Question
How to Prepare a 50 mM pH 6 Citrate buffer starting from citric acid as the sole source of conjugate base.
Answers
Citric acid has three ionization constants; i.e., k1, k2, and k3.
pK1 = 3.13
pK2 = 4.76
pK3 = 6.39
Your tables may not agree exactly; use the numbers in your text/notes.
The pH you want in the solution should be as close as possible to one of the pKa values; therefore, I would use pK3.
Let's say you want to prepare 1 L of 0.05 M buffer. Weigh 0.05 mols citric acid solid, place in a conveniently sized flask, and add a NaOH solution, dropwise, until the pH reads 6.39, transfer the contents to a 1 L volumetric flask, add DI water to the mark on the flask, stopper and mix thoroughly. When finished you should have a solution that is 0.0355M in citric acid and 0.0145 M in NaH2C (sodium dihydrogen citrate) although the M concentration is not something that must be done before you have prepared the material. I did that just for the fun of it. If you will substitute those concentrations into the Henderson-Hasselbalch equation, you will see the pH is 6.00.
pK1 = 3.13
pK2 = 4.76
pK3 = 6.39
Your tables may not agree exactly; use the numbers in your text/notes.
The pH you want in the solution should be as close as possible to one of the pKa values; therefore, I would use pK3.
Let's say you want to prepare 1 L of 0.05 M buffer. Weigh 0.05 mols citric acid solid, place in a conveniently sized flask, and add a NaOH solution, dropwise, until the pH reads 6.39, transfer the contents to a 1 L volumetric flask, add DI water to the mark on the flask, stopper and mix thoroughly. When finished you should have a solution that is 0.0355M in citric acid and 0.0145 M in NaH2C (sodium dihydrogen citrate) although the M concentration is not something that must be done before you have prepared the material. I did that just for the fun of it. If you will substitute those concentrations into the Henderson-Hasselbalch equation, you will see the pH is 6.00.
I am confused on the fact that if I am going to use the concentration that I got from the HH equation which was 0.0355M, then did Dr.Bob222 said that you weight 0.05 moles to add?
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