According to general chemistry mathematics based on Association constants for deprotonation of acids, buffers are only valid at pH range of pKa ± 1. Outside of that pH range the buffering capacity of buffers is terrible. One drop of acid or base, drives the pH off-scale.

The best biological buffers for proteins are the Good's buffers which include Tris, Hepes, MOPS, CAPS, etc ...

CAPS buffer has a pKa of 10.40 so it is buffering capacity should be pHs of about 9.40 to 11.40. So it is a good buffer at pH 10, but not 12.

At pH < 4 or pH > 10 most proteins are destroyed by pH as in the proteins precipitate and likely unfold. Unfolded proteins have no secondary structure. If this doesn't happen to your particular protein then it is none of your concern. If it does happen then the enzymatic activities of your enzyme are ruined.

Of the thousands of the proteins that have been crystallized, this the distribution of the pH ranges at which proteins have crystallized which is some measure of the acceptable limits of protein stability.

Each protein is unique, some more hardy than others, but most proteins are within this distribution.

The salt concentration that crystallizes most proteins is usually a unique salt like sodium chloride, Magnesium chloride, sodium acetate, etc. The concentration is variable but typically the optimal concentration is usually about 200 mM most of the time. I am not exactly sure why, but some scientists have speculated. Each enzyme is variable. The enzyme I study precipitates after two days at salt concentration less than 400 mM>

Source:

https://journals.iucr.org/f/issues/2015/10/00/nj5235/index.html

Kirkwood, J., Hargreaves, D., O'Keefe, S., & Wilson, J. (2015). Analysis of crystallization data in the Protein Data Bank. Acta Crystallographica Section F: Structural Biology Communications, 71(10), 1228-1234.

Adron's recommendations are extremely useful for protein crystallization. However, most of the Good Buffers will interfere significantly in circular dichroism studies. Perhaps a buffer that could be used within the high pH range is borate. If salt is required, you may use NaF instead of NaCl, particularly when employing wavelengths lower than 200 nm. Please refer also to the following paper:

http://ctrstbio.org.uic.edu/manuals/kellybba.pdf

About Na3PO4 + NaOH high pH solutions:

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 at high pH.

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. We can expect considerable pH stability within the pKa3 ± 1 range.

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-]}}

VIII. pH can be predicted by solving last equation:

[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))

IX. Specifically: we may want to consider the half-equivalence point for the titration of NaOH with H3PO4. Then, for each initial 2x mol of NaOH, x mol would have been converted accordingly:

xNaOH + (x/3)H3PO4 → (x/3)Na3PO4 + xH2O.

Then CNaOH = 3CNa3PO4 and last equation (§ VIII) simplifies to:

[OH-] ≈ - (1/2)·(Kw/Ka3 - CNaOH) + √{(2CNaOH)·Kw/Ka3 + (1/4)·(Kw/Ka3 - CNaOH)2}

X. With CNaOH = 3CNa3PO4, last equation (§ VI) could alternatively be solved to yield CNaOH in terms of [OH-]:

CNaOH ≈ [OH-]/{2 - [OH-]/{Kw/Ka3 + [OH-]}}

where: [OH-] = 10- pOH M; pOH = 14 - pH

XI. It would be advisable to protect the prepared solution from carbonation by atmospheric CO2.

A practical note on using highly alkaline solutions, they may cause etching of quartz cuvettes which are required for use in CD specrometers