A reaction (not necessarily a displacement reaction) between a substance and water that leads to water decomposition.
Salts composed of weak acid radicals or weak base ions have two kinds of hydrolysis:
(1) weak acid radicals combine with H+ in water to form weak acid, and the solution is alkaline, such as sodium acetate aqueous solution;
←═→ CH3COOH + OH-
(2) Weak alkali ions combine with OH- in water, and the solution is acidic, such as ammonium chloride aqueous solution:
NH4+ + H2O ←═→ NH3 H2O + H+
Four common salts:
1. Strong acid and alkali salts do not hydrolyze, because the anions and cations they ionize cannot destroy the ionization balance of water, so they are neutral.
Secondly, strong acid and weak base salt, we call the weak base part as weak cation, and the weak cation can contain hydroxyl ions ionized from water, which destroys the ionization balance of water and makes the ionization of water move forward. As a result, the concentration of hydrogen ions in the solution is greater than that of hydroxyl ions, which makes the aqueous solution acidic.
Third, strong alkali and weak acid salt, we call the weak acid part weak yin. Similarly, weak yin holds hydrogen ions ionized from water, so that the concentration of hydroxide ions in the solution is greater than that of hydrogen ions, making the solution alkaline.
Four, weak acid and weak base salt, weak acid part holds hydrogen, weak cation part holds hydroxide to generate two kinds of weak electrolytes, and then compare their ionization constants Ka and Kb (not the degree of hydrolysis). At a temperature, the ionization constant (also called ionization equilibrium constant) of weak electrolyte is a constant value. This comparison shows what salt is, who is strong and whose ionization constant is 65438. In a word, the reaction of anion and cation in salt solution ionizing hydrogen ions or hydroxide ions in water to form weak electrolyte is called salt hydrolysis.
The weaker the acidity (or alkali) of weak acid (or alkali), the stronger the hydrolysis tendency of weak acid radical (or weak alkali ion).
For example, the hydrolysis tendency of sodium borate is stronger than that of sodium acetate, and the pH value of the former is higher under the same solution concentration.
The acidity and alkalinity of weak acid and weak base salt solution depends on the hydrolysis tendency of weak acid radical and weak base ion.
For example, the hydrolysis tendency of weak acid radicals in ammonium bicarbonate is stronger than that of weak base ions, and the solution is alkaline;
Weak alkali ions in ammonium fluoride have a strong tendency to hydrolyze, and the solution is acidic.
If they have the same hydrolysis tendency, the solution is neutral, which is an individual case, such as ammonium acetate.
Compared with the hydrolysis of corresponding strong acid weak base salt or strong base weak acid salt,
The degree of hydrolysis of weak acid and weak base salt is large, and the pH value of the solution is closer to 7 (room temperature).
For example, the hydrolysis degree of 0. 10 mol/L Na2CO3 is 4.2%, and the pH value is 1 1.6.
However, the same concentration of (NH4)2CO3 has a hydrolysis degree of 92% and a pH value of 9.3.
Some hydrolyzable salts are used as acids (such as aluminum sulfate) or bases (such as sodium carbonate).
Analysis of Chemical Classification of College Entrance Examination (12) —— Hydrolysis of Salt
● Interpretation of test sites
1. Understand the principle of brine hydrolysis, and master the law and application of brine hydrolysis.
2. Knowing the pH of the salt solution will compare the ion concentration in the salt solution.
● Preparation of test questions
(1) multiple choice questions
1.(Spring 2003, 10) The following are the pH ranges of several acid-base indicators: ① methyl orange 3. 1~4.4 ② methyl red 4.4~6.2 ③ phenolphthalein 8.2~ 10, now 0.1. When L- 1 NaOH solution titrates formic acid with similar concentration, the above indicator
A.both will do. B. only ③ can be used.
C.you can use ① or ② d. You can use ② or ③.
2. (National Synthesis in 2002, 10) At room temperature, formic acid and sodium hydroxide solution are mixed to obtain a solution with pH=7, so in this solution,
Ac (HCOO-) > AC (Na+) AC (HCOO-)c(H2CO3) D.c (sodium+) AC (Cl-) > AC (H+) > AC (CH3COOH)
BC (CH3COO-) > C (Cl-) > C (CH3COOH) > C (H+)
c . c(ch 3c oo-)= c(Cl-)> c(H+)> c(ch 3c ooh)
d . c(Na+)+c(H+)= c(ch 3c oo-)+c(Cl-)+c(OH-)
5. (nationwide in 2000, 10) At room temperature, in the solution with pH = 12, the C (OH-) precipitated by water is
A. 1.0× 10-7 mol? l- 1 b . 1.0× 10-6mol？ L- 1
C. 1.0× 10-2 mol? l- 1d . 1.0× 10- 12mol？ L- 1
6. (Spring 2000, 5)0. 1 mol? L- 1 Na2CO3 and 0. 1 mol? The pH value of L- 1 sodium bicarbonate solution, and the relationship between the two quantities is
A. greater than B. be equal to
C. less than D. uncertain
7. (In the spring of 2000, 18) the concentrations were all 0. 1 mol? After L- 1 formic acid and sodium hydroxide solution are mixed in equal volume, the following relationship is correct.
Ac (sodium ion) > AC (bicarbonate-) > AC (hydroxide-) > AC (hydrogen ion)
Ammonium bicarbonate (HCOO-) > ammonium bicarbonate (Na+) > ammonium bicarbonate (OH-) > ammonium bicarbonate (H+)
C.c(HCOO-)=c(Na+)>c(H+)=c(OH-)
D.c(Na+)=c(HCOO-)>c(OH-)>c(H+)
8. (Shanghai, 2000, 16) The following groups of ions can coexist in large quantities in solution.
A. calcium ion, HCO, chloride ion and potassium ion
B.Al3+、AlO、HCO、Na+
C.Fe2+、NH、SO、S2-
D.Fe3+、SCN-、Na-、CO
9. (Guangdong,17,2000) Adding a certain amount of NaOH solution to formic acid solution can completely react. For the generated solution, the following judgment must be correct.
Communication (HCOO-) communication (HCOO-) > communication (Na+)
c . c(OH-)> c(HCOO-)d . c(OH-)< c(HCOO-)
10.( 1999 Shanghai, 10) In an evaporating dish, heat the solution of the following substances to dryness, and then burn it (below 400℃) to obtain the solid of the substance.
A. Aluminum chloride B. Sodium bicarbonate
C. magnesium sulfate and potassium permanganate
1 1.( 1999 Shanghai, 22) put 0.02 mol? L- 1ch3cooh solution and 0.0 1 mol? When NaOH solution of L- 1 is mixed in equal volume, the correct relationship of particle concentration in the mixed solution is as follows
Communication (CH3COO-) > communication (Na+)
BC (CH3COOH) > BC (CH3COO-)
C.2c(H+)=c(CH3COO-)-c(CH3COOH)
d . c(ch 3c ooh)+c(ch 3c oo-)= 0.0 1mol？ L- 1
12. (National 1998, 1 1) After mixing Mohs strong alkali solution and Hartmann weak acid solution, the concentration of related ions in the mixed solution should satisfy the following relationship.
Ac (M+) > AC (OH-) > AC (A-) > AC (H+)
B.c(M+)>c(A-)>c(H+)>c(OH-)
C.c(M+)>c(A-)>c(OH-)>c(H+)
D.c(M+)+c(H+)=c(A-)+c(OH-)
13.( 1998 Shanghai, 16) The following groups of ions can coexist in large quantities in aqueous solution.
A. sodium ions, anions, copper ions and chloride ions
B. Anions, sodium ions, hydroxyl ions and potassium ions
C. Potassium, cobalt, bromine and aluminum
D. Hydrogen ion, chloride ion, sodium ion and sulfate ion
14. (National 1997, 17)0. 1 mol After fully stirring in 1 liter of water, the most anions in the solution are
A.KCl magnesium hydroxide
C. sodium carbonate and magnesium sulfate
15. (National 1997, 14)0. 1 mol? L- 1 NaOH and 0. 1 mol? After l- 1nh4cl solution is mixed in equal volume, the correct ion concentration order is as follows
Alternating current (sodium ion) > alternating current (chloride ion) > alternating current (hydrogen ion) > alternating current (hydrogen ion)
B.c(Na+)=c(Cl-)>c(OH-)>c(H+)
C.c(Na+)=c(Cl-)>c(H+)>c(OH-)
Direct current (Cl-) > carbon (Na+) > carbon (OH-) > carbon (H+)
16. (National1997,9) The following groups of ions can coexist in large quantities in strong alkali solution.
A. Iodine, Allo, Chlorine, S2
B. sodium ion, potassium ion, ammonia ion and barium ion
C. Bromine, S2, chlorine and cobalt
D. So, no, so, HCO
17.(65438+Shanghai, 0997, 22) was set in NaOH and CH3COONa solution with pH equal to 9, and the concentration of OH- generated by water ionization was mol respectively. L- 1 and b mol? L- 1, then the relationship between a and b is
A.a & gtb B.a= 10-4b
C.b= 10-4a D.a=b
18.( 1996 nationwide, 8) The ferric chloride solution is evaporated and burned, and the final solid product is
A. anhydrous ferric chloride and ferric hydroxide
C. iron oxide
19. (National1996,25) After the hydrochloric acid solution with 25)pH = 3 and ammonia water with pH= 1 1 are mixed in equal volumes, the relationship between ion concentrations in the solution is correct.
Ammonia > chlorine > hydrogen > hydrogen
Ammonium bicarbonate > ammonium bicarbonate > ammonium bicarbonate > ammonium bicarbonate
C.c(Cl-)>c(NH )>c(H+)>c(OH-)
Direct current (Cl-) > ammonium bicarbonate > ammonium bicarbonate (OH-) > ammonium bicarbonate (H+)
20.( 1996, Shanghai, 14) In the solution with the same quantity and concentration of the following substances, which contains the most kinds of particles?
A. calcium chloride B.CH3COONa
C.NH3 D.K2S
2 1.( 1996 Shanghai, 2 1), the largest concentration of NH is
A.NH4Cl B.NH4HSO4
C.CH3COONH4 D.NH4HCO3
22.( 1996 Shanghai, 22) In FeCl3 solution with pH 4 and K2CO3 solution with pH 10, the ionization degrees of water are α 1 and α2, respectively, and their relationships are as follows.
A.α 1>α2 B.α 1 c (na+), then the following relationship is correct.
Ac (hydrogen+) > AC (hydrogen-)
B.c(H+)c(CH3COO-)
d . c(ch 3c ooh)+c(ch 3c oo-)= 0.02mol？ L- 1
24. (National 1995, 15) The following substances can react with magnesium to generate hydrogen.
A. Formic acid solution B. Sodium hydroxide solution
C. ammonium chloride solution D. sodium carbonate solution
25. (National 1995,10)100 ml 0.1mol? L- 1 acetic acid and 50ml 0.2mol? L- 1 sodium hydroxide solution, in the obtained solution
Ac (sodium ion) > AC (acetic acid) > AC (hydroxide) > AC (hydroxide)
b . c(Na+)> c(ch 3c oo-)> c(H+)> c(OH-)
c . c(Na+)> c(ch 3c oo-)> c(H+)= c(OH-)
d . c(Na+)= c(ch 3c oo-)> c(OH-)> c(H+)
26. (National 1995, 1 1) In a solution with pH= 1, a large number of ionic groups can coexist.
A. sodium ion, potassium ion, S2 ion and chloride ion
B. aluminum ions, magnesium ions, sulfate ions and chloride ions
C. potassium ion, sodium ion, AlO, NO
D. Potassium, sodium, sulfur and sulfur oxide
27.( 1995, Shanghai, 2 1) The pH of the dilute solution prepared by mixing the amounts of CH3COOH and CH3COONa is 4.7. The following statement is wrong.
The ionization of A.A.CH3COOH is greater than the hydrolysis of CH3COONa.
The hydrolysis of B.B.CH3COONa is greater than the ionization of CH3COOH.
The existence of C.CH3COOH inhibited the hydrolysis of CH3COONa.
The existence of D.CH3COONa inhibits the ionization of CH3COOH.
28.( 1995 Shanghai, 20) After adding alkali MOH to HA solution, the solution was neutral. The following judgment is correct.
A. excessive addition of alkali
B. The amount of solute in acid and alkali is equal before mixing.
C, the generated salt does not hydrolyze.
D, after the reaction, the amount and concentration of A- and M+ substances in the solution are equal.
29. (National 1994, National 1 1) In ammonium chloride solution, the following relationship is correct.
Ac (chlorine-) > AC (ammonia) > AC (hydrogen+) > AC (hydroxyl)
Ammonium bicarbonate > ammonium bicarbonate > ammonium bicarbonate > ammonium bicarbonate > ammonium bicarbonate > ammonium bicarbonate
C.c(Cl-)=c(NH )>c(H+)=c(OH-)
D.c(NH )=c(Cl-)>c(H+)>c(OH-)
30. (National 1994, 13) The ionic groups that cannot coexist in large quantities in a colorless and transparent solution with pH= 1
A. aluminum ion, silver ion, nitric oxide and chloride ion
B.Mg2+、NH、NO、Cl-
C. Barium, potassium, S2 and chlorine
D.Zn2+、Na+、NO、SO
3 1.( 1994 Shanghai, 10) Heating, dropwise adding saturated sodium acetate solution of methyl orange, the color of the solution is
A. the yellow color remains unchanged. B. Yellow becomes orange
C. Yellow turns red. Orange remains the same.
32.( 1994 Shanghai, 20) In order to prepare a solution with the ratio of NH concentration to Cl- concentration of 1: 1, an appropriate amount of HCl, NaCl, ammonia water and NaOH can be added to the NH4Cl solution.
A.①② B.③ C.③④ D.④
33.( 1994 Shanghai, 1 1) The relationship between the number and concentration of various ions in alum solution can be determined as follows.
Ac potassium ion > AC aluminum ion
C.c(H+)=c(OH-) D.c(Al3+) 2C (Co), so A is wrong; According to the conservation of matter, C (Na+) =
2 [c (co)+c (HCO)+c (H2CO3)], so d is also wrong; Because the solution is alkaline, C (OH-) should be greater than.
2c (hydrogen +).C (HCO) > C (H2CO3) and C are correct, because the secondary hydrolysis of CO +H2O HCO +OH-and C(HCO)> C(h2co 3+OH- is very weak.
4. answer: BD
Analysis: After the two solutions are mixed, the solution is essentially a mixed solution of acetic acid, sodium acetate and sodium chloride with the same concentration. Because the solution is acidic, it shows that the degree of hydrolysis of Ch3coo- is less than the ionization degree of CH3COOH, so C (Ch3coo-) > C (Cl-) > C (CH3COOH), but the ionization degree of CH3COOH is smaller, and C (Ch3cooh) > C (h+), so options A and C are wrong and B is correct. According to the principle of charge conservation, option D is also correct.
5. answer: CD
Analysis: The solution with pH = 12 is alkaline, in which the total C (H+) is =1.0×10-12mol? If L- 1 is an aqueous solution of alkali and H+ is completely ionized by water, then ionized C (H+) water is equal to C (OH-) water, so D is correct. If it is a saline solution,
When OH- is completely ionized by water, then C (OH-) water = 1× 10-2 mol? L- 1, c is correct.
6. a: a
Analysis: Na2CO3 solution with the same concentration is more hydrolyzed than NaHCO3 solution, and the solution is more alkaline, so the pH value of the former is higher.
7. A: A.
Analysis: The two solutions reacted completely to obtain sodium formate solution. Due to the hydrolysis of formate, the concentration of HCOO-is less than that of Na+, and the solution is alkaline, so A should be selected.
8. A: A.
Analysis: Al3 ++ and AlO in group B will undergo double hydrolysis reaction and cannot coexist; In group C, Fe2+ and S2 formed FeS precipitate and could not coexist. In group D, Fe3+ and SCN-can't coexist because of complexation reaction, and Fe3+ and CO can't coexist because of double hydrolysis reaction.
9. Answer: AD
Analysis: Sodium formate solution is obtained after the complete reaction of the two solutions. Due to the hydrolysis of formate, the concentration of HCOO- is less than that of Na+, and the solution is alkaline, but the degree of hydrolysis of HCOO- is not great, and the concentration of unhydrolyzed HCOO- in the solution is still greater than the concentration of OH- generated by hydrolysis.
10. answer: C.
Analysis: This question examines the thermal stability of solutions and solids of several substances. AlCl3 is easy to hydrolyze, only Al2O3 is decomposed from Al(OH)3, and its solution can be obtained after heating and evaporation. Sodium bicarbonate solid and potassium permanganate solid will decompose when heated, and only magnesium sulfate is the most stable.
1 1. answer: AD
Analysis: 0.02 mol? L- 1ch3cooh and 0.0 1 mol? After L- 1 NaOH are mixed in equal volume, the solutes in the mixed solution are CH3COOH and CH3COONa, and the amounts of these two substances are equal. Because the ionization trend of CH3COOH is greater than Ch3C OO-, c (ch3coo-) > c (na+); But no matter how it changes, we can know from the conservation of matter that c (Ch3C OOH)+c (Ch3C OO-) = 0.01mol? L- 1 .
65438+
Analysis: Because both of them are unitary, they can completely react to generate strong base and weak acid salt MA, and A- hydrolysis is alkaline, so C is correct; The solution is electrically neutral, so d is correct.
13. Answer: C.
Analysis: HS-+Cu2+= = CUS ↓+H+, HS-+OH-= = = S2-+H2O, 2H++SO = = H2O+
SO2↑ = =, so A, B and D cannot coexist in large quantities.
14. answer: C.
Analysis: Mg(OH)2 is insoluble, so the number of anions is the least; CO2-3 has the following balance in solution: CO +H2O HCO +OH-,which increases the number of anions.
15. Answer: B.
Analysis: The reaction is complete after mixing, and the products are NaCl and NH3? H2O is alkaline, so it is B. ..
16. Answer: AC
Analysis: NH+OH-= = = NH3? H2O HCO+OH-= = Co+H2O, so B and D can't coexist in strong alkali solution.
17. Answer: B.
Analysis: C (H+) = 10-9 mol in NaOH solution? L- 1, then c (oh-) produced by water ionization =10-9 mol? l- 1； C (H+) = 10-9 mol in CH3COONa solution? L- 1，c(OH-)= 10-5 mol？ L- 1, this OH- is produced by water ionization, so choose B.
18. answer: D.
Analysis: FeCl _ 3 has a high degree of hydrolysis, FeCl _ 3+3H2O Fe (OH) _ 3+3HCl, and the hydrolysis reaction absorbs heat, so the hydrolysis equilibrium moves forward when heated. Fe(OH)3 solid can be obtained after the solution is evaporated, and Fe(OH)3 decomposes to generate Fe2O3 during combustion, so the final product is Fe2O3.
19. Answer: B.
Analysis: The concentration of hydrochloric acid with pH=3 is 10-3 mol? The ammonia concentration of L- 1 and pH= 1 1 is about 0. 1 mol?
L- 1 is much greater than the concentration of hydrochloric acid, so there is too much ammonia water after equal volume mixing, and some of it is in the mixed solution.
NH4Cl===NH +Cl- ①
NH3？ H2O (excess) NH+OH-②
(1) where c(NH) = C (Cl-), and c(NH) in solution is always the sum of ionized c(NH) in two formulas, so C (NH) > C (Cl-), and the excessive ammonia solution is alkaline, so C (OH-) > C (H+).
20. Answer: D.
Analysis: Particles in solution include molecules and ions. Ionization and hydrolysis should be considered in this problem. S2-in D is hydrolyzed in two steps. The molecules in the solution are H2O and H2S, and the ions are K+, S2-, HS-, OH- and H+, which are seven kinds of particles, with the most kinds.
2 1. answer: B.
Analysis: NH has the following equilibrium in solution: NH +H2O NH3? H2O+H+ and NH4HSO4 in B ionize a lot of H+, which makes the above equilibrium move to the left, so c(NH) in B is greater than C (NH) in A; The CH3COO-ionized by CH3COONH4 in item C and the HCO hydrolyzed by item D are alkaline, which moves the above equilibrium to the right, so the c(NH) in C and D is smaller than that in A. ..
22. answer: C.
Analysis: C (H+) = 10-4 mol in FeCl3 solution? L- 1, C (OH-) = 10-4 mol in K2CO3 solution? L- 1, both of which are separated by water and electricity, and the values are equal, indicating that the ionization degree of water is equal.
23. Answer: AD
Analysis: There are two equilibria in the solution:
CH3COOH CH3COO-+H+ ①
CH3COO-+H2O CH3COOH+OH- ②
The ions in the solution are Na+, Ch3coo-, H+ and OH-, and C (Na+)+C (H+) = C (Ch3coo-)+C (OH-) can be obtained from neutral solution. Because c (ch3coo-) > c (na+) is known, we can get c (h+) > c (oh-), and a is correct. It is concluded that the balance ① is dominant, and C (Ch3C OO-) > C (Ch3C OOH), and c is incorrect; According to the balance ① ②, the sum of c(CH3COOH) and C (Ch3C OO-) is a constant value, that is, 0.02 mol? L- 1, so d is correct.
24. Answer: AC
Analysis: Because magnesium is active, it can react with formic acid, even with hydrolyzed acidic NH4Cl solution to generate H2.
25. A: A.
Analysis: Simply calculate that both acetic acid and sodium hydroxide are 0.0 1 mol, and completely react to generate 0.0 1 mol sodium acetate. While sodium acetate is completely ionized, and ch3coona = = ch3coo-+na+, in which ch3coo- is hydrolyzed to alkaline, so c (na+) > c (ch3coo-), c (oh-) > c (h+). However, C (Ch3C OO-) > C (OH-) is chosen because of its low degree of hydrolysis.
26. answer: B.
Analysis: The solution with pH= 1 is strongly acidic, so both weak acid radical ions S2- and AlO react with h+, and the reaction between S2O and H+ is as follows: 2 h++S2O = = H2O+S+SO2 ↑.
27. answer: B.
Analysis: the following equilibrium exists in the mixed solution: (ignoring the ionization of water)
CH3COOH H++CH3COO- ①
CH3COO-+H2O OH-+CH3COOH ②
Because the amount of substances is mixed and the solution is acidic, it can be judged that A is correct and B is incorrect. Because of the small degree of ionization and hydrolysis, acetic acid molecules mainly exist in acetic acid, which has an inhibitory effect on formula 2; In sodium acetate, the main component is CH3COO-,which can inhibit the formula (1). Therefore, c and d are correct.
28. answer: D.
Analysis: The key is that MOH can be a strong base or a weak base. If it is a strong alkali, it will completely react with HA, and item A is incorrect; If it is weak alkali, the generated brine is hydrolyzed, and the alkali must be excessive, so B and C are incorrect. So option d is correct. In item D, regardless of the strength of MOH, there are C (H+)+C (M+) = C (OH-)+C (A-) (according to the conservation of charge), and the solution is neutral after the reaction, so C (H+) = C (OH-), so C (M+) = C (A-).
29. A: A.
Analysis: Ammonium chloride is easily soluble in water, and completely ionized in solution to generate NH and Cl-. NH is a weak base cation, which is partially hydrolyzed to generate weak base and H+, and the solution is acidic.
30. Answer: AC
Analysis: Regarding whether ions can coexist in large quantities, we should pay attention to the following conditions: ① exclude colored ones to be colorless and transparent; ② exclude groups that cannot exist under acidic conditions to be pH= 1; and ③ exclude groups that cannot coexist. Ag+ and Cl- in A cannot coexist, and S2- and H+ in C cannot coexist.
3 1. answer: a.
Analysis: Due to partial hydrolysis, sodium acetate aqueous solution is alkaline. According to the discoloration range of methyl orange, the saturated solution of methyl orange with sodium acetate drops is yellow. Because the hydrolysis process absorbs heat, heating enhances the alkalinity of the solution, so the yellow color of the solution remains unchanged.
32. Answer: B
Analysis: C (NH) ∶ C (Cl-) < 1 ∶ 1 in NH4Cl solution is partially hydrolyzed by NH, that is,
NH +H2O NH3？ H2O+H+
C (NH) ∶ C (Cl-) = 1 ∶ 1 can be achieved by inhibiting hydrolysis. It is best to add an appropriate amount of NH3. H2O inhibits the hydrolysis of NH. ① Although it can inhibit hydrolysis, increasing Cl-② increases Cl-, introduces Na+④ to promote hydrolysis, and also introduces Na+.
33. A: A.
Analysis: Is alum KAl(SO4)2? 12H2O, kal (SO4) 2 = = k++Al3++2SO in water; However, Al3+ partially hydrolyzes Al3++3H2O Al (OH) 3+3H+, and the solution is acidic. So c (k+) > c (Al3+), c (h+) >
c(OH-)，c(Al3+)>c(H+).
(2) Write an answer.
34. The phenomenon is white precipitation and red fading.
Reason: In soda solution, CO is hydrolyzed, and CO +H2O HCO +OH-,the solution is alkaline, and the solution is red after phenolphthalein is added; After adding BaCl2, Ba2++Co = = baco3 ↓ (white). With the decrease of c(CO), the hydrolysis equilibrium moves to the left, C (OH-) decreases, and phenolphthalein fades.
● Proposition tendency and test-taking strategy
(A) pay attention to the formation of the basis of knowledge laws.
1. Salt water solution law
(1) whoever is weak will hydrolyze, and whoever is strong will show his true nature. The weaker it is, the more hydrolyzed it is. Both weak ones are hydrolyzed, and neither strong one is hydrolyzed.
(2) The weaker the acid (or alkali) corresponding to the salt, the greater the degree of hydrolysis and the stronger the alkalinity (or acidity) of the solution.
(3) The degree of hydrolysis of polybasic weak acid radical and positive acid radical ions is much greater than that of acid radical ions. For example, the degree of hydrolysis of carbon monoxide is several orders of magnitude greater than that of HCO, and the solution is more alkaline.
2. Ion concentration comparison method
Determination of (1) ion concentration
When judging the ion concentration in the hydrolyzable salt solution, we should first make clear that the salt has strong ionization and weak hydrolysis, and then make clear that the hydrolysis of polybasic weak acid salt is carried out step by step, mainly in the first step, and finally don't forget the ionization of water.
① Multi-weak acid solution was analyzed by multi-step ionization. For example, in H3PO4 solution, C (H+) > C (H2PO) > C (HPO) > C (PO).
② According to the gradual hydrolysis of weak acid radical, the normal salt solution of polybasic weak acid was analyzed. For example, in Na2CO3 solution, C (Na+) > C (Co) > C (OH-) > C (HCO).
③ The comparison of the same ion concentration in different solutions depends on the factors that other ions in the solution affect it. For example, in the solution with the same concentration of the following substances: ①NH4Cl, ②CH3COONH4, ③ NH4SO4. The order of c(NH) from large to small is ③ > ① > ②.
(4) The comparison of ion concentrations in mixed solutions should be comprehensively analyzed, such as ionization factors and hydrolysis factors. For example, at 0. 1 mol? L- 1 NH4Cl and 0. 1 mol? In the ammonia mixed solution of L- 1, the order of each ion concentration is C (NH) > C (Cl-) > C (OH-) > C (H+). In this solution, NH3? Ionization of H2O and hydrolysis of NH+4 inhibit each other, NH3? When the ionization factor of H2O is greater than the hydrolysis of NH, the solution is alkaline: C (OH-) > C (H+), C (NH) > C (Cl-).
(2) Law of conservation of charge
No matter how many ions exist in electrolyte solution, the solution is always electrically neutral, that is, the total number of negative charges carried by anions must be equal to the total number of positive charges carried by cations, which is the so-called law of charge conservation. Such as sodium bicarbonate solution.
Na+, H+, HCO, CO, OH-, but they have the following relations:
C(Na+)+c(H+)=c(HCO )+c(OH-)+2c (CO)
(3) Law of Conservation of Matter
In electrolyte solution, because some ions can be hydrolyzed, the types of ions increase, but some key atoms are always conserved. For example, S2- and HS- in K2S solution can be hydrolyzed, so the element S exists in three forms: S2-, HS- and H2S, and there is the following conservation relationship between them:
c(K+)=2c(S2-)+2c(HS-)+2c(H2S).
(B) analysis of hot topics, grasp the trend of proposition
This part is closely related to daily life, agricultural production and laboratory work, so it is the focus of the college entrance examination. Every year, the hot spot is (1) the nature and law of salt hydrolysis, (2) the chemical equation of hydrolysis and ionic equation can be used to represent salt hydrolysis; (3) The knowledge of salt water hydrolysis will be used to explain the phenomena in life and solve some practical problems in production. Such as ① the principle of foam fire extinguishing reaction, ② the use of compound fertilizer, ③ the preparation of FeCl3 _ 3, SnCl2 _ 2 and other solutions in the laboratory, ④ the principle of alum water purification and Na2FeO4 _ 4 water purification, and ⑤ the comparison of particle concentrations in the solutions. Generally in the form of multiple-choice questions or short-answer questions.
Judging from the changing trend of college entrance examination questions, this part mainly applies basic knowledge to solve related problems in production and life, and examines students' application analysis ability. Judging the coexistence of ions, preparing solutions, storing reagents, mixing fertilizers, identifying substances, and separating and removing impurities in some salts are also involved in the college entrance examination. The hot topic of the proposition is the comparison of ion concentrations. In the college entrance examination questions, especially the multiple-choice questions, saline hydrolysis often combines the knowledge of weak electrolyte ionization, acid-base neutralization titration, pH and so on, which has certain comprehensiveness. Looking forward to the future, the test questions will still emphasize the integration of knowledge points, such as the hydrolysis balance of salt and the ionization balance of water, the relationship between the principle of charge conservation and the principle of matter conservation, and physical knowledge. Generally, simple questions are not designed on a single knowledge point, and the difficulty of the questions is moderate.
[Example] (Shanghai, 20, 2002) At room temperature, in the10/0 ml KOH solution with pH=4, the monobasic acid HA solution was added until the pH was just equal to 7 (assuming that the volume was unchanged before and after the reaction), and the description of the solution after the reaction was correct.
A . c(A-)= c(K+)b . c(H+)= c(OH-)< c(K+)< c(A+)
C the total amount of intravenous injection is ≥ 20ml, and the total amount of intravenous injection is ≤ 20ml.
Analysis: This topic examines the analysis of the knowledge about the neutral reaction of two solutions in which the pH of monobasic acid is equal to the pOH of strong alkali. When the amount and concentration of monobasic acid and alkali are the same, it is necessary to distinguish the mixed reactions. Attention should be paid to the application of ion charge conservation when analyzing the concentration relationship between anion and cation in solution.
There is a charge conservation relationship between the concentrations of the four ions in the solution: c (k+)+c (h+) = c (oh-)+c (a-) Because pH=7 means C (H+) = C (OH-) in the solution, there are four ions in the mixed solution after the reaction, so c (a-) = c (k+). If HA is a strong acid, the total V is equal to 20 ml; If HA is a weak acid, the concentration of acid is greater than that of alkali, and the total V is less than 20 ml.
Answer: AD