A) \[2.5\times {{10}^{13}}\]
B) \[5\times {{10}^{-13}}\]
C) \[2.5\times {{10}^{-14}}\]
D) \[4.0\times {{10}^{-13}}\]
E) \[4.0\times {{10}^{-14}}\]
Correct Answer: A
Solution :
\[{{\beta }_{4}}\]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 \[{{k}_{n}}=\frac{{{[M{{L}_{4}}]}^{2-}}}{[{{M}^{2+}}]{{[{{L}^{-1}}]}^{4}}}\] \[{{k}_{n}}={{\beta }_{4}}=2.5\times {{10}^{13}}\]
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