can someome help me do the math, i have the answer cant figureit out!? What is t

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can someome help me do the math, i have the answer cant figureit out!?

What is the maximum solubility ofFeF(2) (Ksp)=2.4x 10(-6)) in a 0.10 M Fe (NO3)2 solution

FeF2 => Fe2+ + 2F-
initial 0.1 0
change + x + 2x
equil 0.1 + x 2x

Ksp = 2.4e-6 = [Fe2+][F-]^2 =
(0.1 + x) (2x)^2



HOW DO YOU GET X??!!!!!!!


So x = 6.0 e-6 M
What is the maximum solubility ofFeF(2) (Ksp)=2.4x 10(-6)) in a 0.10 M Fe (NO3)2 solution

FeF2 => Fe2+ + 2F-
initial 0.1 0
change + x + 2x
equil 0.1 + x 2x

Ksp = 2.4e-6 = [Fe2+][F-]^2 =
(0.1 + x) (2x)^2



HOW DO YOU GET X??!!!!!!!


So x = 6.0 e-6 M

Explanation / Answer

What is the maximum solubility of FeF(2) (Ksp)=2.4x 10(-6)) in a0.10 M Fe (NO3)2 solution

                   FeF2            =>         Fe2+           +    2F-
initial                                                 0.1                 0
change                                              +x                  + 2x
equil                                                0.1 +x              2x

Ksp = 2.4e-6 = [Fe2+][F-]^2 = (0.1 + x) (2x)^2

If you were to try to solve this analytically, you would need tosolve a cubic equation. (2.4e-6 =4*(x2)*(0.1+x) = 4*x3 +0.4*x2 Unless you solved this numerically, youneed to make an approximation)

Assume that x >> 0.1.

Thus 2.4e-6 = (0.1)*(2x)2

2x = (2.4e-6 /0.1)0.5

2x = 4.899e-3

x = 2.45 x 10-3    (Sorry about thewrong answer last time)

Plug this into the original expression: 2.4e-6 =4*(x2)*(0.1+x).

4*(2.45e-3)2*(0.1+2.45e-3) = 2.46e-6 ~ Ksp, which isa good approximation

Thus, x = 2.45 e-3 and not 6e-6. Forgiveness, please!

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