What happens when $A l {(O H)_{4}}^{-}$ is treated with sulphuric acid?

Formation of

A l_{2} O_{3}

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Formation of

A l_{2} O

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Formation of

A l (O H)_{3}

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None of these

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Solution

The correct option is C Formation of $A l (O H)_{3}$
When Al(OH)4− is treated with sulphuric acid, formation of Al(OH)3 occurs. The reaction can be written as-
H2SO4+2Al(OH)4−=2Al(OH)3+2H2O+SO42−

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Q. The value of

Δ_{f} G^{0}

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- 827 k J m o l^{- 1}

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The value of for formation of Cr₂O₃is − 540 kJmol⁻¹ and that of Al₂ O₃ is − 827 kJmol⁻¹. Is the reduction of Cr₂O₃ possible with Al?

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What happens when Al(OH)4− is treated with sulphuric acid?

The correct option is C Formation of Al(OH)3When Al(OH)4−Al(OH)4− is treated with sulphuric acid, formation of Al(OH)3Al(OH)3 occurs. The reaction can be written as-H2SO4H2SO4++2Al(OH)4−2Al(OH)4−==2Al(OH)32Al(OH)3++2H2O2H2O++SO42−

What happens when $A l {(O H)_{4}}^{-}$ is treated with sulphuric acid?

The correct option is C Formation of $A l (O H)_{3}$
When Al(OH)4− is treated with sulphuric acid, formation of Al(OH)3 occurs. The reaction can be written as-
H2SO4+2Al(OH)4−=2Al(OH)3+2H2O+SO42−