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Finding pH using acidity/basicity constants

  1. Oct 23, 2008 #1
    technically it is not homework, more like struggling with textbook material, but something tells me it would end up here anyway so here it is :

    there's something about the method that gives me different results using different paths so i'm going to show both of them and see if you can spot the error in my thinking

    let's say i want to find the pH of the solution of, say, 0.004 M HCOONa, given that for HCOOH [tex] K_a = 1.8\times10^{-4} [/tex]

    For the reaction
    [tex] HCOOH \rightleftharpoons H^+ + HCOO^- [/tex] the constant is [tex] K_a = \frac{[H^+][HCOO^-]}{[HCOOH]} [/tex]

    standard procedure of writing the initial and final concentrations

    [tex]
    \begin{array}{| c|c|c |} \ [HCOOH]&[H^+]&[HCOO^-]\\
    \hline
    \0&10^{-7}&0.004\\
    \hline
    x&10^{-7}-x&0.004-x\\
    \hline
    \end{array}
    [/tex] which gives [tex] K_a = \frac{(10^{-7}-x) (0.004-x)}{x} = 1.8 \times 10^{-4} [/tex]

    solving for x gives [tex] x = 9.569 \times 10^{-8} [/tex] so [tex] [H^+] = 10^{-7} - x = 4.32 \times 10^{-9} [/tex] and [tex] pH = 8.366 [/tex] which sounds OK for HCOONa solution

    the textbook, however, insists on treating the [tex] HCOO^- [/tex] as the conjugate base of HCOOH and using the following reaction

    [tex] HCOO^- + H_2 O \rightleftharpoons HCOOH + OH^- [/tex] and its basicity constant [tex] K_b = \frac{[HCOOH][OH^-]}{[HCOO^-]} [/tex]

    (noting that [tex] K_a \times K_b = \frac{[H^+][HCOO^-]}{[HCOOH]} \times \frac{[HCOOH][OH^-]}{HCOO^-]} = [H^+] [OH^-] = K_w [/tex] so [tex] K_b = \frac{10^{-14}}{1.8 \times 10^{-4}} = 5.56 \times 10^{-11} [/tex]

    and once again the concentrations


    [tex]
    \begin{array}{| c|c|c |} \ [HCOO^-]&[HCOOH]&[OH^-]\\
    \hline
    \0.004&0&10^{-7}\\
    \hline
    0.004-x&x&10^{-7}+x\\
    \hline
    \end{array}
    [/tex] from which [tex] K_b = \frac{x (10^{-7}+x)}{0.004-x} = 5.56 \times 10^{-11} [/tex]

    here the x is found as [tex] x = 4.240 \times 10^{-7} [/tex], therefore [tex] [OH^-] = 10^{-7} + x = 5.24 \times 10^{-7} [/tex] so [tex] pOH = 6.281 [/tex] and [tex] pH = 14 - pOH = 7.719 [/tex]

    as you notice the two results (8.366 and 7.719) are too different to call them equivalent

    so

    any ideas ?


    -- o, and why does my LaTeX appear on white background ?
     
  2. jcsd
  3. Oct 23, 2008 #2

    Borek

    User Avatar

    Staff: Mentor

    Good question, I like it.

    Looks to me like your first approach ignores presence of water. When you remove H+ water dissociation equilibrium shifts right - which you don't take into account. In Kb approach base reacts with water - so its presence is taken into account automatically.
     
  4. Oct 23, 2008 #3
    hmm...

    i suppose so

    i wonder if there's an easy way for modifying the first approach without going into the purely analytical "(1) charge balance, (2) material balance, (3) expressions for equilibrium constants"
     
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