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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
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B
4 times
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C
8 times
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D
16 times
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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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