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Homework Help: Solubility Product Constant Question

  1. Oct 6, 2008 #1
    The solubility-product constant for K2PdCl6 is 6.0 x 10^ -6 (K2PdCl6 ---> 2K+ + PdCl6 2-). What is the K+ concentration of a solution prepared by mixing 50.0 mL of 0.200 M KCl with 50.0 mL of 0.100M PdCl6(2-)?

    My Approach (thus far):

    1) get mmol of KCl and PdCl6(2-) and get difference of them for excess mmol (of what though?).

    2) I'm not really sure what to do from there without knowing which would be in excess...My hunch is to solve whatever is in excess to similar terms to get concentration of K+.

    Anyways, this questions has had me scratching my head for a few days. Thanks for all the help I get!
  2. jcsd
  3. Oct 6, 2008 #2


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    Homework Helper

    - Find the new concentrations.

    - If the solubility product is saturated by incorporating the two concentrations into the solubility product equation then use the equation to find the amount of precipitate formed.

    -From this deduce the actual saturation concentrations of the K + cations.
  4. Oct 6, 2008 #3
    You see, that's my ultimate confusion - the first step.

    I will show you how I'm doing everything mathematically below

    0.200 M KCl x .05 L = .01 mol KCl

    0.100 M PdCl6(2-) x .05 L = .006 mol PdCl6(2-)

    Now I divide each by total volume ( 50 mL + 50 mL = 100 mL)

    .01 mol KCl/.100 L = 0.1 M KCl x 2 = 0.2 M KCl (because of stoichiometric factor from equation above)

    .006 mol PdCl6(2-)/.100 L = 0.06 M PdCl6(2-)

    So, I have the new concentrations...My goal is to get [K+] though; however, unfortunately, this is where I'm stuck...

    I could substitute [PdCl6(2-)] = 2 [K+]

    which would then end up as...

    2 [K+][K+]^2 = 6.0 x 10^-6

    where [K+] = 0.0144 M

    Is that the correct approach? How would you correct it?
  5. Oct 7, 2008 #4


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