Al(OH)3 excess OH^- complex ion Al(OH)4^-

Al(OH)3 excess OH^- complex ion Al(OH)4^-

 

 

Al(OH)3는 과량의 OH^- 이온과 반응하여 착이온 Al(OH)4^-를 형성한다.

 

 

Aluminum hydroxide reacts with an excess of hydroxide ions to form the complex ion Al(OH)4^-.

a) Write an equation for this reaction.

b) Calculate K.

c) Determine the solubility of Al(OH)3 (in mol/L) at pH 12.0.

 

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a) Write an equation for this reaction.

 

Al(OH)3(s) + OH^-(aq) → Al(OH)4^-(aq)

 

 

 

 

b) Calculate K.

 

K = [Al(OH)4^-] / [OH^-]