Chemistry
posted by Saira on .
Given the two reactions
1.PbCl2 <>Pb^2+ + 2 Cl^,
K1= 1.82×10^−10
2. AgCl <> Ag^+ + Cl^,
K2 = 1.15×10−4
what is the equilibrium constant Kfinal for the following reaction?
PbCl2+ 2 Ag^+ <> 2AgCl+ Pb^2+
This is what i have soo far:
PbCl2<>Pb2+ + 2Cl K1= 1.82×10^−10
AgCl <> Ag+ + Cl K2= 1/1.15×10^−4
(AgCl)^2(Pb)/(PbCl2)(Ag)2=
(Pb2+)(Cl)^2/(PbCl2)* (Ag)(Cl)/(AgCl)
I dunt know what to do next because, i did :
= 1.82×10^−10 * 1/1.15×10^−4
and the answer is wrong,
It says When the stoichiometry of a reaction changes, the new equilibrium constant is raised to the same power. Thus, if a reaction is multiplied by 2, the equilibrium constant K is squared.

Right. First, however, check your numbers for the k values. I think they are reversed because Ksp for AgCl is about 10^10 and Ksp for PbCl2 is about 10^4. But if the Ks were put in the problem on purpose, then
K1 = 1.82 x 10^10
and K2 for the reverse direction is 1/1.15 x 10^4 and you square that [that is (1/K2)^2].
Since you added the equations to get the final equation, now you multiply the new Ks. So K for the reaction requested is K1/K2^2. Check me out on that. 
The Values are correct i doubled checked:
But when do
1.82 x 10^10/ (1/1.15×10^−4)^2
= 2.47*10^18
But it says this answer is wrong 
No.
K1 = 1.82 x 10^10
K2 = (1/1.15 x 10^4)^2
Then Keq for the reaction is
K1*(1/K2)^2 =
1.82 x 10^10*(1/1.15 x 10^4)^2 =
1.82 x 10^10*(7.56 x 10^7) =
0.01376. Check it for math. Check it for significant figures. 
I would round it ti 0.014 = Keq.

Here is the error
So K for the reaction requested is K1/K2^2. Check me out on that.
It's K1*(1/K2)^2
1.82 x 10^10*(1/1.15 x 10^4)^2 = 0.014 
That's right thank you