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Solution of salt of a weak base and salt of a weak acid

  1. May 12, 2015 #1
    I am trying to get a better understaning of the pH of a solution containing a salt of a weak base and a salt of a weak acid. I know it depends on which equlibrium constant is greater but I am thinking more of what it is that actually happens.
    For example let say we have a solution containing salt of weak base, NH4+ and a salt of weak acid, CH3COONa. It should make NH3 and Ch3COOH right?
    If NH4 or CH3COONa are not in equal amounts we will at equilibrium have NH4 or CH3COONa and NH3 and CH3COOH. So how will I find the pH?
     
  2. jcsd
  3. May 12, 2015 #2

    epenguin

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    You do not really ever have CH3COONa except as a formal compositional description of sodium acetate in a bottle etc. In solution you will have with the things you mention, the species CH3COOH, CH3COO-, NH3 and NH4+, as well as any counter-ions such as Na+ and Cl- that came with the substances.
    (Of course you can also have armonium acetate without any such extra ions.)

    In ammonium acetate solution, or any equimolar mixture of ammonia/um forms and acetate forms the pH is half way between the two pKa s, off the top of my head.

    When thinking about what happens in such mixtures you can if you like dispense with the independent H+ or H3O+ that figures in the usual equations and consider the equilibrum as the following proton transfer.

    CH3COOH + NH3 ⇔ CH3COO- + NH4+

    This is quite realistic - it represents most of what is going on! (Most often you can think of [H+] to which we attach such importance as a kind of reporter on the state of play between the major players present!)

    The equilibrium constant is a ratio of the two Ka s, and there is a sort of pKeq, difference of the two pKas.

    I think you can do calculations to get the ratio of any conjugate pair in a mixture, and then that ratio and either Ka can give you [H+].

    It is better to try and work all these things out yourself first and whether you succeed completely or nor you may find the treatment of your textbook less heavy going.
     
  4. May 12, 2015 #3
    I don't think it's always so easy to know 'what happens'. As a rule of thumb, the weakest acid and the weakest base must prevail. So if a mixture of CH3COO- + NH4+ contains 'stronger' species than a mixture of CH3COOH + NH3, then the system will tend to get closer to the latter.

    But you seem to be looking for a quantitative answer. Here it is.

    If you neglect the fact that activities are different from concentrations, in theory the pH of any conceivable mixture of salts, acids, bases, strong or weak, etc can be calculated, knowing the initial amount of each substance and the equilibrium constants. In fact you can also calculate the equilibrium concentration of all species.

    So in your example (ammonium - chloride I guess - and sodium acetate), you have two dissociation constants:
    Ka(CH3COOH) = [CH3COO-] * [H+] / [CH3COOH]
    Ka(NH4+) = [NH3] * [H+] / [NH4+]

    four mass balances:
    C0(CH3COONa) = [CH3COO-] + [CH3COOH]
    C0(NH4Cl) = [NH3] + [NH4+]
    C0(CH3COONa) = [Na+]
    C0(NH4Cl) = [Cl-]

    electroneutrality (sum of all positively charged species = sum of all negatively charged species, each multiplied by its charge):
    [NH4+] + [Na+] + [H+] = [CH3COO-] + [OH-] + [Cl-]

    and the ionic product of water:
    Kw = [H+] * [OH-]

    You have 8 unknown concentrations and 8 independent equations; you know the constants from the literature.
    You can submit these equations to a numerical solver (e.g. the free package Maxima) and you'll get all the concentrations.
    From -log10([H+]) you'll have the pH.
    Or you can eliminate all unknowns except [H+]. You get a polynomial equation in [H+], which you can solve numerically for any value of the initial concentration of salts.

    People usually simplify the system by removing some terms from the mass balances, etc.
    That gives you an approximate formula for a mixture of a weak acid and a weak base, for instance, which I can't remember at the moment.
    However, such simplified formulae are only valid in specific cases. The method above works in all cases.
     
  5. May 12, 2015 #4
    Sorry epenguin, I saw your reply just after I posted mine.
     
  6. May 12, 2015 #5

    Borek

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  7. May 12, 2015 #6

    James Pelezo

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    This sounds like you are interested in hydrolysis of salt ions in water. NaOAc (Sodium Acetate) => Na+(aq) + OAc+(aq), The sodium Ion will not hydrolyze but the Acetate Ion (OAc-(aq) + H-OH(l) => HOAc(aq) + OH-(aq)) The excess OH- leaves the solution alkaline (basic pH). An ammonium salt (e.g., NH4Cl(aq) => NH4 + (aq) + Cl-(aq) ). The chloride ion will not hydrolyze but the ammonium ion (NH4+(aq) + H-OH(l) => NH4OH(aq) + H+(aq)). The excess H+(aq) [ = Hydronium ions ] leaves the solution acidic. The concentrations of hydroxide and/or concentration of hydronium ion can be determined from equilibrium concentrations followed by pH = - log[H+(aq)]. I can provide examples of calculations if you wish.
     
  8. May 13, 2015 #7
    Okok let me try this...
    Let's say we have a 1.00 L solution containing 3 moles of NaCH3COO and 2 moles of NH4Cl.
    These reaction will occur:
    CH3COO-+H2O --> CH3COOH + OH- Kb=5.56e-10
    NH4++H2O -->NH3+H3O+ Ka=5.56e-10

    3 moles of CH3COO- will give 4.08e-5 moles OH-
    2 moles of NH4+ will give 3.33e-5 moles H3O+

    4.08e-5 moles OH- will react with 3.33e-5 moles H3O+ and there will be excess OH-, 7.5e-6 moles.
    7.5e-5 moles gives pH=8.9.

    Did i get it right?
     
  9. May 13, 2015 #8

    Borek

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    Don't you think once H+ reacted with OH- the equilibrium of both hydrolysis reactions will shift?

    I am afraid it is not that easy as you are trying to make it, you need much more sophisticated approach (the one that was already mentioned earlier).

    BTW, by my calculations (not that I did them manually) pH of your solution should be 7.1, so almost neutral.
     
  10. May 13, 2015 #9

    James Pelezo

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    Borek, could you show your calculations? I'd like to see houw you got 7.1 pH. Thank you
     
  11. May 13, 2015 #10

    Borek

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  12. May 13, 2015 #11

    epenguin

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    No it is not on the right lines. IMO you would get towards an understanding faster if you tried the simplest problem first, where total of all forms of ammonia and all forms of acetates were of equal concentration as set out in #2. You have to use equations for equilibrium which I think you know plus mass conservation.
     
  13. May 13, 2015 #12

    James Pelezo

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    You are right ... That is confusing, but looks very interesting. I'd like to know more about the calculator. Anyways, looking at Staralfur's calculation, I'd say it seems to be right on point. If I may, here's my analysis ... You are welcome to critique ...

    Given 3 mole NaOAc and 2 mole NH4Cl dissolved into 1.00 Liter water, both ionize 100% to give Na+(aq), OAc-(aq), NH4+(aq) and Cl-(aq) ions in solution. The Na+ and Cl- will not hydrolyze as they would theoretically produce a strong base (NaOH) and strong acid (HCl); i.e., no reaction. However, OAc-(aq) and NH4+(aq) do react with water (hydrolyze) per the following:

    OAc-(aq) + H-OH(l) => HOAc(aq) + OH-(aq); Kb(OAc-) = (Kw / Ka(HOAc)) = (1x10-14/1.8x10-5) = 5.6x10-10

    At equilibrium Kb = ([HOAc][OH-]/[OAc-]) = [ x2/(molar conc of salt anion)] = [(x2/(3M)] = 5.6x10-10 => x = [OH-] = (√(3)(5.6x10-10)M = 4.1x10-5M in OH-(aq)



    NH4+(aq) + H-OH(l) => NH4+ OH(aq) + H+(aq); Ka(NH4+) = (Kw / Kb(NH4+OH)) = [(1x10-14/1.8x10-5) = 5.6x10-10

    At equilibrium Ka = [NH4+][H+]/(molar conc of salt cation)] = [ x2/(molar conc of salt anion))] = [(x2/(2M)] = 5.6x10-10 => x = [H+] = (√(2)(5.6x10-10)M = 3.4x10-5M in H+ (aq)

    Now, since H+ + OH- => H-OH in 1:1 rxn ratio, then 3.4x10-5M H+ + 4.1x10-5M OH- => 3.4x10-5M H-OH(l) with 7.3x10-6M in OH- ions remaining in excess; pOH = - log[7.3x10-6M] = 5.12

    pH + pOH = 14 => pH + 5.12 = 14 => pH = 14 - 5.12 = 8.9 indicating alkaline solution based on amount of excess OH-(aq) ions remaining unreacted. I would support Staralfur's results. All the best. jp:wink:
     
    Last edited: May 13, 2015
  14. May 13, 2015 #13
    @James Pelezo : I don't think you can treat each equilibrium separately and then make hydroxide and hydronium ion react. All these equilibria take place in the same solution, and influence one another. I think this was also Borek's point. The only case where your approach could work (approximately) is when you mix a very strong base and a very strong acid in water, I suppose.
    I'm afraid this is a case where some slightly tougher maths is needed. Luckily we have computers that do it for us.
     
  15. May 13, 2015 #14

    James Pelezo

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    A strong acid + strong base in the context of this problem would, in fact, be a limiting reagent type problem, and you may be 100% right about equilibrium interfering with each other, but in the absence of additional information, separating the calculations as such and treating it as a simple limiting reagent problem seemed to be a plausible start. This is very interesting issue. I'll be in my lab later this week and plan to try a titration type mixing of a 3M acetate salt into a 2M ammonium salt solution to see how the pH trend changes on mixing. I have equipment that can monitor trends like this. It will be interesting. I'll keep you posted. Later. jp
     
  16. May 13, 2015 #15

    Borek

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    You better try much more diluted solutions. At 2/3 M ionic strength will be so high results of measurements will be at least several tenth of a pH unit off. And this is one of these cases where the ratio of concentrations is much more important than the concentrations itself (at least for a reasonably concentrated solutions - I have checked that 0.03/0.02 M solution gives the same calculation result).

    An interesting line of attack could be based of the fact pH of a salt of a weak acid and a weak base is more or less average of both pKa values (which means pH of ammonium acetate is almost exactly 7.0). I would start with the derivation of the pH of a salt formula (compare http://www.chembuddy.com/?left=pH-calculation&right=pH-salt-simplified, eq 13.16) to see whether it can be modified to help solve this particular problem. Doesn't have to work.

    Follow the link, you will find everything you need.
     
  17. May 13, 2015 #16

    James Pelezo

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    Absolutely, I agree... I plan to start with moderately dilute solutions 0.03/0.02M is a good place to start and work up the concentrations of salts (of course, keeping the ratios constant) to see if there is a point of limiting returns. After thinking about the system, the questions I'd like to answer is how much does the Ammonium Hydroxide + Acetic Acid neutralization 'after' hydrolysis affect the pH character. That's kind of hard to visualize (well, it is to me) in the dynamics of the hydrolysis effect and neutralization occurring at the same time. I've done a number of salt hydrolysis experiments for my chem classes on single salt solutions, and they all pretty much match calculation expectations (excluding student accuracy factors, that is. Ha!). But, I've never considered an issue like mixing two salts as proposed in this exchange. Anyways, I'm looking forward to a head-scratch'n good time on this one... I'll review the link you've suggested (thanks) and keep all posted on progress. I know it's going to generate some interesting exchange. All the best... jp
     
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