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!