Answer to Question #185400 in General Chemistry for Jusna

Question

Calculate the concentration of free Ni²⁺ that is present in equilibrium with 1.5 x10^-3 M [Ni(NH₃)₆²⁺ and 0.243 M NH₃.    [Ni(NH₃)]₆²⁺K_f = 1.8 x 10⁸ 

Give your concentration as p[Ni²⁺] (-log[Ni²⁺])

Answer 1

the reactino is Ni2+ + 6NH3 = Ni(NH3)62+

Initial conditions : Ni+2 =0.045M, NH3= 1M Ni(NH3)6+2=0

Ni+ 2 is limiting

hence final =0.045-0.045= 0 NH3= 1-0.045×6= 0.73 and complex =0.045

Ni(NH3)6+2------->Ni+2 +6NH3

let x= drop in concentration of Ni+2

so at equilibrium [NiNH3]6+2 =0.045-x , [Ni+2] =x and NH3=0.27+6x

1/Kf= [Ni+2] [NH3]6/ [Ni(NH3)6+2= 1/5.5×108 =x×(0.73+6x)6/(0.045-x)= 5.5×108

when x is small, x×0.151334/ 0.045 =1/5.5×108= 1.82×10-9

(since 0.73+x =0.73)

x= 5.4×10-10

x= 1.82×10-9×0.045/0.000387