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Coordination Compounds question

2021 · 26 Aug · Shift 2 · Q19
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Coordination Compounds question

2021 · 26 Aug · Shift 2 · Q19

JEE MainChemistryCoordination CompoundsNumerical+4 / −1
The overall stability constant of the complex ion [Cu(NH3)4]2+[Cu(NH_3)_4]^{2+}[Cu(NH3​)4​]2+ is 2.1 ×\times× 1013. The overall dissociations constant is y ×\times× 10 −-− 14. Then y is ‾\underline{\hspace{2cm}}​. (Nearest integer)
Numerical answer
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Correct answer: 5

  1. For the complex formation

Cu2++4NH3⇌[Cu(NH3)4]2+Cu^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+}Cu2++4NH3​⇌[Cu(NH3​)4​]2+

the overall stability constant is

β4=[Cu(NH3)42+][Cu2+][NH3]4=2.1×1013\beta_4 = \frac{[Cu(NH_3)_4^{2+}]}{[Cu^{2+}][NH_3]^4} = 2.1 \times 10^{13}β4​=[Cu2+][NH3​]4[Cu(NH3​)42+​]​=2.1×1013

  1. The overall dissociation constant is the reciprocal of the stability constant:

Kd=1β4K_d = \frac{1}{\beta_4}Kd​=β4​1​

So,

Kd=12.1×1013K_d = \frac{1}{2.1 \times 10^{13}}Kd​=2.1×10131​

  1. Calculate it:

Kd=12.1×10−13K_d = \frac{1}{2.1} \times 10^{-13}Kd​=2.11​×10−13

Kd≈0.476×10−13K_d \approx 0.476 \times 10^{-13}Kd​≈0.476×10−13

Kd=4.76×10−14K_d = 4.76 \times 10^{-14}Kd​=4.76×10−14

  1. Comparing with the given form:

Kd=y×10−14K_d = y \times 10^{-14}Kd​=y×10−14

we get

y=4.76y = 4.76y=4.76

Nearest integer:

y=5y = 5y=5

Therefore, the required integer is 5.

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