What is Molar solubility of PbBr2 (Ks= 4.67*10^-6) in 0.050M MgBr2?
The answer that was given was x= 4.59*10^-4 M
How did they find that answer?
MgBr2 here is Strong electrolyte
It will dissociate completely to give [Br-] = 2*[MgBr2]
= 2*0.05 M
= 0.1 M
At equilibrium:
PbBr2 <----> Pb2+ + 2 Br-
x 0.1 + 2x
Ksp = [Pb2+][Br-]^2
4.67*10^-6=(x)*(0.1+ 2x)^2
Since Ksp is small, x can be ignored as compared to 0.1
Above expression thus becomes:
4.67*10^-6=(x)*(0.1)^2
4.67*10^-6= (x) * 10^-2
x = 4.67*10^-4 M
Answer: 4.67*10^-4 M
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