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PH of 6,0*10^-4 M NaNO2 solution

  1. Nov 28, 2015 #1
    1. The problem statement, all variables and given/known data

    It's all in the title. Correct answer: pH = 7,20. [itex]K_a (HNO_2) = 4,0 \cdot 10^{-4}[/itex]

    2. Relevant equations

    [tex]K_b = \dfrac{K_w}{K_a}[/tex]
    [tex]K_b = \dfrac{HB^+ \cdot OH^-}{B}[/tex]

    3. The attempt at a solution

    NaNO2 fully dissolves in water giving [itex]6,0 \cdot 10^{-4} M \, NO2^-[/itex]. Now water forms an equilibrium with NO2- as it gives H+ to NO2-:

    [tex]NO_2 ^- + H_2 O \leftrightharpoons HNO_2 + OH^-[/tex]

    c & NO_2^- & HNO_2 & OH^-\\
    At \, start & 6 \cdot 10^{-4} & 0 & 0 \\
    At \, equilibrium & 6 \cdot 10^{-4}-x & x & x


    [itex]K_b = \dfrac{x^2}{6 \cdot 10^{-4}-x} \leftrightarrow x^2 + K_b x - 6K_b \cdot 10^{-4} = 0 \\

    \rightarrow x = \dfrac{-K_b \stackrel{+}{(-)} \sqrt{(K_b)^2 - 4(1)(-6K_b \cdot 10^{-4})}}{2}
    = 1,22462 \cdot 10^{-7} = [HO^-]\\

    \rightarrow pH = 14 - pOH = 14 + lg[OH^-] = 7,088

    This is not the correct answer (7,20). What am I missing? Water? I only know the iterative process for determining the correct [H+] for low concentrations H+ (when water has an effect on the result as Zumdahl describes it), not OH-.

    Can someone point me in the right direction?
    Last edited: Nov 28, 2015
  2. jcsd
  3. Nov 29, 2015 #2


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    Staff: Mentor

    Looks like it is one of these cases where water autodissociation can't be ignored. 7.20 is an answer I got for an exact solution (done with BATE, pH calculator), 7.09 is what the ICE table produces.

    In general this requires solving a 3rd degree polynomial.
  4. Nov 29, 2015 #3
    Ok, now I'm baffled, since unless the polynomial is in product form (=0) I don't think I know how to solve such a thing. So the formula [tex]K_a = \dfrac{[H^+]^2 - K_w}{[HA]_0 - \dfrac{[H^+]^2 - K_w}{[H^+]}}[/tex] isn't going to help me here? What I thought of doing is finding out the [H+] from the calculated pH and using the above formula, but it would require knowing [itex] [HA]_0 [/itex], meaning the amount of acid put into the solution before it dissociated. Sadly that is information I don't have (and there was no acid to begin with, just a salt of an acid).
  5. Nov 29, 2015 #4


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    Staff: Mentor

  6. Nov 29, 2015 #5
    Turns out it was much simpler and I didn't have to go beyond what the book teaches. I tried the iteration process to find out [OH-], but with Kb, [ B ]0 and [OH-] instead of the values used in the formula for acids in the book as follows

    [tex] K_b = \dfrac{[OH^-]^2 - K_w}{[ B ]_0 - \dfrac{[OH^-]_0^2 - K_w}{[OH^-]_0}} [/tex]

    , and got the right answer. I suppose I could have derived the formula for bases myself by applying mass-, charge- and other conservation laws, just like the iteration formula for acids([H+]) was derived. Still, this conversation helped my thinking, so thanks for that.
  7. Nov 29, 2015 #6


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    Homework Helper
    Gold Member

    I'm getting pH 7.2 as near as makes no difference., with a square rooting but no polynomial equations

    [OH-] - [H+] = [Na+] - [NO2-] = [HNO2] = [H+][NO2-] /Ka

    I then make the approximation [NO2-] = [Na+] which, since we have salt of strong base and weak acid and we must be above pH 7 is good to better than 1/1000, and thence get, as the numbers happen to be easy

    [HNO2] = (3/2) [H+]]

    and thence from first line

    (5/2) [H+] = [OH-]

    and thence via the Kw equation [H+] = 0.4×10-7 and square-rooting etc. finally pH = 7.1989 insignificantly different from 7.2.

    I hope you can see these are fairly natural steps.
    Last edited: Nov 29, 2015
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