题目

将0.1molAl投入3mol/LNaOH溶液100mL中,充分反应后再滴入1mol/L H2SO4溶液,若要得到全部沉淀,需要加入多少1mol/L H2S04溶液?

提问时间：2021-03-04

答案

2Al + 2OH- + 6H2O ----> 2[Al(OH)4]- + 3H2
to completely react with 0.1 mol of Al,0.1 mol OH- is required.There is still excess OH- left behind,of which the mol = 0.2 mol.
To neutralize 0.2 mol OH- requires 1 M H2SO4
V = 0.2/2 =0.1 L = 100 mL
Since the final precipitate is Al(OH)3
[Al(OH)4]- + H+ ----> Al(OH)3 + H2O
to completely convert [Al(OH)4]- to Al(OH)3 requires
V'=0.1/2 =0.05 L = 50 mL
so V(total) = V + V' = 150 ml of H2SO4