Question

A solution is prepared having having the following initial concentrations: [Fe3+]=[Hg2 2+] = 0.5000M; [Fe3+]= [Hg2+]=0.03000M *The following reaction occurs among the ions at a certain temperature. 2Fe3+ (aq) +Hg2 2+ (aq)<-->2Fe2+(aq) +2Hg2+(aq) Kc=9.14*10^-8 What will be the ion concentrations when equilibrium is established? answer: [Fe3+]=0.5193M, [Hg2+]=0.5096M, [Fe2+]=0.0107M, [Hg2+]= 0.0107 M

Answer 1

2Fe³⁺(aq) + Hg₂²⁺(aq) <-------> 2Fe²⁺(aq) + 2Hg²⁺(aq)

K_C = [ Fe²⁺]²[Hg²⁺]²/[Fe³⁺]²[Hg₂²⁺] = 9.14 *10^-8

at equillibrium

[Fe³⁺] = 0.500 - 2x

[ Hg₂²⁺] = 0.500 - x

[ Fe²⁺] = 0.0300 - 2x

[ Hg²⁺] = 0.0300 - 2x

therefore

(0.0300- 2x)²(0.0300- 2x) / (0.500 -2x)²(0.500 - x) = 9.14*10^-8

solving the equation

x = -0.00965

therefore, at equillibrium

[Fe³⁺] = 0.500 - 2(-0.00965) = 0.5193M

[ Fe²⁺] = 0.500 - (-0.00965) = 0.5096M

[ Hg₂²⁺] = 0.0300 +( 2(-0.00965)) = 0.0107M

[ Hg²⁺] = 0.0300 + (2(-0.00965) = 0.0107 M