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PH of mixed solution

  1. Dec 15, 2015 #1
    1. The problem statement, all variables and given/known data

    500 mL KOH solution (0.1 M) is mixed with 500 mL HNO2 (0.1 M) so that the reaction becomes:
    ##KOH + HNO_2 \rightarrow KNO_2+H_2O##

    If Ka of HNO2 is ##5 \times 10^{-4}##, determine the pH of the solution

    2. The attempt at a solution


    ##KOH + HNO_2 \rightarrow KNO_2+H_2O##

    Both KOH and HNO2 are in the same number of moles, which is 0.5 * 0.1 = 0.05 mol
    So, the product (KNO2) is also 0.05 mol due to the same coefficient

    And, the ionization reaction of KNO2 is
    ##KNO_2 \rightarrow K^+ + NO_2^-##

    NO2 is also 0.05 mol due to same coefficient.

    But, how to determine the pH??
    I don't see how to get H+ or OH- into the reaction so I'll be able to calculate the pH...
    I don't understand what the Ka of HNO2 matters since KOH and HNO2 has fully react becoming KNO2 and H2O (In other words, there are no more KOH and HNO2)
     
  2. jcsd
  3. Dec 15, 2015 #2

    Borek

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    NO2- + H2O ↔ HNO2 + OH-
     
  4. Dec 15, 2015 #3
    Aha!!
    But, I have no idea about the Kb of NO2-
     
  5. Dec 15, 2015 #4

    Borek

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    Ka×Kb = ...
     
  6. Dec 15, 2015 #5
    Ka×Kb = 10-14

    But, I also don't know the Ka of NO2-
    Is it the same as Ka of HNO2? How come it be the same?
     
  7. Dec 15, 2015 #6

    Borek

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    Ka is a property of HNO2, not of NO2-. It is the ACID dissociation constant and it is HNO2 that is an acid.
     
  8. Dec 15, 2015 #7

    epenguin

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    ##KOH + HNO_2 \rightarrow KNO_2+H_2O##

    Both KOH and HNO2 are in the same number of moles, which is 0.5 * 0.1 = 0.05 mol
    So, the product (KNO2) is also 0.05 mol due to the same coefficient

    And, the ionization reaction of KNO2 is
    ##KNO_2 \rightarrow K^+ + NO_2^-##

    Kind of confusing yourself with words and formulae here. You've got a solution of what we call potassium nitrite (0.05 molar). And what is that physically? Solution of salt of a strong base and weak acid. So 0.05 M K+ which just stays that, and then you would have exactly the same concentration of nitrite ions except some of them get protonated so you have got both NO2- and HNO2

    NO2 is also 0.05 mol due to same coefficient.

    Sorry to me that doesn't mean anything I think we can forget it.

    But, how to determine the pH??
    I don't see how to get H+ or OH- into the reaction so I'll be able to calculate the pH...
    I don't understand what the Ka of HNO2 matters since KOH and HNO2 has fully react becoming KNO2 and H2O (In other words, there are no more KOH and HNO2)


    You have omitted the usual "relevant equations" of the template and these are needed to be able to solve the problem.

    These equations are always the same in all these pH problems:

    Mass conservation equations
    . Electroneutrality equation.. Just one.
    Equilibrium equations. I'll tell you - the equation for acid dissociation and the equation for water dissociation.

    These will need to be used, but probably how to proceed will look complicated to you. What to have to do, and this happens more often than not, is to make some simplifying assumption or approximation. In what pH zone do you expect a salt of a weak acid and strong base to be? Can that reduce the complication?
     
  9. Dec 15, 2015 #8
    NO2- + H2O ↔ HNO2 + OH-
    Kb HNO2 = [HNO2][OH-] / [NO2-]
    Kw/Ka = [HNO2][OH-] / [NO2-]
    10-10 / 5 = [HNO2][OH-] / [NO2-]

    [NO2-] = 0.05 mol / 1 L = 0.05 M
    But, what about [HNO2] ??
     
  10. Dec 16, 2015 #9

    Borek

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    This is not Kb of HNO2, this is Kb of NO2-. Ka×Kb=Kw combines dissociation constants of an acid and its conjugate base. Can you name the acid and its conjugate base here?

    Do you know what an ICE table is? That's the best approach to this kind of problems. Concentrations of all substances involved in the equilibrium are connected by the reaction stoichiometry and the ICE table is a nice way of keeping track of it. It won't work for very diluted solutions (then the only correct approach is the one signaled by epenguin) but 0.05M is quite safe.
     
  11. Dec 16, 2015 #10

    epenguin

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    Look back to my post and you will see and you will see that [NO2-] is not exactly 0 .05 M but something else is.

    You ask what about [HNO2]? You have written equilibrium equations that involve it, but these are not sufficient, you need to write those other equations I mentioned.

    (Even in pH calculations people have different styles. I never remember what an ICE table is from one time to another, but it is probably a equivalent to what I'm suggesting. Actually I manage perfectly well with never using Kb's. My way.)
     
  12. Dec 16, 2015 #11

    The acid is HNO2 and the conjugate base is NO2-
    Ka of acid * Kb of conjugate base is 10^-14
    So, Kb of the conjugate base (NO2-) is 10^(-14) / (5 * 10^-4) = 10^(-10) / 5 , right??

    Yeah, NO2- is not 0.05 mol
    This is the ICE table of KNO2 ionization reaction

    2889w82.jpg

    Then, how to know the [NO2-] since Kc is unknown??

    It really makes me confused now
     
  13. Dec 16, 2015 #12

    Borek

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    Yes.

    It is not KNO2 that you have to deal with, but just NO2-. We assume KNO2 to be fully dissociated, and we are interested only in the reaction I put in my very first post in this thread.
     
  14. Dec 16, 2015 #13
    e9f1q8.png

    I omitted the H2O in the ICE table since it's not used in the Kb
    In the equilibrium, the concentration of NO2- is (0.05 mol - x)/ 1 L = ( 0.05 -x ) M

    ##K_b = \frac{[HNO_2][OH^-]}{[NO_2^-]}##
    ##\frac{10^{-10}}{5}=\frac{x^2}{5*10^{-2}-x}##

    Since x is very small, I neglect it so [NO2-] ≈ 0.05 M
    ##\frac{10^{-10}}{5}=\frac{x^2}{5*10^{-2}}##
    ##x = \sqrt{5*10^{-2}*\frac{10^{-10}}{5}}##
    ##x = 10^{-6}##

    [OH-] = x = 10^-6
    pOH = - log [OH-] = 6
    pH = 14 - 6 = 8

    Thanks for your help
     
  15. Dec 16, 2015 #14

    epenguin

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    As far as I understand your ICE table you have written [K+] as something unknown. This is something you do know at all times as I explained in #7.
     
  16. Dec 16, 2015 #15

    Borek

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    While this is a common thing to do it is always wise to check - after finishing calculations - whether the assumption that x is very small was a correct one. Rule of thumb says if x is lower than 5% of the original concentration it is OK to neglect it.
     
  17. Dec 16, 2015 #16

    epenguin

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    Our posts crossed. Your solution is fine, congratulations!

    In my way, if you write out the equations I said, and, seeing that the solution will be basic, make of the approximation [H+]] = 0 you find, as you did that
    [HNO3] ≈ [OH-].
    Then if I just write it out the usual expression for Ka, use the above results, the expression for Kw, and the same approximation as you I get the same result.
     
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