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How to Prepare 20 mM Potassium Phosphate Buffer pH 6.8?
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+ Ph
+ Phosphates
+ Potassium
+ Biochemistry
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Mubarak Abdullahi Khaleed
How to Prepare 20 mM Potassium Phosphate Buffer pH 6.8?
Oluwatobi T. Somade Thanks for the detailed method. I exactly used the same method, but everytime I found confusion between my results and the results of softwares like Buffer-maker for exemple. I don't know why?!!
Oluwatobi T. Somade Thanks for the detailed method. I exactly used the same method, but everytime I found confusion between my results and the results of softwares like Buffer-maker for exemple. I don't know why?!!
Use Henderson Hasselbalch equation pH = pKa + log([salt]/[acid]) pKa nearest to 6.8 is 7.2, therefore, 6.8 = 7.2 + log ([salt]/[acid]) -0.4 = log([salt]/[acid]) Antilog (-0.4) = [salt]/[acid] 0.398 = [salt]/[acid] [Salt] = 0.398[acid]..........….........equation 1 Also, from your question, [Acid] + [salt] = 0.02......................equation 2 [Acid] + 0.398[acid] = 0.02 1.398[acid] = 0.02 [Acid] = 0.02/1.398 [Acid] = 0.0143M From equation 1, [Salt] = 0.398 * 0.0143 [Salt] = 0.00569M Then you calculate the mass concentrations of the acid and salt Mass concentration of acid (KH2PO4) = molar concentration * molar mass = 0.0143 * 136 = 1.945 g/dm3 Mass concentration of salt (K2HPO4) = 0.00569 * 174 = 0.990 g/dm3 The respective mass concentrations are to prepare 1000 ml of the buffer, But to prepare 100 ml, divide the mass concentrations by 10, to give 0.1945 g/dm3 and 0.099 g/dm3 for acid and base respectively. Finally, to prepare 100mls of 20mM potassium phosphate buffer pH 6.8, Weight the respective grammes of acid and salt in a beaker, and measure 100 ml of distilled water to dissolve it. Measure the pH, and adjust if need be. I believe this is explanatory enough
Use Henderson Hasselbalch equation pH = pKa + log([salt]/[acid]) pKa nearest to 6.8 is 7.2, therefore, 6.8 = 7.2 + log ([salt]/[acid]) -0.4 = log([salt]/[acid]) Antilog (-0.4) = [salt]/[acid] 0.398 = [salt]/[acid] [Salt] = 0.398[acid]..........….........equation 1 Also, from your question, [Acid] + [salt] = 0.02......................equation 2 [Acid] + 0.398[acid] = 0.02 1.398[acid] = 0.02 [Acid] = 0.02/1.398 [Acid] = 0.0143M From equation 1, [Salt] = 0.398 * 0.0143 [Salt] = 0.00569M Then you calculate the mass concentrations of the acid and salt Mass concentration of acid (KH2PO4) = molar concentration * molar mass = 0.0143 * 136 = 1.945 g/dm3 Mass concentration of salt (K2HPO4) = 0.00569 * 174 = 0.990 g/dm3 The respective mass concentrations are to prepare 1000 ml of the buffer, But to prepare 100 ml, divide the mass concentrations by 10, to give 0.1945 g/dm3 and 0.099 g/dm3 for acid and base respectively. Finally, to prepare 100mls of 20mM potassium phosphate buffer pH 6.8, Weight the respective grammes of acid and salt in a beaker, and measure 100 ml of distilled water to dissolve it. Measure the pH, and adjust if need be. I believe this is explanatory enough
Oluwatobi T. Somade Thanks for the detailed method. I exactly used the same method, but everytime I found confusion between my results and the results of softwares like Buffer-maker for exemple. I don't know why?!!
Oluwatobi T. Somade Thanks for the detailed method. I exactly used the same method, but everytime I found confusion between my results and the results of softwares like Buffer-maker for exemple. I don't know why?!!
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you need weak acid/base and its salt to prepare a buffer
you need weak acid/base and its salt to prepare a buffer
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If you are lazy like me you can use a buffer calculator.
https://www.liverpool.ac.uk/buffers/buffercalc.html
If you are lazy like me you can use a buffer calculator.
https://www.liverpool.ac.uk/buffers/buffercalc.html
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Oluwatobi T. Somade For example with this case, the software give 1.6 g/l (KH2PO4) and 1.44 g/l (K2HPO4)
Oluwatobi T. Somade For example with this case, the software give 1.6 g/l (KH2PO4) and 1.44 g/l (K2HPO4)
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It is from the question, which request preparation of 20mM buffer. 20mM is same thing as 20/1000 = 0.02M
It is from the question, which request preparation of 20mM buffer. 20mM is same thing as 20/1000 = 0.02M
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Use Henderson Hasselbalch equation
pH = pKa + log([salt]/[acid])
pKa nearest to 6.8 is 7.2, therefore,
6.8 = 7.2 + log ([salt]/[acid])
-0.4 = log([salt]/[acid])
Antilog (-0.4) = [salt]/[acid]
0.398 = [salt]/[acid]
[Salt] = 0.398[acid]..........….........equation 1
Also, from your question,
[Acid] + [salt] = 0.02......................equation 2
[Acid] + 0.398[acid] = 0.02
1.398[acid] = 0.02
[Acid] = 0.02/1.398
[Acid] = 0.0143M
From equation 1,
[Salt] = 0.398 * 0.0143
[Salt] = 0.00569M
Then you calculate the mass concentrations of the acid and salt
Mass concentration of acid (KH2PO4) = molar concentration * molar mass
= 0.0143 * 136
= 1.945 g/dm3
Mass concentration of salt (K2HPO4) = 0.00569 * 174
= 0.990 g/dm3
The respective mass concentrations are to prepare 1000 ml of the buffer,
But to prepare 100 ml, divide the mass concentrations by 10, to give 0.1945 g/dm3 and 0.099 g/dm3 for acid and base respectively.
Finally, to prepare 100mls of 20mM potassium phosphate buffer pH 6.8,
Weight the respective grammes of acid and salt in a beaker, and measure 100 ml of distilled water to dissolve it. Measure the pH, and adjust if need be.
I believe this is explanatory enough
Use Henderson Hasselbalch equation
pH = pKa + log([salt]/[acid])
pKa nearest to 6.8 is 7.2, therefore,
6.8 = 7.2 + log ([salt]/[acid])
-0.4 = log([salt]/[acid])
Antilog (-0.4) = [salt]/[acid]
0.398 = [salt]/[acid]
[Salt] = 0.398[acid]..........….........equation 1
Also, from your question,
[Acid] + [salt] = 0.02......................equation 2
[Acid] + 0.398[acid] = 0.02
1.398[acid] = 0.02
[Acid] = 0.02/1.398
[Acid] = 0.0143M
From equation 1,
[Salt] = 0.398 * 0.0143
[Salt] = 0.00569M
Then you calculate the mass concentrations of the acid and salt
Mass concentration of acid (KH2PO4) = molar concentration * molar mass
= 0.0143 * 136
= 1.945 g/dm3
Mass concentration of salt (K2HPO4) = 0.00569 * 174
= 0.990 g/dm3
The respective mass concentrations are to prepare 1000 ml of the buffer,
But to prepare 100 ml, divide the mass concentrations by 10, to give 0.1945 g/dm3 and 0.099 g/dm3 for acid and base respectively.
Finally, to prepare 100mls of 20mM potassium phosphate buffer pH 6.8,
Weight the respective grammes of acid and salt in a beaker, and measure 100 ml of distilled water to dissolve it. Measure the pH, and adjust if need be.
I believe this is explanatory enough
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If you are lazy like me you can use a buffer calculator.
https://www.liverpool.ac.uk/buffers/buffercalc.html
If you are lazy like me you can use a buffer calculator.
https://www.liverpool.ac.uk/buffers/buffercalc.html
More
VOTE
How can you prepare this buffer if you have only KH2PO4?
How can you prepare this buffer if you have only KH2PO4?
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Thanks a lot. I could not undertand how did you write equation 2 . Does it mean the sum of acid and baic is 0.02M ?
Thanks a lot. I could not undertand how did you write equation 2 . Does it mean the sum of acid and baic is 0.02M ?
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@Abegal Then, this link will help, "https://youtu.be/B6ADtyWu6i4"
The video is explicitly clear.
@Abegal Then, this link will help, "https://youtu.be/B6ADtyWu6i4"
The video is explicitly clear.
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@Zohra Kebili, buffer softwares can be unreliable. The programmed molecular masses or the components of the buffer may be different.
@Zohra Kebili, buffer softwares can be unreliable. The programmed molecular masses or the components of the buffer may be different.
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@Oluwatobi. I checked the WM of K2HPO4 and KH2PO4 putted by the software. They are correct!
@Oluwatobi. I checked the WM of K2HPO4 and KH2PO4 putted by the software. They are correct!
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