Hydrolysis of chemical salts Ask the teacher to answer the problem

A: CH3COONA = CH3COO- +Na+ (complete), CH3COO- +H2O = CH3COOH + OH- (reversible), promotes the ionization of water, and the solution is alkaline.

B: NH4Cl = NH4+ +Cl- (complete), NH4+ +H2O = NH3·H2O + H+ (reversible), promote the ionization of water, and the solution is acidic.

C: HCl = H+ +Cl- (complete), ionization inhibition of water, solution acidic.

d: nano3 = na+ +no3- (complete), which has little effect on the ionization of water.

So the answer: A is NH4Cl, B is Nano3, C is hydrochloric acid, and D is sodium acetate.

According to the conservation of charge (i.e., the solution is electrically neutral. The total charge of the cation and the total charge of the anion, the amount of electricity is equal, and the electrical properties are opposite): C(H+) C(NH4+) = C(OH-) C(Cl-), you can easily deduce the correct relationship, which must be incorrect.

According to the expression of conservation of charge written in , it can be known that the solution of C(H+) = C(OH-) is neutral when only C(NH4+) = C(Cl-) is present.

A is NH4Cl, B is Nano3, C is hydrochloric acid, and D is sodium acetate.

2 is definitely not true.

Cl-NH4+ H+ OH- neutral.

Summary. Can you type me your question?

Can you type me your question?

Why is it written like 8-35 in the book, but the conservation of protons that I solved based on the conservation of charge and materials is not the one in the book.

1.The ratio of the concentration of moh and ha is equal to the ratio of the concentration of m* and the concentration of h*: c(moh) c(ha) = c(m*) c(h*) 2

The expression for the hydrolytic balance constant kmonth (m*) is: k month auspicious roll (m*) = c(moh)*c(h*) c(m*).

Because 8-35 in the book refers to the equilibrium constant k months (m*) of the hydrolysis reaction of the book, it is the conservation of protons that you can solve according to the conservation of charge and materials, which are two different concepts.

There are more acids than alkalis.

At 5 Fe3+ is completely precipitated.

Al2(SO4)3 has gas formation.

1.There is an ionization equilibrium in weak acids, and the ionizable H+ is more than the actual H+, and because of 3+11 14, the acid is more than the alkali after the same volume is mixed, so it is acidic.

2.No, the pH should be increased, which can be considered from the hydrolysis equilibrium and the dissolution equilibrium3and Ba(NO3)2 to form a precipitate, and Al2(SO4)3 to form a precipitate and gas CO2

4.Aluminum ion double hydrolysis to produce aluminum hydroxide.

In how mixed it can not be alkaline.

1. Acetic acid is a weak acid, only a small part of it can be ionized (about 10%), and the concentration of acetic acid at pH 3 is about 10 times that of NaOH, so it will definitely be acidic after neutralization.

2. Putting MGC3 in an acidic solution is actually the same as putting in MGO, because MGC3 will combine with H+ to form MGO and CO2. To increase the HP, the iron hydroxide can be more fully precipitated in the alkaline solution.

3. Ba(NO3)2 is directly white precipitate, while Al3+ in Al2(SO4)3 and CO32- in Na2CO3 will be double hydrolyzed to form aluminum hydroxide precipitate and CO2 gas.

4. The double hydrolysis produces aluminum hydroxide.

Dihydrolysis is a reaction of two ions, a weak acid and a weak base, and there is generally precipitation or gas formation.

Question 4: Since it is an acidic solution, the concentration must be H>OH. Since the anion and cation are conserved, the concentration order is H>OH >CH3CoO-C(Na).

CH3CoO- is dissociated from OH-, hence C(OH)C(Na). The solution is acidic and can only be mixed by the equal volume of CH3COOH of pH 3 and the NaOH solution of pH 11, and will not be mixed by the amount and concentration of the same substance, the same volume of NaOH solution and CH3COOH solution So A pair.

Question 8: Magnesium chloride solution with pH = 5 because it contains free H. In order to create a precipitation environment for Fe3+, which is a pH=2-3 environment, H+ must be neutralized without introducing impurities

So both A and B are OK. The answer online is yes.

Question 10: The reaction of BA(NO3)2 and B will only precipitate, while the reaction of Al2(SO4)3 and B will precipitate and gas CO2 (aluminum ions and CO3 ions should not be discarded by double hydrolysis).

Let's start with the answer (1) K2CO3 (2) KAL (SO4)2 (3) BAC3 (4) Na2SO4 (5) AL(OH)3, and AL2O3 (6) H2SO4 if burned

Explanation: 1-5 are all hydrolyzable salts. However, there are strong acids or strong bases that are difficult to volatilize in the hydrolysate products of 1-4, so they are still themselves after evaporation.

For example, the hydrolysis of K2CO3 yields KOH and H2CO3, KOH is a strong base that is difficult to volatilize, and it has always existed in the solution during the evaporation process, so H2CO3 cannot be converted into CO2 and detached from the system in this atmosphere, and it will still exist in the form of K2CO3 after evaporation. Ba(HCO)3 in question 3 will decompose into BAC3 when evaporated and heated, and Na2SO3 in question 4 is easily oxidized, and heating accelerates its oxidation process, and finally Na2SO4, but this has nothing to do with hydrolysis.

In fact, there are 5 questions that are often tested, and the reason is answered: there is a reaction in the aluminum chloride solution AlCl3 + 3H2O = (reversible number) Al(OH) 3 + 3HCl, the hydrolysis equilibrium moves forward when heated, HCl volatilizes, and it is separated from the solution (this sentence must be there), so only Al(OH)3 solid remains after evaporation.

Can you see the difference between 1-4 and 5 now? 1-4 One of the hydrolysate products is a precipitate or gas, while the other is a strong acid or base that is difficult to volatile; 5 The hydrolysate is a precipitate or gas, and the other is a volatile strong acid. Therefore, the composition changes after evaporation.

Question 6 is the volatilization problem of HCL.

gas) +6F- +SiO2==SIF6 2- +OH-===NH3+H2O (ammonium nitrogen fertilizer contains ammonium root (NH4), plant ash is (K2CO3)).

The addition of dilute sulfuric acid = the addition of H+, which can inhibit its hydrolysis.

1, D2, (1) CH3COO- and CH3COOH

2) If I'm not mistaken, it should be.

Let's answer the landlord.