Question

What is the equilibrium Fe³⁺ concentration when 2.65 L of a 0.149 M iron(III) chloride solution are mixed with 2.51 L of a 0.434 M potassium sulfide solution?

Answer 1

Fe+3 and S-2

Fe2S3(s) <-> 2Fe+3 + 3S-2

Identify the limiting reactant

[Fe+3] = M1V1 /(V1+V2) = 2.65*0.149 / (2.65+2.51) = 0.076521 M

[S-2] = M2V2 /(V1+V2) = 2.51*0.434/ (2.65+2.51) = 0.2342666 M

note that

1 mol of Fe+3 requires -- > 3/2 mol of S-2 --> 3/2*0.076521 = 0.1147815 M, we have 0.2342666 M so S-2 is in excess

[S-2] left = 0.2342666 -0.1147815 = 0.1202666

[Fe+3] left = 0.076521 -0.076521 = 0

so

Ksp = [Fe+3]^2[S-2]^3

Ksp = 2.1*10^-18

2.1*10^-18 = (Fe+3)(0.1202666)^3

[Fe+3] = (2.1*10^-18 ) / ((0.1202666)^3)=1.2072*10^-15