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}} $