What is the concentration of Fe2 in a solution that is initially

Question

What is the concentration of Fe2 in a solution that is initially 0.010M Fe2 and 0.060M CN-?

Fe2 (aq) 6CN-(aq) ↔ [Fe(CN)6]4-(aq)

Kf = 1.0x1037

A. 3x10-20M

B. 6x10-7M

C. 2x104M

D. 3x10-6M

E. 1x10-20M

i remember it was B, but kind of lost my paper in the mess i got. please explain this to me nice n easy. thanks so much

Answer 1

Be Careful! Your question should look like:

What is the concentration of Fe(2+) in a solution that is initially 0.010M Fe(2+) and 0.060M CN-?

Fe(2+)(aq) + 6CN-(aq) ↔ [Fe(CN)6](4-)(aq), Kf = 1.0x10^37

A. 3x10^-20M

B. 6x10^-7M

C. 2x10^4M

D. 3x10^-6M

E. 1x10^-20M

Answer:

Since the initial concentrations [Fe(2+)] = 0.010M, and [CN-] =  0.060M, and since the Kf of [Fe(CN)6](4-) formation is big, almost all Fe(2+) and CN- in the solution are reacted, thus the final [Fe(CN)6](4-) concentration is 0.010M, and the final concentrations of [CN-] must still be 6 times of [F2(2+)].  Assume x (in unit M) be the final concentration [F2(2+)].  According to:

Kf = 1.0*10^37 = [[Fe(CN)6](4-)] / {[Fe(2+)]*[CN-]^6}

= 0.010/(x*(6x)^6) = 0.010/(x^7*6^6), we have:

x = (0.010/(1.0x10^37*6^6))^(1/7) = 5.8*10^-7 (M)

Thus the answer is indeed B.