You can use the Henderson-Hasselbalch equation to make buffer solutions at a desired pH value.
https://www.thoughtco.com/make-a-phosphate-buffer-solution-603665
You can prepare this buffer. However, you must ensure to use NaH2PO4 salt and H3PO4. The first Ka is close to 4 and using the pH meter and above solutions you should be able to do it.
Assume that you want 1 M buffer. Prepare two solutions of H3PO4 and NaH2PO4 and mix them carefully addition one into other until you have pH 4.5
0.05 M Phosphate buffer solution pH 4.5. 4009000. Dissolve 6.80 g of potassium dihydrogen phosphate R in 1000.0 ml of water R. The pH (2.2.3) of the solution is 4.5.
Let us refer to the titration curve for H3PO4 with NaOH, presented in some chemistry textbooks. The acid dissociation constants for phosphoric acid correspond to pKa's of 2.15, 7.20, and 12.15. The first equivalent point of phosphoric acid can be detected by about pH = (2.15 + 7.20)/2 = 4.7, depending somewhat on concentration, where 1 mol of H3PO4 is neutralized, to its first equivalent point, by 1 mole of NaOH. The second equivalence point of phosphoric acid occurs by about pH = (7.20 + 12.15)/2 = 9.7. From the first equivalent point of phosphoric acid to the second, just he second proton of H3PO4 is titrated.
About speciation of the phosphate species ― Generically speaking, we expect aq. solutions where both NaH2PO4 and Na2HPO4 are present to have pH between approx. 4.7 and 9.7. The contributions of Na3PO4 and H3PO4 can in principle be neglected within this pH range. Below approx. pH 4.7 the contribution of Na2HPO4 can be neglected, but that of H3PO4 should be considered. Above approx. pH 9.7 the contribution of NaH2PO4 can be neglected, but that of Na3PO4 should be considered.
About pH of a NaH2PO4 aq. sol. ― By approx. pH 4.7 the contributions of H3PO4, Na2HPO4 and Na3PO4 can in principle be neglected, and the solution is essentially of just NaH2PO4, hence weakly acidic. The exact pH depends somewhat on concentration.
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