Question

2. You mix a 105.0 mL sample of a solution that is 0.0114 M in Ni(NO3)2 with a 190.0 mL sample of a solution that is 0.300 M in NH3. After the solution reaches equilibrium, what concentration of Ni2+ remains? The value of Kr for [Ni(NH3)6]2+ is 2.0x108.

Answer 1

After mixing

Concentration of Ni²⁺ = 0.0114M/( 295ml/105ml) = 0.004058M

Concentration of NH3 = 0.300M/(295ml/190ml) = 0.1932M

Let assume complete formation of Ni(NH3)6

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

stoichiometrically, 1mole of Ni²⁺ react with 6moles of NH3 to produce 1mole of Ni(NH3)6²⁺

0.004058M of Ni²⁺ react with 0.02435M of NH3 to give 0.004058M of Ni²⁺

After completion of reaction remaining concentration of NH3 = 0.1932M - 0.02435M = 0.1689M

Now , consider the dissociation equilibrium of Ni(NH3)6²⁺

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

K_d = [Ni²⁺][NH3]⁶/[Ni(NH3)6]

K_d = 1/K_f = 1/ 2.0 ×10⁸ = 5.0 ×10^-9

Initial concentration

[Ni(NH3)6²⁺] = 0.004058

[Ni²⁺] = 0

[NH3] = 0.1689

change in concentration

[Ni(NH3)6²⁺] = - x

[Ni²⁺] = + x

[NH3] = + 6x

equilibrium concentration

[ Ni(NH3)6²⁺] = 0.004058 - x

[Ni²⁺] = x

[NH3] = 0.1689 + 6x

so,

x(0.1689 + 6x)⁶/ 0.004058 - x) = 5.0 ×10^-9

we can assume , 0.1689 + 6x = 0.1689 and 0.004058 - x = 0.004058 because x is small value

x( 0.1689)⁶/ 0.004058 = 5.0 ×10^-9

x 0.005721= 5.0 ×10^-9

x = 8.74 ×10^-7

Therefore, at equilibrium

Concentration of Ni²⁺ remains = 8.74 ×10^-7M

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