Mix dilute NaOH solution with dilute CH3COOH solution at room temperature

Drops of CH3COOH are dropped into NaOH, and the mixed solution is mainly NaOH with a small amount of NaAC

c(na+)>c(oh－) c(ch3coo－)>c(h+)

2. CH3COOH is dropped into NaOH and just reacted, and the mixed solution is mainly NaAC

c(na+)>c(ch3coo－)>c(oh－) c(h+)

3. CH3COOH is dropped into NaOH, just neutral, and the mixed solution is mainly NaAC and a small amount of HAC

c(na+)= c(ch3coo－)>c(oh－) =c(h+)

4. CH3COOH is dropped into NaOH just neutral, and then a small amount of CH3COOH is dropped, the solution is acidic, and the mixed solution is mainly NAAC and a small amount of HAC

c(ch3coo－) c(na+)>c(h+)>c(oh－)

5. A large amount of CH3COOH is added, and the mixed solution is mainly HAC, with a small amount of NAAC (equivalent to 1 drop of NAOH into CH3COOH).

c(ch3coo－) c(h+)>c(na+)>c(oh－)

Obviously not possible and C(OH)>C(Na+)>C(H+)C(CH3COO).

The reason is that both the OH- and H+ reactions will decrease.

i.e. oh- will not be more than na+.

OH- reacts with H+ to produce water and the concentration is reduced.

Acetic acid is a weak electrolyte that cannot be completely ionized, so sodium ions acetate hydroxide hydrogen ions.

1 All CH3COOH ionization to form CH3COO- and H+, which is a reversible reaction, NaOH is completely ionized, OH- consumes H+ to produce H20, which is difficult to ionize, and the ionized CH3COO- is one to one with H+, so C(CH3CoO-)>C(H+).

pH<7, that is, acidic after mixing, CH3COOH is a weak acid, so the amount of acid used is more than that of alkali.

C(H+) C(CH3COO-) C(H+) C(Na+) C(OH-) I hope I can help you

C(H+) and C(Na+) cannot be compared.

Analysis: When CH3COOH solution is added dropwise to NaOH solution, the conservation of charge is satisfied in any case, that is, C(Na+)+C(H+) C(CH3CoO-)+C(OH-) is established.

The possible outcomes are:

Equivalent. 1. NaOH solution and CH3Coona solution.

pH 7 and C(Na+) C(OH-) C(CH3COO-) C(H+).

2. CH3Coona solution.

pH 7 and C(Na+) C(CH3CoO-) C(OH-) C(H+).

3. CH3COONA solution and a small amount of CH3COOH solution.

pH 7 and C(Na+) C(CH3CoO-) C(OH-) C(H+).

4. CH3COONA solution and an appropriate amount of CH3COOH solution.

pH<7 and C(CH3CoO-) C(Na+)>C(H+)>C(OH-).

5. CH3COONA solution and a large amount of CH3COOH solution.

pH<7 and C(CH3CoO-) C(H+)>C(Na+)>C(OH-).

So the answer is AD. I just talked about it.

Because pH 7 is acidic, the acetate ion is greater than the sodium ion, the hydrogen ion is greater than the hydroxide ion.

pH 7 is acidic and does not exhibit C(CH3CoO-) C(Na+).

A false alkali overload i.e. a mixture of CH3Coona and NaOH C(H+)b correct Conservation of charge.

C acid excess is a mixture of CH3COONA and CH3COOH.

D error pH = 7 C(CH3COO-) = C(Na+).

C, Na+ ion is definitely the one with the most among all ions.

The first is an excess of bases, the second is conservation of charge, and the third is an excess of acids. The last one is not right. The first one should be: =