There is NO good buffer with phosphate for pH = 4.5, because pKa-value's differ too much from 4.5: pKa = 2.13 and 7.21 for H3PO4 and H2PO4- respectively.
A good alternative would be Acetic acid / Acetate in molar ratio of about 2 : 1, because pK(acetic acid) = 4.77 .
More precisely you can use the formula, based on the Henderson-Hasselbalch equation:
log[acid]/[conjug.base] = = 4.77 - 4.5 = 0.27
or
[acid]/[conjug.base] = 10(pKacid -pHbuffer) = 100.27 = inv.log(0.27) = 1.82
Doing the same for a H3PO4 / H2PO4- buffer, this results in
[H3PO4/H2PO4-] = 10(pKacid -pHbuffer) = 10(2.13-4.5) = inv.log(-2.37) = 4.3*10-3
or for H2PO4-/HPO42- buffer, this results in [H2PO4-/HPO42-] = 10(7.21-4.5) = 102.71 = inv.log(2.71) = 5.1*102
Both ratio's are far beyond values for optimal buffercapacities: 0.1 < ratio < 10
See the Related Links for "Columbia.edu: Phosphate buffer automatic calculator" to the bottom for the answer.
See the Related Links for "Columbia.edu: Phosphate buffer automatic calculator" to the bottom for the answer.
1. Why is the phosphate buffer made up by using the Henderson-Hasselbalch equation not the expected pH?
PBS (phosphate buffer saline) is used to maintain osmolarity of cells i.e. maintaining an isotonic environment. it is used to maintain pH of proteins at which its native structure could be maintained.
10ml of 0.4M Citric acis solution 90 ml of 0.4M sodium phsophate dibasic solution 12.22g NaCl (tomake 150mM ionic solution) or 6.38g to make 100mM ionic strength. make up volume upto 2000ml with water. You should not need to pH this buffer . :)
See the Web Links to the left of this answer.I especially like the Smith.edu link -- it has complete and very useful description of how to prepare a buffer.Use the Henderson-Hasselbach equation:pH = pKa + log [A-]/[HA]where HA is the protonated form of the weak acid, A- is the salt (dissociated acid, or in other words, its conjugate base), and the pKa is the strength of the acid.What this says is that the pH that you want your buffer to be depends on two things:-- the pKa of the weak acid you are using (see reference tables under the Web Links to the left)-- and the RATIO of the concentration of the acid and the salt that you add to the solution.The pH of the buffer does not depend on the actual concentration of the buffer, but on the ratio of the two parts.The buffer capacity depends on two things -- how close to the pKa the pH of the buffer actually is (it should be within 1-2 pH units), and what the total concentration of the buffer is.For instance if you have 0.001 M acetic acid and 0.001 M sodium acetate, the resulting buffer will have the exact same pH as a buffer made with 0.1 M acetic acid and 0.1 M sodium acetate (because the ratio is 1 to 1, the pH = pKa = 4.76). However, the 0.1 M buffer will have a much larger buffer capacity, and will much better resist changes in pH upon the addition of a strong acid or base.
preparation of 5.8 ph phosphate buffer
5 mM phosphate buffer (4.82 g/l monohydrate, monosodium phosphate, pH 6.5).
See the Related Links for "Columbia.edu: Phosphate buffer automatic calculator" to the bottom for the answer.
dissolve the 12 g of crystals of sodium phosphate in water to make 1oo ml
1. Why is the phosphate buffer made up by using the Henderson-Hasselbalch equation not the expected pH?
Add 250 ml of 0.2M KH2PO4 to 393.4 ml of 0.1M NaOH
Cfjn
It changed pH. I have to adjust back with NaOH
A buffer is something that regulates or maintains the pH in the body. In the human body, carbonate is the main buffer in the blood and phosphate is the main buffer within cells.
Dibasic potassium phosphate has the chemical formula is KH2PO4. The standardized solutions has a pH between 6 and 8.
There is information lacking in the question. Either ýou have to specify which application you want to use the buffer for so I can make an estimate of the buffer capacity you need or, you have to specify directly which ionic strength you need.
Decide on the concentration of the buffer, use 1L to be simple PH for your buffer should be within one pH unit from the pKa of the acid/conjugate base use Henderson Hasselbalch Equation pH = pKa + log ([Base]/[Acid]) For a 1 M buffer [Acid] + [Base] = 1