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 =...
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...