A) 16 times
B) 64 times
C) 4 times
D) 8 times
Correct Answer: B
Solution :
\[{{K}_{e}}=[F{{e}^{3+}}]{{[O{{H}^{-}}]}^{3}}\] If\[\,[O{{H}^{-}}]=\frac{1}{4}\times [O{{H}^{-}}]\] initial, then \[[F{{e}^{3+}}]\] \[=64{{[F{{e}^{3+}}]}_{\text{initial}}}\]to have\[{{K}_{c}}\]You need to login to perform this action.
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