[imath]n_{OH^-}[/imath]=0,2
[imath]n_{OH^-}[/imath]=0,025
[imath]OH^- + Al(OH)_3 \to AlO_2^- + 2H_2O[/imath]
0,025--------0,025
=>[imath][Al^{3+}][/imath]=0,1 M
[imath]H^+ OH^- \to H_2O[/imath]
[imath]Mg^{2+} + 2OH^- \to Mg(OH)_2[/imath]
[imath]Al^{3+} + 3OH^- \to Al(OH)_3[/imath]
0,025--------0,075-------0,025
kết tủa sau pư là [imath]Mg(OH)_2[/imath]=>n=0,05=>[imath][Mg^{2+}][/imath]=0,2 M
BTĐT :[imath]n_{H^+}[/imath]+[imath]2n_{Mg^{2+}}[/imath]+[imath]3n_{Al^{3+}}[/imath]=[imath]n_{OH^-}[/imath]
=>[imath]n_{H^+}[/imath]+2.0,05+3.0,025=0,2=>[imath]n_{H^+}[/imath]=0,025=>[imath][H^+][/imath]=0,1 M
pH=[imath]-log([H^+])[/imath]=-log(0,1)=1
Bạn tham khảo nhé, chúc bạn học tốt !