need to separate protein different types of proteins from a sample . are there is a difference between using ammonium sulfate and using sodium sulfate in this separation.
Amm. Sulphate precipitation is a method used to purify proteins by altering their solubility
For a RG discussion on the pH of pure sodium sulfate (Na2SO4) aq. sol.: https://www.researchgate.net/post/What_do_you_call_the_electrolyte_of_sodium_sulfate_a_base_an_acid_or_neutral
We can predict the pH of a pure (NH4)2SO4 aq. sol. as follows:
As a salt of a strong acid and a weak base, the solution is acidic.
NH4+ contributes to pH by its hydrolytic equilibrium:
For NH4OH dissociation: NH4OH⇌ NH4+ + OH-, Kb = [NH4+]·[OH-]/[NH4OH] = 1.80·10-5 M; [NH4OH] = [NH4+]·[OH-]/Kb
SO42- contributes to pH accordingly the second ionization of H2SO4:
HSO4- + H2O ⇌ SO42- + H3O+; Ka2 = [SO42-]·[H3O+]/[HSO4-] = 1.2·10−2 M; [HSO4-] = [SO42-]·[H3O+]/Ka2
Water contributes to pH by its autoionization: Kw = 1.0·10-14 M2
Let C(NH4)2SO4 denote the nominal molar concentration of dissolved (NH4)2SO4 (formality, to be precise).
Sulfur balance is: C(NH4)2SO4 = [SO42-] + [HSO4-] = [SO42-]·{1 + [H3O+]/Ka2};
[SO42-] = C(NH4)2SO4/{1 + [H3O+]/Ka2}
Ammonium balance is: 2C(NH4)2SO4 = [NH4+] + [NH4OH] = [NH4+]·{1 + [OH-]/Kb}; [NH4+] = 2C(NH4)2SO4 /{1 + [OH-]/Kb}
Charge balance writes: [NH4+] + [H3O+] = 2[SO42-] + [HSO4-] + [OH-]
after substitution:
2C(NH4)2SO4 /{1 + [OH-]/Kb} + [H3O+] = C(NH4)2SO4·{2 + [H3O+]/Ka2}/{1 + [H3O+]/Ka2} + [OH-]
or:
{[H3O+] - [OH-]}·{1 + [H3O+]/Ka2}·{1 + [OH-]/Kb}/C(NH4)2SO4 = 2{1 + [OH-]/Kb} + [H3O+]·{1 + [OH-]/Kb}/Ka2 - 2{1 + [H3O+]/Ka2}
i.e.:
{[H3O+] - [OH-]}·{Ka2 + [H3O+]}·{[H3O+] + Kw/Kb}/C(NH4)2SO4 = 2KwKa2/Kb + (Kw/Kb)·[H3O+] - [H3O+]2
Since Kw/Kb = (1.0·10-14/ 1.80·10-5) M = 5.56·10-10 M 100Kw; [H3O+] > 10√Kw = 1.0·10-6; pH < 6:
Ka2·[H3O+]2 + [H3O+]3 = 2C(NH4)2SO4·(Kw/Kb)·Ka2 + C(NH4)2SO4·(Kw/Kb)·[H3O+] - C(NH4)2SO4·[H3O+]2
or:
{[H3O+] + (Ka2 + C(NH4)2SO4)}·[H3O+]2 - C(NH4)2SO4·(Kw/Kb)·[H3O+] = 2C(NH4)2SO4·(Kw/Kb)·Ka2
with (Ka2 + C(NH4)2SO4) > 100[H3O+]:
(Ka2 + C(NH4)2SO4)·[H3O+]2 - C(NH4)2SO4·(Kw/Kb)·[H3O+] = 2C(NH4)2SO4·(Kw/Kb)·Ka2
or:
[H3O+]2 - X·[H3O+] = 2·Ka2·X
with:
X ≡ C(NH4)2SO4·(Kw/Kb)/(Ka2 + C(NH4)2SO4)
this solves as:
[H3O+] = ½X + √{2·Ka2·X + (1/4)·X2}
or:
[H3O+] = ½X·{1 + √{1 + 8·Ka2/X}}; pH = - log10{[H3O+]/M}
__
Example: with C(NH4)2SO4 = Ka2 = 1.2·10−2 M; X = ½(Kw/Kb);
[H3O+] ≈ (1/4)·(Kw/Kb)·{1 + √{1 + 16·Ka2·Kb/Kw}} ≈ (1/4)·(Kw/Kb)·{1 + 4√(Ka2·Kb/Kw)} ≈ (1/4)·(Kw/Kb)·{4√(Ka2·Kb/Kw)} = √{Kw·(Ka2/Kb)} =2.58·10-6; pH = 5.59.
- Badges
- Science topic
- Similar topics
- Counseling Psychology
- Confusion