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Acid-Base Equilibria - Ka constant

  1. Jun 13, 2009 #1
    Ok, I'm having a hard time understanding the concept behind finding the dissociation constant. The question is:

    A 0.1 mol/L aqueous solution of weak monoprotic acid has a hydrogen ion concentration of 0.001 mol/L. The value of the ionization, Ka, for this acid is:

    a) 10-6
    b) 10-2
    c) 10-3
    d) 10-5

    Ok, I'm not necessarily interested in the answer, I just need a really good explanation of how to go about solving this question.

    I'm assuming that I would first write out a balanced equation, however, the question is asking for the dissociation constant, therefore I'm not sure if the equation should be:

    HX + H2O --> H3O + X

    or

    HX --> H+ + X- <-- my bet is on this equation because it says nothing about reacting with water.

    The HX represents the weak acid, and X represents its anion.

    The dissociation constant can be found as:

    Ka = [H+][X-] / [HX]

    However, this is where I get lost. If I had to go further, I would substitute the concentrations given, into the equation:

    Ka = (0.001 mol/L)(X) / (0.1 mol/L)

    The balanced equation shows that [H+] = [X-] = 1:1

    Therefore, Ka = (0.001 mol/L)(0.001 mol/L) / (0.1 mol/L) = 10-5

    This last little bit of the questions was somewhat of a guess, but some guidance would be greatly appreciated! Thanks in advance.
     
  2. jcsd
  3. Jun 13, 2009 #2

    symbolipoint

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    You expressed the concept well. You were more intent on the concept than the calculation so you can rethink the calculation according to your best judgement.
    The simpler approach is to ignore the dissociation of water; as you advance in this study, you will learn how to handle situations in which ignoring the dissociation of water is not advised.
     
  4. Jun 13, 2009 #3
    Great. Thanks a lot.
     
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