Consider the reaction, $F e^{3 +} (a q) + S C N^{-} (a q) ⇌ F e S C N^{2 +} (a q)$
How will the equilibrium position shift if $N a O H (a q)$ is added?
[ $F e (O H)_{3}$ is insoluble]

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Solution

$F e^{3 +} (a q) + S C N^{-} (a q) ⇌ F e S C N^{2 +} (a q) x x - ⟶ At t = 0 x - y x - y y ⟶ At equilibrium$
$K_{e q} = \frac{y}{{(x - y)}^{2}}$
Upon addition of $N a O H_{,} F e (O H)_{3}$ form, which takes $F e^{+ 3}$ from the solution and the remaining moles of $F e^{+ 3}$ be $x^{'} < x$ .
$F e^{+ 3} + S C N^{-} ⇌ F e S C N^{2 -}$
$x - y$ $x - y$ $y$ New equilibrium.
$Q_{c} = \frac{y}{(x - y) (x^{'} - y)}$
as $x^{'} < x$ so $x^{'} - y < x - y$
Thus, $Q_{c} > K_{e q}$
So, the reaction will go in backward direction to attain equilibrium.

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