[Analyticalstudent]

Well I took it and got it back with a 1.5 out of a possible 6 pts.

The solubility product constant for K₂PdCl₆ is 6.0 x10^-6

The Rxn:
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

(a) 0.0500 M PdCl₆^2-

(b) 0.200 M PdCl₆^2-


My answers were

(a) K+ = 0.0155 M

(b) K+ = 0.008 M

i dont know how to incorporate the 50.0 mLs please help

[lemonoman]

50 mL of one solution + 50 mL of another solution = 100 mL of the total solution.

So if you originally had 0.0500 M PdCl₆^2-, then now you have

(0.0500 M PdCl₆^2-)(50 mL/100 mL) = 0.0250 M PdCl₆^2-
(original conc.)    (dilution factor) = (new concentration)

[Analyticalstudent]

so would it be

[K⁺] = (6.0 x10^-6  / 0.025 M) ^(1/2)

= .015492 for the first one?

so i wouldnt use the 0.2 M from the KCl?

or is there a ratio involved?
                             

[Borek]

1. 0.2M gets diluted before precipitation.

2. K⁺ concentration is either limited by Kso and excess PdCl₆^2- or (once all PdCl₆^2- is precipitated) by the K⁺ excess.

Start with limiting reagent stoichiometry.

[Analyticalstudent]

so what do i do from there?

[AWK]

In the case a) you have an excess of KCl - ignore Ksp - this is a simple stoichiometric calculation.
In the case b) you have the excess of PdCl6 2-. Put its  concentration to Ksp equation and calculate concentration K+ from them.