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 $A g N O_{3} (a q)$ is added? ( $A g S C N$ is insoluble)

Open in App

Solution

$K_{e q} = \frac{[{F e (S C N)}^{2 -}]}{[{S C N}^{-}] [{F e}^{3 +}]}$ , On adding ${A g N O}_{3}$ $[{S C N}^{-}]$ will diminish because of the precipitation of $A g S C N$ , then the equilibrium will shift to the left.

Suggest Corrections

Similar questions

F e^{3 +} (a q) + S C N^{-} (a q) ⇌ F e S C N^{2 +} (a q)

N a O H (a q)

F e (O H)_{3}

F e^{3 +} (a q) + S C N^{-} (a q) ⇌ F e S C N^{2 -} (a q)

F e^{3 +} (a q) + S C N^{-} (a q) ⇌ F e S C N^{2} (a q)

F e (N O_{3})_{3} (a q)

F e^{3 +} (a q) + S C N ⇋ F e S C N^{2 +} (a q)

F e^{3 +} (a q) + S C N^{-} (a q) ⇌ F e S C N^{2 +} (a q)

2 F e^{3 +} (a q) + 2 S C N^{-} (a q) ⇌ 2 F e S C N^{2 +} (a q)

Join BYJU'S Learning Program