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← JEE Main 2021 · 24 Feb · Shift 1
ChemistryInteger+4 / -1Question 60 of 90

JEE Main 2021 · 24 Feb · Shift 1, Chemistry question 60

The stepwise formation of [Cu(NH3_3)4_4]2+^{2+} is given below :

Cu2++NH3⇌K1[Cu(NH3)]2+\mathrm{Cu^{2+}+NH_3\overset{K_1}{\rightleftharpoons}[Cu(NH_3)]^{2+}}
[Cu(NH3)]2++NH3⇌K2[Cu(NH3)2]2+\mathrm{[Cu(NH_3)]^{2+}+NH_3\overset{K_2}{\rightleftharpoons}[Cu(NH_3)_2]^{2+}}
[Cu(NH3)2]2++NH3⇌K3[Cu(NH3)3]2+\mathrm{[Cu(NH_3)_2]^{2+}+NH_3\overset{K_3}{\rightleftharpoons}[Cu(NH_3)_3]^{2+}}
[Cu(NH3)3]2++NH3⇌K4[Cu(NH3)4]2+\mathrm{[Cu(NH_3)_3]^{2+}+NH_3\overset{K_4}{\rightleftharpoons}[Cu(NH_3)_4]^{2+}}

The value of stability constants K1_1, K2_2, K3_3 and K4_4 are 104^4, 1.58×1031.58\times10^3, 5×1025\times10^2 and 102^2 respectively. The overall equilibrium constants for dissociation of [Cu(NH3_3)4_4]2+^{2+} is x×10−12x\times10^{-12}. The value of xx is _______. (Rounded off to the nearest integer)

Answer and worked solution

Official answer: 1

Overall formation K1K2K3K4=104×1.58×103×5×102×102=7.9×1011K_1K_2K_3K_4=10^4\times1.58\times10^3\times5\times10^2\times10^2=7.9\times10^{11}; dissociation is the reciprocal, 1.27×10−121.27\times10^{-12}, so x≈1x\approx1.

Answer: 1

Topics:Stability and applications
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