In the presence of CN-, Fe3 forms the complex ion Fe(CN)63-. The equilibrium concentrations of Fe3 and Fe(CN)63- are 8.5 x 10-40
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Answer : The equilibrium concentration of at
is,
.
Solution : Given,
Equilibrium constant,
Initial concentration of = 0.260 m
Let, the 'x' mol/L of are formed and at same time 'x' mol/L of
are also formed.
The equilibrium reaction is,
Initially 0.260 m 0 0
At equilibrium (0.260 - x) x x
The expression for equilibrium constant for a given reaction is,
Now put all the given values in this expression, we get
By rearranging the terms, we get the value of 'x'.
Therefore, the equilibrium concentration of at
is,
.