The equilibrium constant for each reaction of lead sulfate dissolving in water is 1.5 x 10-8. If enough PbS04 is added for equilibrium to be reached what will be the equilibrium concentration of Pb2=?

Question

WickedHeart · Answer

Hi, what would you say the answer would be?
@cbenavidez

Cbenavidez · Answer

Hi  
I picked C. 
[pb^2+][so4^2-]
1.5 X10^-8
1.5X10^-8=x^2
I squared both sides but something doesn't seem right. 
My options are:

A. 2.25 x 10-16M
B. 7.0 x 10-9M
C. 1.22 x 10-4M
D. This cannot be determined without knowing the initial concentration of PBS04.

WickedHeart · Answer

Keq = [Pb^2+][SO4^2-]
PbSO4 is not important as it is a solid and it will not affect the equation

we know Keq= 1.5x10^-8

1.5x10^-8 = [x][x]

1.5x10^-8 = x^2

square both sides and use the positive x value for your answer.

x is equivalent to both Pb^2+ and SO4^2+ so you have found the concentration of both with one calculation.

WickedHeart · Answer

what would you get?

Cbenavidez · Answer

Honestly, I don't know. I keep coming up with different answers. I even came up with 2.25x 10-16M

Rosee5656 · Answer

.