Help: Chemistry Solubility?
A sample of solid bromide is placed in a 0.2 molar NaBr solution and allowed to reach equilibrium. What is the concentration of Pb ions at equilibrium? (ksp= 4.0 x 10^ -4 of PbBr2)
...so..this is what i tried to do...
getting the molar solubility of PbBr2...
PbBr2 = [Pb][Br2]^2
4.0x10^-4 = [x][2x]^2
you solve for x...
okay...back to the original question.
how come its
4.0 x 10^-4 = [Pb][.2]^2
then solve for Pb, rather than
4.0 x 10^-4 = [Pb][2 x .2]^2?
Answer:
The equilibrium
PbBr2 <> Pb2+ + 2Br-
requires that
Ksp = 4.0 x 10^-4 = [Pb2+][Br-]^2
But we have arleady some Br- in the solution corresponding to 0.2 M NaBr
At equilibrium
[ Pb2+] = x
[ Br-] = 2x + 0.2.
Since 2x is small compared to 0.2 we can assume
[Br-] = 0.2
4.0 x 10^-4 = x (0.2)^2
x = [Pb2+] = 0.01 M
I am very sorry.V weak in chemistry.
where do the Pb comes from?
U told us that
solide Bromide + NaBr ...
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