Q. What is the overall formation equilibrium constant for the ion$\left\lbrack ML_{4} \right\rbrack^{2 -}$ 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) $4.0 \times 10^{- 13}$
Explanation

(a)
$\beta$_{4} for $\left\lbrack ML_{4} \right\rbrack^{2 -}$can be written as
$\beta$_{4}=$\frac{\left\lbrack ML_{4} \right\rbrack^{2 -}}{\left\lbrack M^{2 +} \right\rbrack\left\lbrack L^{-} \right\rbrack^{4}}$ $= 2.5 \times 10^{13}$
The overall formation equilibrium constant can be written as
k=$\frac{\left\lbrack ML_{4} \right\rbrack^{2 -}}{\left\lbrack M^{2 +} \right\rbrack\left\lbrack L^{- 1} \right\rbrack^{4}}$
k=$\ \beta_{4} = 2.5 \times 10^{13}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.