Asked by Karissa
What [I-] should be maintained in KI(aq) to produce a solubility of 1.4×10−5 mol PbI2/L when PbI2 is added?
Ksp of PbI2 = 7.1 x 10^-9
PbI2 <-> Pb + 2I
i: - (1.4x10^-5) 2(1.4x10^-5)
c: - +x "
I solved for x.
Ksp = [Pb][I]^2
(7.1x10^-9) = (1.4x10^-5 + x)((2)1.4x10^-5))
I got 9.056 as an answer, but it's wrong.
Why??
Ksp of PbI2 = 7.1 x 10^-9
PbI2 <-> Pb + 2I
i: - (1.4x10^-5) 2(1.4x10^-5)
c: - +x "
I solved for x.
Ksp = [Pb][I]^2
(7.1x10^-9) = (1.4x10^-5 + x)((2)1.4x10^-5))
I got 9.056 as an answer, but it's wrong.
Why??
Answers
Answered by
Karissa
Nvm, I added +x to [Pb] instead of to [I-]
I got the right answer now.
I got the right answer now.
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