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Hydrolysis of salt containing amphiprotic anion

  1. Sep 4, 2015 #1

    Titan97

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    My teacher was teaching me salt hydrolysis today and I understood everything up to hydrolysis of salt of weak acid and weak base. Then he introduced amphiprotic anion in salt.
    He took NaHCO3
    This is what he wrote:
    HCO3-+H2O ##\leftrightarrows## H2CO3+OH-

    H2CO3##\leftrightarrows##H++HCO3-
    HCO3-##\leftrightarrows##CO32-+H+

    $$k_{a1}=\frac{[H^+][HCO_3^{-}]}{[H_2CO_3]}$$
    $$k_{a2}=\frac{[H^+][CO_3^{2-}]}{[HCO_3^{-}]}$$

    He then assumed that concentration of carbonate ion and concentration of
    H2CO3 to be same and after multiplying the ka values he got

    $$pH=\frac{1}{2}(pk_{a1}+pk_{a2})$$

    I don't understand that assumption. Also, why did he assume that concentration of hydronium ions in both first and second step to be equal? Is the final result correct? Where can I read more of this? He also discussed about salts where both ions are amphiprotic. I did not understand anything.
     
    Last edited: Sep 4, 2015
  2. jcsd
  3. Sep 4, 2015 #2

    Borek

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  4. Sep 4, 2015 #3

    Titan97

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    @Borek, these are my doubts after reading that article.

    Firstly,
    Capture1.PNG
    this like saying x=a+x+b! also, how did they even get CHA-?

    Secondly:
    Capture2.PNG
    Its A2- and HA- thats being hydrolysed right? And also how does it yield H2A?

    Thirdly:
    Capture3.PNG
    How can yous say that concentration of H+ is same from both reactions?
     
  5. Sep 4, 2015 #4

    Borek

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    This is a mass balance - sum of all forms of carbonates present in the solution must equal initial amount of carbonate introduced (so called analytical concentration, denoted here by CHA-, as we start with the solution containing only HA-; note that the analytical concentration CHA- is a different thing than the equilibrium concentration [HA-]). We assume analytical concentration is given. In this particular case it will cancel out in the approximate final formula, but it doesn't happen often.

    Compare introduction to the general method of pH calculation here: http://www.chembuddy.com/?left=pH-calculation&right=general-pH-calculation

    Nope, we start with a solution containing HA- only, and it reacts with water producing both A2- and H2A.

    How could it be different? There are no different H+ reacting in different reactions, and solution can't contain different concentrations of the same ion at the same time. For every reaction involved [H+] (or pH) is identical.
     
  6. Sep 4, 2015 #5

    Titan97

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    The two reactions taking place are ##HA^-+H_2O\rightleftarrows H_2A+OH^-## and ##HA^-+H_2O\rightleftarrows A^{2-}+H_3O^+##
    Let initial number of moles of HA- be n and volume of vessel be 1l. At any point, it will become n-x-y. Number of moles of H2A=x and number of moles of A2-=y
    ##C_{HA^-}=n-x-y+x+y=n##
    Is this correct?
    Also, [H+]=y-x=[A2-]-[H2A]

    But my second doubt is:
    taking one reaction ##HA^-+H_2O\rightleftarrows H_2A+OH^-##, H2A formed can dissociate to give H+ and HA- .
    Therefore, ##[H_2A]=\frac{[H^+][HA^-]}{k_a}##. But this does not make sense to me because the actual equation involves [OH-] but no H+.
     
  7. Sep 4, 2015 #6

    Borek

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    Yes. Basically you have just proven that 1=1 :wink:

    Yes, that's equation 12.4 from the page I linked to earlier.

    You can derive the final approximation using the hydrolysis equation and taking into account fact changes in H+ and OH- concentration are not independent - they are linked by the water autodissociation, Kw=[H+][OH-]. You will need to start with Kb2 (which describes the hydrolysis), and use the Ka1×Kb2=Kw relationship.
     
  8. Sep 4, 2015 #7

    Titan97

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    kb for an acid?
    So I have to write it as $$[H_2A]=\frac{k_w.[HA^-]}{k_a.[OH^-]}$$
    But how is there a $k_b$ for H2A?
     
  9. Sep 4, 2015 #8

    Borek

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  10. Sep 4, 2015 #9

    Titan97

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    Is it the reaction ##HA^-+H^+\rightleftarrows H_2A##?
    the equilibrium constant is 1/ka
     
  11. Sep 4, 2015 #10

    Borek

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    HA ↔ H+ + A-

    [tex]K_a = \frac {[H^+][A^-]}{[HA]}[/tex]

    A- + H2O ↔ HA + OH-

    [tex]K_b = \frac{[HA][OH^-]}{[A^-]}[/tex]

    [tex]K_a \times K_b = K_w[/tex]
     
  12. Sep 4, 2015 #11

    epenguin

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    To your question about what you thought the teacher assumed, you may find the following a short cut.

    HCO3- is a weak acid and a weak base. Weak acids and bases produce low H+ and OH- concentrations.
    So these concentrations are negligible compared to those of the main species there in solution which here are Na+, CO32-, HCO3-, and H2CO3.

    Students don't get this because [H+] and [OH-] are all-important, and in this case what you are trying to find. But they are important for the equilibria they participate in, mediating as it were between the big players, but not for the amounts they contribute to the mix. Or to put it another way, small numbers have an effect when they multiply or divide much bigger numbers, but not when they add or subtract to them.

    Thus, suppose you add NaHCO3 salt to water, to make it say 0.1 or 0.01 M. You can imagine it starts as that concentration of Na+ and an equal concentration of HCO3-. Now if a significantly large number of protons dissociate from the HCO3- turning it into CO32- and we've said that [H+] is negligible, where can they have gone? Only to another HCO3- turning them into H2CO3. Thus for every CO32- created a H2CO3 is created. Result: what the teacher said in red in your OP
     
  13. Sep 4, 2015 #12

    Titan97

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    @Borek or @epenguin , see of I am correct.
    The meaning of ##k_a=\frac{[HA^-][H^+]}{H_2A]}## is the dissociation of ##H_2A## which is being formed by hydrolysis of HA-. It's "almost" the reverse reaction of hydrolysis of HA- but gives an extra product H+.
    So there are totally four reactions taking place:
    2 Hydrolysis reactions of HA-, dissociation of of H2A and formation of HA- from A2-
    My only doubt remaining is that dissociation of H2A. Does it give H+ ions just enough to maintain autoprotolysis equilibrium?
     
  14. Sep 5, 2015 #13

    Borek

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    You can write even more reaction equations, but there are only three independent equilibria in the solution (and their choice is somewhat arbitrary).

    H2A ↔ H+ + HA-

    HA- ↔ H+ + A2-

    H2O ↔ H+ + OH-

    Equilibrium constant for every other reaction you can come up with can be expressed using equilibrium constants for these three reactions, and doesn't add any new information to the problem.

    I feel like half of your problem stems from the fact you don't know how to deal with the general case of pH calculation, and you are rather shaky about details (both in terms of math and chemistry), so you are being thrown off by minutia that don't matter.
     
  15. Sep 5, 2015 #14

    Titan97

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    Shouldn't it be HA-+H2O↔H3O++A2-
    $$K_a=\frac{[H_3O^+][A^{2-}]}{[HA^-]}$$
    I got confused with hydrolysis constant and dissociation constant
    The second reaction is HA-+H2O↔H2A+OH-
    $$K_b=\frac{[H_2A][OH^-]}{[HA^-]}$$
    Thank you for that constructive feedback.
    Also, its teacher's day today in India. Happy teacher's day. (You might not have taught me in person. But I have learned many things from you)
     
    Last edited: Sep 5, 2015
  16. Sep 5, 2015 #15

    epenguin

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    That is the same thing as Borek's second equation. H+ and H3O+ are different representations of the same thing. Any textbook should tell you this.
     
  17. Sep 5, 2015 #16

    Borek

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    In most cases concentration of water is for all practical purposes constant, so whether we include it in the reaction equation or not doesn't matter. Note that in your formula for Ka you use only H3O+, but you ignore the presence of water (you don't put it in the denominator, even if that means lack of consistency).

    Treating them as separate reactions is just making it more confusing. It doesn't matter what is really happening in the solution, whether it is a conjugate base reacting directly with water and liberating OH-, or water dissociating to produce H+ (consumed by the conjugate base that gets protonated) and OH- (that becomes a product of the whole process). What matters is that there is in fact just one reaction involving the acid and its conjugate base and one describing water behavior.
     
  18. Sep 5, 2015 #17

    Titan97

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    • I got the same equation.
    Multiplying and dividing by [H+],
    $$k_b=\frac{[H_2A]k_w}{[HA^-][H^+]}$$
    $$\frac{k_b}{k_w}=\frac{[H_2A]}{[HA^-][H^+]}=\frac{1}{k_a}$$
    $$[H_2A]=\frac{[HA^-][H^+]}{k_a}$$
     
  19. Sep 5, 2015 #18

    Borek

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    Hardly surprising, as Ka×Kb=Kw.
     
  20. Sep 5, 2015 #19

    Titan97

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    Thank you.
     
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