Question

You may want to reference (Pages 821 - 825) Section 17.8 while completing this problem. You mix a 120.0 −mL − m L sample of a solution that is 0.0123 M M in NiCl2 N i C l 2 with a 185.0 −mL − m L sample of a solution that is 0.300 M M in NH3 N H 3 .

After the solution reaches equilibrium, what concentration of Ni2+(aq)Ni2+(aq) remains? The value of KfKf for Ni(NH3)62+Ni(NH3)62+ is 2.0×1082.0×108.

Express the concentration to two significant figures and include the appropriate units.

Answer 1

Reaction involved is:
NiCl2 + 6NH3 [Ni(NH3)6]Cl2

ionic equation can ve represented as:

Ni2+ + 6NH3 [Ni(NH3)6]2+
Given that
[Ni2+] = 0.0123 M
Volume of Ni2+ solution = 120.0 mL = 0.12 L
moles of Ni2+ = molarity * volume
moles of Ni2+ = 0.0123 * 0.12 = 1.476 * 10-3 moles
[NH3] = 0.300 M
Volume of NH3 solution = 185 mL = 0.185 L
moles of NH3 = molarity * volume
moles of NH3 = 0.300 * 0.185
moles of NH3 = 55.5 * 10-3 moles

thus NH3 is in excess

Kf = 2*108

Considering the disssociation reaction as:

[Ni(NH3)6]2+ Ni2++ 6NH3

for the above reaction equillibrium constant is:
Kd = [Ni2+][NH3]6 /[Ni(NH3)6]2+ = 1/Kf

Initially, 1.476 * 10-3 moles of Ni2+ will give reacts to give 1.476 * 10-3 moles of [Ni(NH3)6]2+ in 0.305 L of reaction mixture

thus [Ni(NH3)6]2+ = number of moles/volume = 4.839 * 10-3 M

concentration, M	[Ni(NH3)6]2+	Ni2+	6NH3
Initial	4.839 * 10-3	0	0
change	-x	x	+6x
equillibrium	4.839 * 10-3 -x	x	6x

substituting the equillibrium concentration in Kd expression we get:

1/(2* 108) = x.(6x)6/(4.839 * 10-3 -x)

5*10-9 = 46656x7/(4.839 * 10-3-x)

since Kd is very small thus 4.839 * 10-3-x 4.839 * 10-3

5*10-9 = 46656x7/(4.839 * 10-3)

x7 = 5*10-9 *4.839 * 10-3/46656

x7 = 5.186*10-16

x = 6.552 * 10-3 M

thus equillibrium concentration of Ni2+ is 6.5 mM