Depends on how broad you define "reaction" and which conditions you apply. In aqueous solution I would say no reaction would occur. A bit hydrolysis with HCl (aq.) (-> HSO4-).
Depends on how broad you define "reaction" and which conditions you apply. In aqueous solution I would say no reaction would occur. A bit hydrolysis with HCl (aq.) (-> HSO4-).
Dear Andreas Leineweber, all of the reactions are at room conditions / room temperature. Is this the equation ? 2 HCl + Na2SO4 ----> 2 NaCl + H2SO4
And in diluted aqueous solution?! You can write this equation, but it is an equilibrium reaction, which is more something formal. Look at the pKa values.....
Hi Hor, I suggest you check the thermodynamic feasibility at room temperature.... using the delta G ....
As long as you are dealing with dilute solutions, you will have a variety of different dissolved ions along with some very small level of atomic pairs (non-dissociated molecules). As Andreas Leinweber points out, you could use a series of simultaneous equations using the pKas to figure out exactly what the concentration of each species will be. Where that all adds up depends upon the pH and the concentration of the added concentrations of each species (you said room T).
The big question is whether you really care about 10^-10 or lower concentrations on non-dissociated HCl or H2SO4 and is this really a "reaction"?
Thanks for your helping, ANdreas Leinweber, Ayodele GEORGE Olumide Bolarinwa and Stephen N Smith. Actually, I have 2 cases/experiments here.
1st, I will try to titrate my sample solution (consists of pyridine , acetic acid, my sample and imidazole as catalyst) with sodium hydroxide. I worry if my sample consist of Na2SO4 as impurity, it will influence my titration result (either react with sodium hydroxide or acetic acid).
2nd, I wil try to titrate my sample solution (consists of KOH, p-hydroxybenzoic acid and my sample) with HCl. I also worry if my sample consist of Na2SO4 as impurity, it will influence my titration result (react with HCl).
No it will not affect things in either case. During the titrations Na2SO4 will effectively act as spectator ions. In a large enough concentration the ionic activity will alter the pH slightly but negligibly. pKa2 of H2SO4 is unlikely to have any effect.
You are unlikely to be able to deconvolute useful data from complex mixtures of weak acids and bases. Why are you titrating them?
No, there is no chemical reaction appear between these compounds. The sodium salt of all your mentioned compound completely ionized in aqueous solutions. Therefore by adding Na2SO4 to any of these solution you only change the ionic strength of solution.
Sayyed is right completely. In dilute solutions Na2SO4 will not disturb your titrations due to reasons given by above. If you are doubt, you can add a known amount of Na2SO4 and you will see its effect. I would be surpsied if you could observe any. It is easier than modelling with various calculations.
Sodium sulfate is salt of stong acid (sulfuric acid) and strong base (NaOH) and so no one of the weak acids (acetic and hydraulic acid, both are weak compared to sulfuric acid) can remove sulfuric acid from its salts (as sodium sulfate here). Also sodium sulfate and sodium hydroxide are both can not react together because both are salts of sodium and hydroxide can not produce acid as you expect.
I. Aqueous solutions of pure sodium sulfate (Na2SO4) can be predicted to be (weakly) basic ― I have discussed this subject elsewhere at this forum:
https://www.researchgate.net/post/What_do_you_call_the_electrolyte_of_sodium_sulfate_a_base_an_acid_or_neutral
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.
incorrect considerations, although I cannot follow the calculations. Where do the protons come from if you only introduce SO42-?!!
Andreas Leineweber ― concerning to your post above:
Brønsted–Lowry acid-base theory does not require an acid to dissociate and directly donate protons to the solvent or solution. Sodium sulfate contributes to pH by means of the hydrolytic equilibrium, chiefly through the sulfate/bisulfate conversion.
Cf. also my posts at: https://www.researchgate.net/post/What_do_you_call_the_electrolyte_of_sodium_sulfate_a_base_an_acid_or_neutral
SO42- can at maximum take up protons, since it HSO4- is only a moderately strong acid (pKS of +2). So such a solution (no contribution from Na+) will at maximum be slightly basic. There is no source of H+ in excess to the autoprotolysis of H2O. Hence pH > 7. 1st year Chemistry.
It is very different if you dissolve NaHSO4.
Andreas Leineweber ― concerning to your last post:
It does not seem straightforward at all to qualitatively access if a pure Na2SO4 aq. sol. can be predicted to be acidic, neutral, or basic. My first post here discusses that quantitatively: https://www.researchgate.net/post/What_do_you_call_the_electrolyte_of_sodium_sulfate_a_base_an_acid_or_neutral
It is straightforward by intuition in this case and it is basic (pH > 7). The attachment contains the calculation demonstrating this rigorously.
Andreas Leineweber ― comment on your last post:
I regret to note that the file you have inserted at above post (*) is not rigorous at all, but plagued with error.
i) 4rd and 5th lines:
«In the following c(X) is the concentration of the species X, and c0 is its initial concentration due to addition of Na2SO4»
ii) equation (iv):
«c(SO42-) + c(HSO4-) = c0(SO42-) = ½c0(Na+)»
Comment to (i) and (ii):
a) «initial» is a misleading qualification here ― because nothing changes; the solution is never initial or final. The related subscript (0) is also equivocal here.
b) Because nothing changes, and following your own notation, you should recognize that c(SO42-) = c0(SO42-); even if this equation conveys no useful information. So; first equality above is both redundant and incorrect.
Further comments:
c) According to conventional notation, the species molar concentration is typically expressed with square brackets.
d) Still according to conventional notation, widely adopted with regards to pH prediction and other speciation problems, we should preferably reserve the c (or C) symbol to denote formal concentration ― which does not account for speciation and is typically restricted to neutral compounds. For instance, for 1 M Na2SO4 aq. sol.; CNa2SO4 = 1 M (the F symbol sometimes substitute for M, here; same units). However [Na2SO4] ≈ 0 M, because the salt is a strong electrolyte. These are distinct concentration concepts; not to be confused. Your notation is propense to such confusion.
e) It is not clear how you have denoted salt concentration (CNa2SO4 at above §). I simply could not find it. This awkwardness seems to derive from your peculiar notation.
f) Neither is clear how you have related salt concentration (seemingly absent) with that of the relevant ionic species. This is crucial for any meaningful speciation reasoning.
g) You missed to clearly identify the physical/chemical meaning of all first introduced equations; namely of balance equations. This also precludes clarity.
h) The text is plagued with error (cf. § b), imprecise notation, if not conceptual confusion.
―
(*) Your text, that I have commented at this post:
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