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Finding the value of the ionization for an acid

  1. Jul 7, 2005 #1
    So this is probably going to look like a ridiculously easy question, but I'm stumped. I have searched for hours looking for a formula to use, and I've given up on my text.

    Question: 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

    Now I want to use the formula Ka=[H+] [A-] / [HA] whcih would give me
    Ka= [0.001] [A-] / [0.1]. The answer makes no sense. I have no idea where to go next. Can anyone point me in the right direction???

  2. jcsd
  3. Jul 7, 2005 #2


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    The Ka represents concentrations at equilibrium. the original concentration of the acid is .100mol/L, find the equilibrium concentration. Assume that .001 mol/L of acid has dissociated.
  4. Jul 12, 2005 #3
    Moderator Note : Nerro, do not post complete solutions to homework problems. That rarely helps.
    Last edited by a moderator: Jul 12, 2005
  5. Jul 12, 2005 #4


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    To the OP : Start with the balanced equation. You can not do much without the balanced equation down on paper (or in your head). From there, use the stoichiometric proportions (as GCT suggested) to find the value of [A-] at equilibrium.

    If the final answer you get is a little off from one of the options, pick the closest choice (there should be one that's within about 2% of your calculation).
  6. Jul 12, 2005 #5
    Since my complete answer is apparently a little over the top (I figured after a week he should have figured it out anyway). Here's a hint: If there is only one acid in solution and a certain amount of H+ is in solution. How much of the acid will have deprotonated to supply those hydrogen ions?
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