Question

A 0.140 mole quantity of NiCl₂ is added to a liter of 1.2 M NH₃ solution. What is the concentration of Ni²⁺ ions at equilibrium? Assume the formation constant of Ni(NH₃)₆²⁺ is 5.5x10^8.

Answer 1

The equilibrium of Ni-ammonia complex formation is

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

The equilibrium constant of formation is:

Kf = [Ni(NH3)6(2+)]/[Ni(2+)]*[NH3]^6

[Ni(2+)] = [Ni(NH3)6(2+)]/Kf*[NH3]^6

Kf = 5.5*10^8.

The initial concentrations of Ni(2+) ions and NH3 in solution are respectively 0.160 M and 1.20 M
Ammonia is in excess vs. Ni2+, based on reaction stoichiometry.
So we can assume that Ni2+ is almost totally converted into the ammonia complex, whose concentration will be 0.160 M.

The concentration of free NH3 in solution will be, therefore, 1.20 - 0.160*6 = 0.24 M

If we put these values in the above equation, we get:

[Ni(2+)] = 0.160/5.5*10^8*(0.24)^6 = 1.52*10^-6 M

*-*-*-
The half reaction for a standard hydrogen electrode (SHE) is:

2H(+) + 2e(-) <----> H2(g)

The Nernst equation for SHE at 25°C (298 K) is:

E = -0.059/2*log{p(H2)/[H+]^2}

We have:

[H+] = 0.31 M
p(H2) = 3.7 atm

If we put these values in the equation, we get:

E = -0.059/2*log{3.7/(0.31)^2} = - 0.047 V

Claudio · 3 years ago

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• 

  If you put 0.160 moles of NiCl2 in 1 L of solution, then [Ni2+] = 0.160 M before it reacts with NH3 to form the Ni(NH3)6 2+ complex ion. With such a large Kf value, we can say that the reaction is essentially quantitative, forming 0.160 M Ni(NH3)6 2+. There will be a VERY small amount of Ni2+ ion left in solution.
  
  Molarity . . . . . .Ni2+ + 6NH3 ==> Ni(NH3)6 2+
  Initial . . . . . . . 0.160 . . .1.20 . . . . . . . . .0
  Change . . . . .0.160-x . .6(0.160-x) . . . .0.160
  Final . . . . . . . . .x . . . . .0.24-6x . . . . .0.160
  
  Kf = [Ni(NH3)6 2+] / [Ni2+][NH3]^6 = 0.160 / (x)((0.24-6x)^6) = 5.5 x 10^8
  
  We know that x will be very small, so 6x will be very small compared to 0.24, so we delete the -6x term to simplify the math.
  
  0.160 / (x)((0.24)^6) = 5.5 x 10^8
  0.160 / 0.00019x = 5.5 x 10^8
  0.160 = (0.00019x)(5.5 x 10^8)
  0.160 = 105,000x
  x = 1.5 x 10^-6 M = [Ni2+]
  
  ======================================...
  
  The redox half-reaction for the SHE is
  
  2H+(aq) + 2e- → H2(g)
  
  and its potential is defined to be zero at 298 K.
  
  Using the Nernst equation to correct for the nonstandard [H+] and P H2,
  
  E cell = Eo cell - 0.059/n log Q
  E cell = 0.00 V - 0.059/2 log (P H2 / [H+]^2) = 0.00 V - 0.059/2 log (3.7 / (0.31)^2)) = -0.05 V