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s2008m [1.1K]
2 years ago
13

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} .
Chemistry
1 answer:
love history [14]2 years ago
7 0

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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Answer:

The enthalpy change (∆H) for the reaction is -108.7 kJ

Explanation:

Hess's law can be stated as: when the reactants are converted to products, the enthalpy change is the same, regardless of whether the reaction is carried out in one step or in a series of steps. Then, Hess's Law states that the enthalpy of one reaction can be achieved by algebraically adding the enthalpies of other reactions.

So,  to calculate the ∆H (heat of reaction) of the combustion reaction, that is, the heat that accompanies the entire reaction, you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient ( number of molecules of each compound participating in the reaction) and finally subtract them.

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2 F₂ + O₂ →2 F₂O ∆H=-43.4kJ

Reactants and products are added or canceled, taking into account that certain substances sometimes appear as a reagent and others as a product, so they are totally eliminated (there is nothing left of them anywhere in the reaction, if the same amount in reagents and products) or partially (this substance remains, in less quantity, only on one side), obtaining:

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Adding the enthalpies algebraically, and dividing by 2, because to get the "data" reaction you had to multiply by two, you get:

ΔH= [167.4 kJ - 341.4 kJ - 43.3 kJ]÷2

ΔH= -108.7 kJ

<u><em>The enthalpy change (∆H) for the reaction is -108.7 kJ</em></u>

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