Question

# If the concentration of $${OH}^{-}$$ ions in the reaction:$$Fe{ (OH) }_{ 3 }(s)\rightleftharpoons { Fe }^{ 3+ }(aq)+3{ OH }^{ - }(aq.)$$is decreased by $$1/4$$ times, then equilibrium concentration of $${Fe}^{3+}$$ will increase by:

A
64 times
B
4 times
C
8 times
D
16 times

Solution

## The correct option is A $$64$$ times If the concentration of $${OH}^{-}$$ ions in the reaction:$$Fe{ (OH) }_{ 3 }(s)\rightleftharpoons { Fe }^{ 3+ }(aq)+3{ OH }^{ - }(aq.)$$is decreased by $$1/4$$ times, then equilibrium concentration of $${Fe}^{3+}$$ will increase by $$64$$ times.$$\displaystyle K=[{ Fe }^{ 3+ }]_i[{ OH }^{ - }]_i^3$$....(1)$$\displaystyle K=[{ Fe }^{ 3+ }]_f[{ OH }^{ - }]_f^3$$....(2)$$\displaystyle [{ Fe }^{ 3+ }]_i$$ and  $$\displaystyle [{ OH }^{ - }]_i^3$$ are the inital concentrations.$$\displaystyle [{ Fe }^{ 3+ }]_f$$ and  $$\displaystyle [{ OH }^{ - }]_f^3$$ are the final concentrations.From (1) and (2) $$\displaystyle [{ Fe }^{ 3+ }]_i[{ OH }^{ - }]_i^3=[{ Fe }^{ 3+ }]_f[{ OH }^{ - }]_f^3$$.....(3)The concentration of $${OH}^{-}$$ ions is decreased by $$1/4$$ times$$\displaystyle [{ OH }^{ - }]_f=\dfrac {[{ OH }^{ - }]_i}{4}$$....(4)Substitute equation (4) in equation (3)$$\displaystyle [{ Fe }^{ 3+ }]_i[{ OH }^{ - }]_i^3=[{ Fe }^{ 3+ }]_f (\dfrac {[{ OH }^{ - }]_i}{4})^3$$$$\displaystyle [{ Fe }^{ 3+ }]_i=[{ Fe }^{ 3+ }]_f (\dfrac {1}{4})^3$$$$\displaystyle 4^3 \times [{ Fe }^{ 3+ }]_i =[{ Fe }^{ 3+ }]_f$$$$\displaystyle 64 \times [{ Fe }^{ 3+ }]_i =[{ Fe }^{ 3+ }]_f$$Chemistry

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