Question

You mix a 125.0 ml sample of a solution that is 0.0117 M in NiCl₂ with a 175.0 ml sample of a solution that is 0.250M in NH_3. After the solution reaches equillibrium, what concentration of Ni²⁺ remains?

please show work

Answer 1

Given,

Concentration of NiCl₂ Or Ni²⁺ solution = 0.0117 M

The volume of NiCl₂ Or Ni²⁺ solution = 125.0 mL x ( 1L /1000 mL) = 0.125 L

The concentration of NH₃ solution = 0.250 M

The volume of NH₃ solution = 175 mL x ( 1L /1000 mL) = 0.175 L

Calculating the number of moles of Ni²⁺ and NH₃,

= 0.0117 M x 0.125 L = 0.001462 mol of Ni²⁺

Similarly,

= 0.250 M x 0.175 L = 0.04375 mol of NH₃

Now, the reaction between Ni²⁺ and NH₃ is,

Ni²⁺(aq) + 6NH₃(aq) [Ni(NH₃)₆]²⁺(aq)

Drawing an ICE chart,

	Ni2+(aq)	6NH3(aq)	[Ni(NH3)6]2+(aq)
I(moles)	0.001462	0.04375	0
C(moles)	-0.001462	-6x(0.001462)	+0.001462
E(moles)	0	0.03498	0.001462

Now, the new concentrations of [NH₃] and [Ni(NH₃)₆]²⁺

Total volume = 0.125 L + 0.175 L = 0.3 L

[NH₃] = 0.03498 mol / 0.3 L = 0.1166 M

[Ni(NH₃)₆]²⁺] = 0.001462 mol / 0.3 L = 0.004873 M

Now, the dissociation reaction for  [Ni(NH₃)₆]²⁺ is,

[Ni(NH₃)₆]²⁺(aq) Ni²⁺(aq) + 6NH₃(aq)

	[Ni(NH3)6]2+(aq)	Ni2+(aq)	6NH3(aq)
I(M)	0.004873	0	0.1166
C(M)	-x	+x	+6x
E(M)	0.004873-x	x	0.1166+6x

Now, the K_d expression is,

K_d = [Ni²⁺][NH₃]⁶ / [[Ni(NH₃)₆]²⁺]

We know,

K_f for [Ni(NH₃)₆]²⁺ = 2.0 x 10⁸

K_d = 1 / K_f

K_d = 1 / 2.0 x 10⁸

K_d = 5.0 x 10^-9

Now,

5.0 x 10^-9 = [x] [0.1166+6x] / [0.004873-x]

5.0 x 10^-9 = [x] [0.1166] / [0.004873] ------- Here, [0.1106+6x] 0.1166 and [0.004873-x] 0.004873, since x<<0.116 and 0.004873

x = 2.1 x 10^-10

[Ni²⁺] = x = 2.10 x 10^-10 M