Question

A 40.0 mL of 2.00 M Fe^{3+ (aq)} solution are mixed with 60.0 mL of 3.00 M SCN^- (aq) solution. What is the equibrium concentratio of FeSCN²⁺ (aq), if the complex formation constanst is 189 for the reaction: Fe³⁺(aq) + SCN^- (aq) -----> FeSCN^{2+ (aq)} , Kf=189

Answer 1

Answer –

We are given, 40.0 mL of 2.00 M Fe³⁺ (aq) solution and 60.0 mL of 3.00 M SCN^- (aq) solution

First we need to calculate moles of each

Moles of Fe³⁺ (aq) = 2.00 M * 0.040 L

                               = 0.080 moles

Moles of SCN^- (aq) = 3.00 M * 0.060 L

                                = 0.18 moles

Now we need to put ICE chart –

     Fe³⁺(aq) + SCN^- (aq) -----> FeSCN²⁺ (aq)

I     0.080        0.180                    0

C     -x               -x                     +x

E 0.080-x      0.180-x                +x

Kf = [FeSCN²⁺ (aq) ] / [Fe³⁺(aq)] [SCN^- (aq)]

189 = x / (0.080-x)(0.180-x)

189 [(0.080-x)(0.180-x)] = x

189x² -49.14x +2.72 = x

189x²-49.14x-x + 2.72 = 0

189x²-50.14x + 2.72 = 0

a= 189 , b = -50.14 , c = 2.72

We know the quadratic formula

x = -b+/-√b²-4ac / 2a

by placing the value in it and calculate x value

x= 0.189 and x = 0.0760

So, x = 0.0760, because we take lowest value from both x values

So at equilibrium,

Moles of FeSCN²⁺ (aq) = x = 0.0760 moles

Total volume = 40+60 = 100 mL

[FeSCN²⁺ (aq)] = 0.0760 moles / 0.100 L

                         = 0.760 M