Question

You mix a 105.0 −mL sample of a solution that is 0.0130 M in NiCl2 with a 185.0 −mL sample of a solution that is 0.300 M in NH3.

After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? The value of Kf for Ni(NH3)62+ is 2.0×108.

Answer 1

Ni²⁺ + 6NH₃ ---> [Ni(NH₃)₆]₂⁺

^{mmol of Nickel ion =105 x0.013=1.365}

^{mmol of ammonia =185x0.3=55.5}

^{total vol=290ml}

^{this means 1 mmol of Ni ion react to 6 mmol of ammonia.}

^{so 1.365 mmol will react to=1.365x6=8.19mmol}

^{as ammonia is in excess so 8.19mmol complex is form=8.19/290=0.0282M}

^{[Ni(NH₃)₆]}2+ ^{---> Ni}2+ ^{+ 6NH₃}

^_means 2x10⁸ = C-x/x .x⁷ =c/x⁸ =0.0282/x⁸

it will give value of x which is Ni ion conc finally