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Question: The overall complex dissociation equilibrium constant for the complex \(\lbrack Cu(NH_{3})_{4}\rbrac...

The overall complex dissociation equilibrium constant for the complex [Cu(NH3)4]2+\lbrack Cu(NH_{3})_{4}\rbrack^{2 +}ion will be

(β4\beta_{4}for this complex is 2.1×10132.1 \times 10^{13})

A

4.7×10−144.7 \times 10^{- 14}

B

2.1×10132.1 \times 10^{13}

C

11.9×10−211.9 \times 10^{- 2}

D

2.1×10−132.1 \times 10^{- 13}

Answer

4.7×10−144.7 \times 10^{- 14}

Explanation

Solution

Overall dissociation constant

=12.1×1013=4.7×10−14= \frac{1}{2.1 \times 10^{13}} = 4.7 \times 10^{- 14}