Answer to Question #185400 in General Chemistry for Jusna

Question #185400

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+Kf = 1.8 x 108 

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


1
Expert's answer
2021-04-25T23:11:48-0400

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


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