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Weak acid-base solution assumption

  1. May 16, 2017 #1
    I have been using a way to verify the assumptions.

    For example lets say, If we add 200 mL of NaOH to 400 mL of HNO2 2M, The pH become 1.5 + (##pH_i##)
    calculate the concentration of NaOH.

    The usual way of doing is this:
    ## K_a = (0.02+x)[H^+]/(4/3-x) ## then find the x which most of the time is equal to the initial concentration of OH- after addition which you can convert back to the original solution and find the required information.

    But the way I see how these approximations work is that you have to also think about how the additional OH can interact with H+ (or H3O+ if you like) to form water. So you really need simultaneous equations
    ##K_w = ([H+]_i -y)([OH-]_i-x-y) ##
    Where X is from the original acid reaction and Y comes from forming water reaction.

    Now usually Y is negligible so I can assume the OH- fully reacts with the acid but here in this situation Y isn't

    I did the math and found that x = 0.41345 and y = 0.01917811 which is not negligible ( about 4.63% mistake, However I am talking about more extreme version of this) . Is this way of thinking is true?
  2. jcsd
  3. May 16, 2017 #2


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    Staff: Mentor

    No idea where these equations and numbers come from nor what the i means, care to elaborate?

    In general when it comes to dealing with such calculations it is best to approach the system in a way described here: http://www.chembuddy.com/?left=pH-calculation&right=toc
  4. May 16, 2017 #3
    Sure sorry,
    First the concentration of NO2- in the acid solution alone is
    ## Ka * [HNO3]_i = x^2 ## solve for x, I get x = 0.03 (assuming [HNO3] doesn't change)

    initial [NO2-] in the combined solution is 0.02 M
    Initial [HNO3] lets assume it doesnt change so it is 4/3 M
    Now when you add OH- the reaction will go to the right:
    HNO2 (-----) H+ + NO2-
    4/3-x ##[H+]_f## 0.02+x
    Using the equilibrium constant equation we get the first equation.
    After finding x which is 0.41345 M, It means that we had 0.41345 M of OH- from NaOH in the combined solution. Going back to the NaOH solution alone we get 0.620175

    Note: Prepare for the mess! I know there are much simpler ways.

    The other way that I am concerned about is this:
    ##[OH-]_i## represents the initial concentration of OH- that comes from NaOH in the combined solution
    ##[H+]_i## represents the initial concentration of H+ that comes from the disassociation of HNO2 in the combined solution before any reaction happen in the combined solution
    If I form two simultaneous equation concerning that OH- might react with H+ or HNO2, I get
    ##\frac{K_a}{K_w} = \frac{(0.02+x)}{(4/3-x)([OH-]_i -x -y)}##
    Where x results from HNO2 and OH- reaction

    and Y results from this

    ##K_w = ([H+]_i -y)([OH-]_i-x-y) ##

    Usually, Y is negligible but in this specific question it is not entirely negligible.
    I found X = 0.41345 and Y = 0.019178
    First, Is all this true?
    and I am asking about if there was a extreme version of this, Where Y is not negligible at all. I have to do this in order to get an exact solution right.
  5. May 16, 2017 #4


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    Staff: Mentor

    That is already based on an approximation, and you are mixing HNO3 with NO2- as if these were related - basically they are not.

    I strongly suggest you read the linked pages before digging deeper, as it seems like you are just juggling equations, no wonder you are getting strange results.
  6. May 16, 2017 #5
    Ah, I meant HNO2 with a ka of 4.8 * 10^(-4)
    I made 2 parts one with an approximation and one without

    Please see the second part. It must be true because I used it multiple times on very weak acid reaction and it gave exact answer. I want to know if this approach is true.

    Yes I have seen some of topics you provided before. I also said in the beginning there is another way
    Thanks in advance
  7. May 16, 2017 #6


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    Staff: Mentor

    You can't assume your[OH-]i (nor [H+]i) to be the only source of OH- (H+), once their concentrations go down acid (and base) will dissociate further (even if you ignore base dissociation here).

    I told you to approach the problem in a more rigorous way, you are wasting time ATM.
  8. May 17, 2017 #7
    I tried both ways and it gave the same answer.
    So I guess both ways are true

    I also found a way to simplify all what I wrote to 2 lines.
    and I derived a equation to solve these kind of questions from your link

    Thanks borek
    Last edited: May 18, 2017
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