I did try phosphate using Na2HPO4.7H20 and NaOH but buffering capacity was very low especially upon dilution.
Dear Sir,
You can prepare pH 13 buffer using either 45% (w/w) Potassium hydroxide-50% (w/w) sodium hydroxide solution mixture or ammoniacal buffer.
Please refer the attched documents,
Thank you
I. Strictly; it is generally accepted that pH buffers consist on weak acid with its conjugate base, or on weak base with its conjugate acid.
II. Nonetheless, a solution of trisodium phosphate (Na3PO4) and NaOH, possibly obtained by titration of phosphoric acid with NaOH, has a fairly stable pH by pH = 13.
III. For the 3rd dissociation equilibrium of phosphoric acid, HPO42- + H2O ⇌ PO43- + H3O+; pKa3 = 12.37; Ka3 = [H3O+]·[PO43−]/[HPO42-] = 2.14·10−13 M. This equilibrium is largely displaced to the right by NaOH addition while some OH- is consumed. Trisodium phosphate is basic salt at its own solution, but behaves as acidic with regards to the stronger base NaOH. The pKa3 ± 1 range covers pH = 13, what should favour pH stability.
IV. Phosphorus balance writes: CNa3PO4 ≈ [HPO42-] + [PO43-]; species H2PO4- and H3PO4 can be neglected for the mentioned alkaline solution. Sodium balance is: CNaOH + 3CNa3PO4 = [Na+]. C denotes nominal (i.e. formal) concentrations.
V. Hence: [H3O+] = Ka3·[HPO42-]/[PO43−] = Ka3·{CNa3PO4/[PO43−] - 1}; or: [PO43−] = CNa3PO4/{{(Kw/Ka3)/[OH-]} + 1}; Kw = 1.00·10-14 M2; [PO43−] ≠ 0 M.
VI. Charge balance writes ― neglecting H3O+ and HPO42- contributions: [Na+] ≈ 3[PO43−] + [OH-]; or: [OH-] ≈ [Na+] - 3[PO43−]; i.e. (with CNa3PO4 ≠ 0 M at sufficiently alkaline solution):
[OH-] ≈ CNaOH + 3CNa3PO4·{1 - [OH-]/{Kw/Ka3 + [OH-]}}
VII. Here Kw/Ka3 = 0.0467 M. For the considered pH = 13, pOH = 14 - pH = 1; so [OH-] = 1.0·10-1 M and Kw/Ka3 + [OH-] = 0.147 M. Hence, 0.1 M ≈ CNaOH + 3CNa3PO4·(1 - 0.1/0.147), or:
0.1 M ≈ CNaOH + 0.959·CNa3PO4
VIII. More generally, pH can be predicted by solving the equation obtained at § VI:
[OH-] ≈ - (1/2)·(Kw/Ka3 - CNaOH) + √{(3CNa3PO4 + CNaOH)·Kw/Ka3 + (1/4)·(Kw/Ka3 - CNaOH)2}
with, obviously: pH = 14 - pOH; pOH = - log10(|[OH-]|) = - log10([OH-]/(1 M))
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