Dear Javad,

In order to interpret the results of the PH for the sodium sulfate electrolyte you need only to refer the PH scale of the hydrogen ion concentration in aqueous solutions. It is so that PH =7 is neutral , PH< 7 acidic and PH > 7 is basic or alkaline. All what has been changed observed by the PH meter is the H+ concentration.

For more information please refer to the link: https://www.thoughtco.com/definition-of-ph-in-chemistry-604605

Best wishes

The use of PH or ph is to be deplored. The correct expression is pH.

Slightly acidic.. If I'm not mistaken it is also called natural pH

Javad Hikmati Not sure if this is a real question as there are a number of red flags for me here. Obviously you are not a chemist and I can't see any indication of the school you are in. If your DI water is fresh then it ill be neutral (pH = 7). If there's dissolved CO2, then the pH will drop (slightly acidic to around 5.5 -6.0). Sodium sulfate is the salt of a strong acid and a strong base and would be expected to be neutral. However, absorbed CO2 will reduce this somewhat. Electrolysis will produce NaOH at one electrode (cathode) and this will cause the pH to go basic (> 7 perhaps to 10 or higher). Around the anode we'd have H+ ions and it will be acidic there. We also expect gas evolution (exactly as the electrolysis of water) taking place.

We can predict the pH of pure sodium sulfate (Na2SO4) aq. sol. as follows:

The salt is formed from a strong base and a strong acid; hence the pH should be close to neutral.

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

Na+ hydrolysis does not contribute significantly to pH (*).

Water contributes to pH by its autoionization (Kw = 1.0·10-14 M2).

Let CNa2SO4 denote the nominal molar concentration of dissolved Na2SO4 (formality, to be precise).

Sulfate balance is: CNa2SO4 = [SO42-] + [HSO4-] = {1 + [H3O+]/Ka2}·[SO42-];

[SO42-] = CNa2SO4/{1 + [H3O+]/Ka2}

Sodium balance is: CNa2SO4 = ½[Na+]

Charge balance writes: [Na+] + [H3O+] = 2[SO42-] + [HSO4-] + [OH-]

or, after substitution:

2CNa2SO4 + [H3O+] = CNa2SO4·{2 +[H3O+]/Ka2}/{1 + [H3O+]/Ka2} + [OH-]

or:

[OH-] - [H3O+] = CNa2SO4·{1 - 1/{1 + [H3O+]/Ka2}} > 0

Here it can be seen that we can expect pH > 7, yet very close to neutral, as we expect [H3O+]/Ka2 > [H3O+]:

{CNa2SO4·[H3O+] - Kw}·{1 + [H3O+]/Ka2} ≈ CNa2SO4·[H3O+]

or:

[H3O+]2 - 2½(Kw/CNa2SO4)[H3O+] ≈ Kw·Ka2/CNa2SO4

this solves as:

[H3O+] ≈ ½(Kw/CNa2SO4) + √{Ka2·Kw/CNa2SO4 + (1/4)·(Kw/CNa2SO4)2}

or:

[H3O+] ≈ ½(Kw/CNa2SO4)·{1 + √{4CNa2SO4·(Ka2/Kw) + 1}}

with 4CNa2SO4·(Ka2/Kw) = CNa2SO4·4.8·1012 M-1 >> 1:

[H3O+] ≈ ½(Kw/CNa2SO4)·{1 + √4{CNa2SO4·(Ka2/Kw)}}

with √4{CNa2SO4·(Ka2/Kw)} = 106·2.19√{CNa2SO4 M-1} >> 1:

[H3O+] ≈ (Kw/CNa2SO4)·√{CNa2SO4·(Ka2/Kw)}

or (pH > 7):

[H3O+] ≈ √{(Ka2·Kw)/CNa2SO4} < √Kw; with CNa2SO4 > Ka2;

pH = - log10{[H3O+]/M}

―

Example: for CNa2SO4 = 100Ka2 = 1.2 M; [H3O+] ≈ ·√{Kw/100} = 1.0·10-8 M; pH = 8.0

―

Notes:

i) carbonation by atmospheric CO2 may decrease pH: Na2SO4 + CO2 + H2O → NaHSO4 + NaHCO3

ii) Sodium sulfate (Na2SO4) can react with acids to form sodium bisulfate (NaHSO4): SO42- + H3O+ ⇌ HSO4- + H2O. The salt performs as proton acceptor, or Brønsted–Lowry base.

―

(*) See my posts at: https://www.researchgate.net/post/Please_tell_me_more_details_why_pH_in_1_M_KCl_is_higher_than_pH_in_water https://www.researchgate.net/post/How_pH_varies_if_in_a_solution_we_permutate_NH4_cations_with_Mg2_or_K_cations

I. We may either write:

i) Considering the second ionization of H2SO4:

HSO4- + H2O ⇌ SO42- + H3O+; Ka2 = [SO42-]·[H3O+]/[HSO4-] = 1.2·10−2 M

ii) Considering the hydrolytic equilibrium of SO42-:

SO42- + H2O ⇌ HSO4- + OH-; Kh = [HSO4-]·[OH-]/[SO42-] = [HSO4-]·Kw/{[SO42-]·[H3O+]} = Kw/Ka2

Both algebric equations are equivalent in terms of pH prediction. We can use either one or the other.

II. Although the above hydrolytic equilibrium (ii) contributes with OH-, it also contributes with HSO4-. Last species follows the hydrolytic equilibrium (i) that corresponds to the second ionization of H2SO4. Equilibrium (i) contributes with H3O+.

III. Both hydrolytic equilibriums (i and ii) contribute to pH; the contribution of the first one (i) is measured by Ka2; that of the last by Kh = Kw/Ka2.

IV. It seems not straightforward to qualitatively access if pure Na2SO4 aq. sol. solution can be predicted to be acidic, neutral, or basic. My first post shows (quantitatively) that it can be predicted to be basic.

Carlos Araújo Queiroz

After rereading your first post I realised that my post was wrong. Rather than correct it I decided to delete it, and then found your second post too late to undelete my faulty post.

Alastair Bain McDonald Strong warning, Carlos Araújo Queiroz is wrong. I have indicated the solution at https://www.researchgate.net/post/Can-sodium-sulphate-Na2SO4-react-with-acetic-acid-sodium-hydroxide-hydrochloric-acid

Andreas Leineweber:

If you think I was wrong at my early posts to this discussion, you should clearly justify that assessment. Sound criticism is most welcomed. You did not presented a valid 'solution' at the link you signalled above (*). Your 'warning' above is not any 'stronger' than the faulty 'solution' to which it appeals.

(*) https://www.researchgate.net/post/Can-sodium-sulphate-Na2SO4-react-with-acetic-acid-sodium-hydroxide-hydrochloric-acid

https://en.wikipedia.org/wiki/PH

" The pH of a solution containing a weak base may require the solution of a cubic equation. ". I did so. Carlos Araújo Queiroz used approximations. This is dangerous. I have only an old (east) German book on my desk with a "map" in which range of inititial concentration which approximation to the cubic equation is to be applied.

Carlos Araújo Queiroz

I now realise that my deleted post was in fact correct.

I wrote that roughly speaking the solution will be neutral, because Na2SO4 will ionise into Na+ which is the cation of the strong base NaOH, and into SO42- which is the anion of the strong acid H2SO4.

That is not strictly correct; SO42- is the anion of the weak acid HSO4-. But it is a very strong weak acid, pKa = 2.0, making its conjugate base SO42- a very weak base p Kb = 14 - pKa = 12.

What confused me at the time was your reaction which showed hydronium ions being created:

HSO4 + H2O ⇌ SO42- + H3O+; Ka2 = [SO42-]·[H3O+]/[HSO4] = 1.2·10−2 M

Note, HSO4 should read HSO4-.

However, this reaction is only appropriate when sulphuric acid is being added to pure water, but not when sulphate is added, as is the case here.

The correct reaction for sulphate reacting with water is:

SO42- + H2O ⇌ HSO4- + OH-; Kb1 = [HSO4-][OH-]/[SO42-] = 1.0x10-12

where Kb1 = Kw/Ka2 = 1.0x10-14/1.2x10−2 = 8.3x10-13.

This gives an [OH-] of sqrt(8.3x10-13 ) = 9.1x10-7 enough to make the solution basic.

Hope this helps.

Alastair Bain McDonald Agrees with my solution presented in another thread. Your solution is the more common one, which is a special case of solving a cubic equation in [OH-] or [H+]. The solution presented, I think, violates charge neutrality of all ionic species at some places, which can, however, be neglected in the range of pKS values and salt concentration considered.

Alastair Bain McDonald ― concerning to your last post, above:

a) I suggest that you carefully read my second post to this discussion; namely, §.I:

i) Considering the second ionization of H2SO4:

HSO4- + H2O ⇌ SO42- + H3O+; Ka2 = [SO42-]·[H3O+]/[HSO4-] = 1.2·10−2 M

ii) Considering the hydrolytic equilibrium of SO42-:

SO42- + H2O ⇌ HSO4- + OH-; Kh = [HSO4-]·[OH-]/[SO42-] = [HSO4-]·Kw/{[SO42-]·[H3O+]} = Kw/Ka2

Both algebric equations are equivalent in terms of pH prediction. We can use either one or the other.

Until now; the referred post ― actually inspired by your prior post to this discussion; hereafter deleted.Consequently, I have eliminated my prior reference to your deleted post; for the sake of readability.

There is no dicotomy here; between the «correct reaction» an the other «only appropriate when sulphuric acid is being added to pure water» ― quotations from your post above. Either equilibrium correctly translates the hydrolytic equilibrium between the species HSO4- and SO42-.

b) I always wellcome corrections, but no sign is or was missing in my post referred to above. I use brackets for the species concentrations and superscripts for signs in ionic species; the minus sign may possibly have been visually confused with the contiguous upper part of the right bracket.

c) [OH-] is not given by √(Kw/Ka2). Even just dimensional analysis shows that: the units of Kw = [OH-]·[H3O+] are M2, those of Ka2 are M; √(Kw/Ka2) has M½ units.

Carlos Araújo Queiroz

Re a) I agree that your reaction aided by 2H2O = H3O+ + OH- can be derived from mine, but you have to use the correct version for the circumstances where it is be used. The reacton of a base with water is described by mine the reaction of and acid with water by yours.

Carlos Araújo Queiroz Re a) I agree that your reaction aided by 2H2O = H3O+ + OH- can be derived from mine, but you have to use the correct version for the circumstances where it is be used. The reacton of a base with water is described by mine the reaction of and acid with water by yours.Re b) I suggest that you carefully read your first post to this discussion where you wrote:" HSO4 + H2O ⇌ SO42- + H3O+; Ka2 = [SO42-]·[H3O+]/[HSO4] = 1.2·10−2 M" which I have cut and pasted here. HSO4 occurs twice without a negative sign. As a result the charge does not balance i.e LHS = 0, RHS = -2 + 1 = -1. Similarly, in your formula for Ka2.To quote Lady Bracknell (from Oscar Wilde's play 'The Importance of Being Earnest"), ‘To lose one parent, Mr Worthing, may be regarded as a misfortune; to lose both looks like carelessness.’ Re c) You are correct. [OH-] is not given by √(Kw/Ka2). It is given by √([SO42-] x Kw/Ka2). I was assuming that [SO42-] = 1. you carefully read your first post to this discussion where you wrote:

" HSO4 + H2O ⇌ SO42- + H3O+; Ka2 = [SO42-]·[H3O+]/[HSO4] = 1.2·10−2 M" which I have cut and pasted here. HSO4 occurs twice without a negative sign. As a result the charge does not balance i.e LHS = 0, RHS = -2 + 1 = -1. Similarly, in your formula for Ka2.

To quote Lady Bracknell (from Oscar Wilde's play 'The Importance of Being Earnest"), ‘To lose one parent, Mr Worthing, may be regarded as a misfortune; to lose both looks like carelessness.’

Re c) You are correct. [OH-] is not given by √(Kw/Ka2). It is given by √([SO42-] x Kw/Ka2). I was assuming that [SO42-] = 1.

--------

The original question is about how Na2SO4 can be acidic, then basic after electrolysis, and then acidic later. I can only surmise that the sodium sulphate was contaminated with sulphuric acid during its preparation. The electrolysis drove off the hydrogen ions leaving the solution basic.

Alastair Bain McDonald ― concerning to your last post, above:

a) About the minus sign I have already explained what might have happened. This is a triviality; you should have accepted my justification. Your post is obnoxious on this matter. What you have pasted is not conform. I am here to discuss chemistry; not courtesy (or the lack of it). I have no patience for derisory or frivolous discussion.

b) The charge balance at my first post to this discussion is (and was) correct.

c) About both equilibrium equations, that correctly translate the hydrolytic equilibrium between the species HSO4- and SO42-. ― I have nothing to add to the explanation given at my previous post and at my second post to this discussion.

d) About your sentence that claims that [OH-] «is given by √([SO42-] x Kw/Ka2)». ― This is still not correct. Note that: √([SO42-]·Kw/Ka2) =√{[HSO4-]·[OH-]} ≠ [OH-], because [HSO4-] ≠ [OH-], in general. See the equilibrium equation at § a.ii. of my last post, or at § I.ii. of my second post to this discussion. Also note that OH- is not just contributed by the following hydrolytic equilibrium: SO42- + H2O ⇌ HSO4- + OH-, as you might (perhaps) have assumed. On one hand, HSO4- also suffers its own hydrolysis: HSO4- + H2O ⇌ SO42- + H3O+. On the other hand, water also contributes to OH- by means of its autoionization equilibrium.

e) About «I was assuming that [SO42-] = 1»: this is (also) dimensionally incorrect. Besides; [SO42-] was never there, at the first place, awaiting to be (awkwardly) substituted by the unit. Neither you have justified this alleged omitted assumption.