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c, strong acid and weak alkali salt solution is acidic; b is obviously wrong, how can the concentration be added in this way; The concentration relationship Cl is greater than NH4 and H is greater than OH,

c is obviously wrong, strong acids can be completely neutralized with the weak bases that have been ionized, but there are still non-ionized, the weak bases that are not ionized after acid neutralization will also be ionized, so they are still alkaline, this question is that the pH value is equal, not the concentration is equal.

Strong alkali and weak alkali with the same pH value, the solution concentration of weak alkali is larger, because the weak alkali can not be completely ionized, the acid solution is the same, so there is still a part of the weak alkali or weak acid to be ionized after mixing, this question is alkaline after mixing, the pH value is greater than 7, A and C are negative, the concentration of CNH4+ must decrease, D is also negative, the concentration of H+ If you add H+ of water ionization, then this equation is not valid, if you do not consider water ionization, it is equal, you can only reluctantly choose B, it is not explained, and there is a problem with the problem.

The selection of BCDB conforms to the conservation of ions.

c, strong acid and weak alkali salt solution is acidic;

d +OH- After dissolving the acid, the hydrogen ions neutralize the hydroxide ions in the ammonia water, and the reaction moves forward, so the ammonium ions in the solution increase.

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b correct. Ammonia is a weak base, and its pH=11 then its concentration is greater than 10-3mol L, so it is excessive when reacting with hydrochloric acid, so the solution after the reaction is alkaline C wrong, and it is deduced that the concentration of oh- is greater than the concentration of H+, and B is wrong according to the conservation of charge. Whereas, kW is only related to temperature.

This question belongs to the common question type of that chapter, first of all, we need to know that the ionization of weak electrolytes such as ammonia and acetic acid is a very small part of a few percent, so at this time, there is pH11 to imagine that the concentration of ammonia is very large, so there is still a lot of ammonia after the hydrochloric acid reaction is added, so the environment is alkaline, so A is wrong, A is neutral B is obviously right, it is the conservation of charge, C is obviously wrong, the ion product mentioned in D is a function of temperature, only affected by temperature, it is also wrong.

First of all, when the pH of ammonia is 11, her concentration must be much greater, when it reacts with HCl, there must be a large amount of surplus, and the remaining ammonia will inhibit the hydrolysis of ammonium ions, so it will be acidic. Determine the presence of particles in the solution, including Cl, NH4, NH3H2O, H, OH, and H2O

Then the second one is correct from the conservation of charge, and other things are wrong.

Ionization constant calculation:

In the Hf solution set by k(hf) = c(f-) c(h+) c(hf), the hf of x mol l is ionized.

x * x - x) = * 10 -4 solve: x

ph = ≈

a. The weak acid pH > 1 can also be known according to chemical knowledge. That's right.

BC, solubility product is the product of the slightly dissolved solid phase and the ion concentration when the corresponding ions in the solution reach equilibrium, which is only related to temperature. The key is to achieve this solubility product constant under the premise of equilibrium, so it has nothing to do with concentration. Exclude.

D. CaCl2 + 2HF CaF2 + 2HCl In this equilibrium, if the solubility of CaF2 is higher than the ionization degree of HF, the equilibrium moves forward and precipitation occurs.

Solubility product constant: k(caf2) = c(ca2+) c(f-) 2] set dissolved caf2 xmol l

x * x^2 = * 10^-10

Solve x = equilibrium right dissolves, left, right > left balance moves positively, precipitation occurs.

The strong acid made the weak acid mentioned upstairs is just a general rule.

So the answer is AD

Answer: The amount of DHY and NaOH substances is 1:1

So the solute is Nay, but due to the hydrolysis of Y-, the solution is alkaline.

y+h2o=hy+oh-

Therefore, pH=9 At this time, the h+ in the solution is 10-9 power, but the oh ionized by water is greater than 10-9 power (because it is alkaline) a, which is not right.

At this time, the equal volume of the mixture yields the amount of Nay substance concentration = mol l.

So C is not right.

The hydrolysis of Nay must be Na+ at most, so B is not right.

d is directly your columnar equation according to the conservation of protons, and you get the answer d after the sub-shift term

A is right, because it is neutral, it means c(h+) c(oh-).

If the reaction of NH4HSO4:NaOH=1:1, the hydrolysis of NH4+ is acidic.

If the NH4HSO4:NaOH=1:2 of the reaction, the ionization is alkaline. (I won't write the equation).

So NaOH is slightly more than NH4HSO4

So C(Na+) C(SO4 2-).

H+ and OH- are very few, NH4+ is behind SO4 2- due to partial hydrolysis, but much more than H+ and OH-.

B is false, NaHCO3 solution is mainly Na+ and HCO3-, the amount of HCO3- hydrolysis is very small, it is impossible OH->HCO3-

c right, proton conservation.

C(H+) C(oh-) ionized by water

The hydrolysis of a CO3 2- is called one OH and HCO3-, and the continued hydrolysis can generate H2CO3 and two OH-

So CO3 2- front Takano noodles to multiply by 2

D false. Conservation of charge: C(H+) C(Na+) C(OH-) C(CH3CoO-)1)

Since the solution is acidic, the ionization of CH3COOH is considered, and the hydrolysis of CH3COO- is ignored.

So C(ch3Cooh) from (1) and (2) gives C(ch3CoO-) C(OH-) C(H+) C(CH3Cooh).

Option d is wrong.

Select c to deform as c(na+)=c(h+)+c(hco3-)+2c(h2co3) This equation is the conservation of protons.

It is obvious that the ions in a are not kept because the cation is greater than the anion.

b:c(na+) c(hco3-)

D: It is impossible to determine the laughing plum, and there are sizes on both sides of the confession bush.