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 water is added, doubling the volume?

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Solution

$K_{e q} = \frac{[{F e (S C N)}^{2 -}]}{[{S C N}^{-}] [{F e}^{3 +}]}$
On adding water the equilibrium will shift to the left so as to

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