Question

What is the overall formation equilibrium constant for the ion ${[M{L}_4]}^{2-}$ ion, given that ${\beta }_4$ for this complex is $2.5\times {10}^{13}$ ?

• A. $2.5\times {10}^{13}$
• B. $5\times {10}^{-13}$
• C. $2.5\times {10}^{-14}$
• D. $4.0\times {10}^{-13}$
• E. $4.0\times {10}^{-14}$

Answer 1

Answer: (A)

${\beta }_4$ for ${[M{L}_4]}^{2-}$ can be written as
${\beta }_4=\frac{{[M{L}_4]}^{2-}}{[M^{2+}]{[L^{-1}]}^4}=2.5\times {10}^{13}$
The overall formation equilibrium constant can be written as
$k_n=\frac{{[M{L}_4]}^{2-}}{[M^{2+}]{[L^{-1}]}^4}$
$k_n={\beta }_4=2.5\times {10}^{13}$