The equilibrium, $A l (O H)_{3} + O H^{-} ⇌ [A l (O H)_{4}]^{-}$
In aqueous medium shifts towards the left in the presences of :

N H_{4} C l

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dilute

H_{2} S O_{4}

(not in excess)

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NaOH

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C O_{2}

gas

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Solution

The correct options are
A $N H_{4} C l$
B dilute $H_{2} S O_{4}$ (not in excess)
D $C O_{2}$ gas
The equilibrium will be shifted towards right in presence of ammonium chloride,dilute sulphuric acid and carbondioxide gas.The reagent which will reduce the hydroxyl ion concentration sufficiently cause the reaction to proceed from right to left with the consequent precipitation of $A l (O H)_{3}$ .
$[A l (O H)_{4}]^{-} + N H_{4}^{+} \to A l (O H)_{3} ↓ + N H_{3} + H_{2} O$
$[A l (O H)_{4}]^{-} + H^{+} \to A l (O H)_{3} ↓ + H_{2} O$
$2 N a [A l (O H)_{4}]^{-} + H^{+} \to A l (O H)_{3} ↓ + H_{2} O$
But the equilibrium will be shifted in reverse direction in presence of of sodium hydroxide as their will be increase in concentration of hydroxyl ions.

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N H_{4} C l

A l (O H)_{3}

N a [A l (O H)_{4}]

A l (O H)_{3}

A l (O H)_{3} (s) ⇌ A l^{3 +} (a q .) + 3 O H^{-} (a q .); K_{s p}

A l (O H)_{3} (s) + O H^{-} (a q) ⇌ A l (O H)_{4}^{-} (a q .); K_{c}

A l (O H)_{3}

A l (O H)_{3} (s) ⇌ A l^{3 +} (a q .) + 3 O H^{-} (a q .); K_{s p}

A l (O H)_{3} (s) + O H^{-} (a q) ⇌ A l (O H)_{4}^{-} (a q .); K_{c}

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