Question

You mix a 110.0 −mL sample of a solution that is 0.0133 M in NiCl2 with a 200.0 −mL sample of a solution that is 0.250 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×$10^{8}$ .

Answer 1

Based on the data provided, at equilibrium, the concentration of Ni2+ that remains is 4.268 ×10^-6 M

What is molar concentration of a solution?

The molar concentration of a solution is the amount in moles of solute dissolved in a given volume of solution in litres.

• Molar concentration = moles/volume in Litres

Total volume of solution after mixing = 110 + 200 = 310 mL

Concentration of Ni2+ = 0.0133M/( 310ml/110ml) = 0.00472M

Concentration of NH3 = 0.250M/(310ml/200ml) = 0.1613 M

Assuming complete formation of Ni(NH3)6, the equation of formation is given as:

• Ni2+(aq) + 6NH3(aq) -------> Ni(NH3)62+ (aq)

From equation of the reaction, 1 mole of Ni2+ reacts with 6 moles of NH3 to produce 1 mole of Ni[(NH3)6]2+

Then,

0.00472 M of Ni2+ react with 0.02832 M of NH3 to give 0.00472 M of Ni2+

Concentration of remaining NH3 = 0.1613 M - 0.02832M = 0.13298 M

The equation of dissociation at equilibrium of Ni(NH3)62+ is given as:

• Ni[(NH3)6]2+(aq) <-------> Ni2+(aq) + 6NH3(aq)

Kd = [Ni2+][NH3]6/[Ni(NH3)6]

Kd = 1/Kf = 1/ 2.0 × 10^8

Kd = 5.0 ×10^-9

Using an ICE chart;

Initial concentration:

[Ni(NH3)62+] = 0.00472

[Ni2+] = 0

[NH3] = 0.

Change in concentration:

[Ni(NH3)62+] = - x

[Ni2+] = + x

[NH3] = + 6x

Equilibrium concentration:

[ Ni(NH3)62+] = 0.00472 - x

[Ni2+] = x

[NH3] = 0.13298 + 6x

From the dissociation constant equation;

x(0.1330 + 6x)^6 / 0.00472 - x) = 5.0 ×10^-9

Assuming x is very small;

• 0.13298 + 6x = 0.13298
• 0.00472 - x = 0.00472

x( 0.13298)^6/ 0.00472 = 5.0 ×10^-9

x × 0.00117= 5.0 ×10^-9

x = 4.268 ×10^-6 M

Therefore, at equilibrium, the concentration of Ni2+ that remains is 4.268 ×10^-6 M

Learn more about molar concentration at: brainly.com/question/26255204