Continue to add more sodium hydroxide until the precipitate dissolves, this reac

Question

Continue to add more sodium hydroxide until the precipitate dissolves, this reaction can be explained by the formation of a hydroxide complex Al^+3(aq) + 4 OH^-1(aq) = Al(OH)^-1 _4(aq) The dissolving of the aluminum hydroxide precipitate demonstrates the behavior of an amphoteric hydroxide. An amphoteric hydroxide is one which can react as either an acid or a base. Behavior as a base is expected. Hydroxides are Bronzed bases: proton acceptors. They react with acid by accepting a proton to produce water. The acid character of an amphoteric hydroxide can be understood in terms of either Lewis acid (electron pair acceptor) or Bronzed acid (proton donor) behavior. Note the relationship between these two explanations: Lewis acid: AI(OH)_3(s) + NaOH(aq) = Na[AI(OH)_4](aq) rightarrow Na^+(aq) + [Al(OH)_4]^-1 (aq) Bronzed acid: Al(OH)_3(s) + 3 NaOH(aq) = Na_3AIO_3(aq) + 3H_2O(I) rightarrow 3 Na^+1 (aq) + AlO^-3 _3 (aq) + 3 H_2O(l) You may write the equilibrium equations for amphoteric Al(OH)_3(s) as follows: AI^+3(aq) + 3 OH^-1 (aq) = Al(OH)_3(s) = 3 H^+1(aq) + A1O^-3 _3 (aq) H_3AIO_3(s) Writing Al(OH)_3 in the form H_3AlO_3 emphasizes the acidic character of the hydroxide. The species AlO^-3 _3 is the "amphoteric anion". Write two possible K_sp equations amphoteric aluminum hydroxide. Write an equation showing the reaction of aluminum hydroxide with an acid. HNO_3. Then write an equation showing its reaction with a strong base. NaOH.

Explanation / Answer

1) Al(OH)3 (s) <---> Al+3 (Aq) + 3(OH-)(aq)

Ksp = [Al+3] [OH-]3

H3AlO3 (s) <------> 3H+ + AlO3-3

Ksp = [H+]3 [AlO3-3]

2) Al(OH)3(s) + 3HNO3( Aq) ---------------> Al(NO3)3 (aq) + 3H2O(l)

Al(OH)3 + NaOH (aq) --------------> Na[Al(OH)4]

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