Calculate the overall complex dissociation equilibrium constant for the $C u (N H_{3})_{4}^{2 +}$ ion, given that $β_{4}$ for this complex is $2.1 \times 10^{3}$ .

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Solution

Dissociation constant is the reciprocal of the stability constant $β_{4}$ .
$∴$ Overall complex dissociation constant $= \frac{1}{β_{4}} = \frac{1}{2.1 \times 10^{3}} = 4.7 \times 10^{- 4}$

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Calculate the overall complex dissociation equilibrium constant for the Cu(NH₃)₄²⁺ ion, given that β₄ for this complex is 2.1 × 10¹³.

[C u (N H_{3})_{4}]^{2 +}

(β_{4})

2.1 \times 10^{13}

[C u (N H_{3})_{4}]^{2 +}

β_{4}

\times 10^{13})

β_{4}

C u (N H_{3})_{4}]^{2 +}

2.1 \times 10^{13}

4.76 \times 10^{- 14}

β_{4}

[C u (N H_{3})_{4}]^{2 +}

2.1 \times 10^{13}

4.76 \times 10^{- 14}

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