Based on the data provided, at equilibrium, the concentration of Ni2+ that remains is 4.268 ×10^-6 M
<h3>What is molar concentration of a solution?</h3>
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
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