Equilibrium - Result Question 9
####10. If the concentration of $OH^{-}$ions in the reaction $Fe(OH)_3(s) \rightarrow Fe^{3+}(aq)+3 OH^{-}(aq)$ is decreased by $\frac{1}{4}$ times, then equilibrium concentration of $Fe^{3+}$ will increase by : [2008]
(a) 8 times
(b) 16 times
(c) 64 times
(d) 4 times
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Solution:
- (c) For this reaction $K _{\text{eq. }}$ is given by
$ K=\frac{[Fe^{3+}][OH^{-}]^{3}}{[Fe(OH)_3]} $
$ .=(Fe^{3+})(OH^{-})^{3} \quad[\therefore \text{ solid }]=1] . $
If $(OH^{-})$is decreased by $\frac{1}{4}$ times then for given reaction, equilibrium constant to remain constant, we have to increase the concentration of $[Fe^{3+}]$ by a factor of $4^{3}$ i.e. $4 \times 4 \times 4=64$. Thus, option (c) is correct answer.
To maintain the equilibrium constant if concentration of one of the product decreases by $\frac{1}{n}$ times then the concentration of another product is increased by the factor of $(n)^{x}$, where $x$ is the stoichiometric coeficient.
