Well, first of all, the front of the image is flat, so it is OH-+H+=H2O with H+

Later, it precipitated, and Al3+, Fe3+ and Mg2+ are all possible. Then it settled a little. This is the bisexual nature of Al(OH), and there must be Al3+.

Look at the lattice number. The previous growth process is 6, and the precipitation reduction process is 1, and Al(OH)3+OH-=Al(OH)4-

It can be seen that 6 parts of OH- have been used before, and the final precipitation is just half of the maximum value, so the amount of Al3+ and another ion is equal. Now we know that the ion is Fe3+, which together with Al3+ consumes 6 parts of OH-.

Later, it was seen that there were anions, so if there were Al3+ and H+, there would be no CO32-, only SO42-.

Oh, by the way, I think the precipitation is still flat, so at this time it is NH4++OH-= NH3 H2O, with NH4+

Then there is absolutely no CO32-, and the uncertainty is Na+

Then look at the options. A said there was Mg2+, but actually there was Fe3+.

B doesn't necessarily have Na+

C correct

D and NH3 H2O?

I hope it helps you.

D There must be Na2SO4 in the end, Na comes from NaOH, and SO42- already exists.

I'm not sure if this is the right thing to do, but I tried my best.

I hope I can help you! !