would you use for both in a 250cm3 buffer?

Got this question on a past paper and just cant figure it out...

(answer is 200cm3 NaNO2, 50cm3 HNO2)

I understand how you work out the ratio of NO2- concentration and HNO2 concentration using your ka and H+ concentration. But, why do you divide the 250cm3 by 5? Isn't five parts the total concentration so why would you divide the volume by the total concentration?!
This is so mathsy its so confusingggg I thought you have to work out moles since they gave cocentration to then work out volume.
(Original post by charco)
So you just need to work out the ratio of moles of each needed in the buffer.

ka = [H+][A-]/[HA]

ka/[H+] = [A-]/[HA]

this means salt/weak acid ratio

pH 4: [H+] = 1 x 10-4
ka = 4 x 10-4

hence ka/[H+] = 4 = [A-]/[HA] = [salt] /[weak acid]

so the salt concentration has to be 4 x the weak acid concentration. As the two solutions are the same concentration then you have to mix four volumes of salt with 1 volume of acid.
I understand how you work out the ratio of NO2- concentration and HNO2 concentration using your ka and H+ concentration. But, why do you divide the 250cm3 by 5? Isn't five parts the total concentration so why would you divide the volume by the total concentration?!

This is so mathsy its so confusingggg I thought you have to work out moles since they gave concentration to then work out volume
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