Calculate the Ni+2 concentration of a solution of a solution made by mixing 0.020 mole of solid Ni(NO3)2 into 1.0 liter of 0.22 M NH40H. Kd for Ni(NH3)4 2+ = 1.1x10^-8
Correct answer is 5.7x10^-7. Please show work as to why this is so. Thank you so much!
Balanced equation for the dissociation of the complex ion that formed in solution is :
[Ni(NH3)4]2+ = Ni2+(aq) + 4NH3(aq)
So, Kd = [Ni2+] [NH3]4 / [Ni(NH3)4]2+ = 1.1*10-8
Assume that all of the Ni2+ ions in solution are in the form of [Ni(NH3)4]2+
So, concentration of [Ni(NH3)4]2+ = 0.020 M
Assume the concentration of complex ion dissociated at equilibrium is x , so x mol/L of Ni2+ are released to solution as are 4x mol/L of NH3.
Set up an ICE table to get the remaining amount of [NH3]
[NH3] = original concentration - 4(0.020 M) = 0.22-0.08 = 0.14 M
Kd = 1.1 *10-8 = [x] [0.14]4 / 0.020
x= 5.7*10-7 M
thus, concentration of Ni2+ = 5.7*10-7 M
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