Prerequisite knowledge point 4: Quantity-related ionic reactions.
One. Continuous reaction.
1. Reaction characteristics: The ions generated by the reaction can continue to react with excess reactants, so the ion equation is related to the amount used.
2 Writing steps and methods.
First of all, the excess reactant is re-reacted with the ions generated by the reaction, and the two reactions are added.
2. Proportional reaction:
When there are two or more constituent ions in a reactant to participate in the reaction, because of its different composition proportions (generally double salt or acid salt), when one constituent ion happens to be completely reacted, the other constituent ion can not be completely reacted (there is residual or insufficient) and is related to the dosage.
1) "Less" is to set the relatively small amount of substances as "1 mol", and in the addition of a small amount of substances, the ions that can react participate in the reaction, 2) "Variable" is the excess reactants, and the stoichiometric number of the ions is determined according to the actual demand of the reaction, which is not subject to the proportion in the chemical formula, and is variable. Such as the reaction of a small amount of NaHCO3 with a sufficient amount of Ca(OH)2 solution:
Less" - that is, the amount of HCO3- is determined to be 1 mol, changeable" - 1 mol HCO3- can react with 1 mol OH to obtain 1 mol H2O and 1 mol CO CO32- and then with 1 such as NaHSO4 solution and BA(OH)2 solution.
When the solution is neutral: 2H SO42- 2OH BA2 ===BA4 2H2O;
When SO42- is completely precipitated: H SO42- Ba2 OH ===H2O BaSO4.
3. Sequential reaction: two or more constituent ions of one reactant can react with the constituent ions of another reactant, but it is related to the dosage due to the different reaction sequences. It can also be called competitive.
1.Alkaline solution plus acid (H).
To the solution containing Na, OH, CO32-, ALO2-, hydrochloric acid is added drop by drop, because of the ability to bind protons: OH ALO2- CO32-, so the sequence of the reaction is:
h＋＋oh－===h2o
h＋＋alo2-＋h2o===al(oh)3↓
co32-＋h＋===hco3-
hco3-＋h＋===co2↑＋h2o
Finally, the Al(OH)3 precipitate is generated and further reacts with H: Al(OH)3H ===Al3 3H2O
2.Acidic solution plus alkali (OH).
To the solution containing Cl-, H+, NH4+, Al3+, sodium hydroxide solution is added drop by drop, because of the ability to react with hydroxide: H+ Al3+ NH4+, so the sequence of the reaction is:
h＋＋oh－===h2o
3oh－＋al3+===al(oh)3↓
nh4+＋oh－===nh3.h2o
Finally, the Al(Oh)3 precipitate is generated and further reacts with OH: Al(OH)3 OH ===ALO2-2H2O
Fourth, the ionic reaction of oxidation and reduction.
For redox reactions, it is written in the order of "strong first and then weak", that is, the reaction of the strong oxidation (or reduction) is preferential, and the reaction of the weak oxidation (or reduction) occurs later, and the writing steps of this type of ion equation are as follows:
Step 1: Determine the sequence of reactions: (oxidation: Hno3 Fe3, Reducibility: I Fe2 br). If Cl2 is introduced into the FEI2 solution, I reacts with Cl2 first.
Step 2: Judge the degree of reaction according to the dosage, such as when a small amount of Cl2 reacts with FEI2 solution, only I reacts with Cl2: 2i Cl2===2Cl I2.
When a sufficient amount of Cl2 reacts with FEI2 solution, both I and Fe2 in the solution react with Cl2:
2fe2＋＋4i－＋3cl2===2fe3＋＋2i2＋6cl－。
Step 3: Write the ion equation using the "small amount of fixed 1 method", that is, set the stoichiometric number of the "small" substance as "1" for writing.
Knowledge comprehension points].
1.Soluble poly-weak acids (or their anhydrides) react with alkaline solutions. If CO2 is passed into the NaOH solution, it becomes carbonate and then regenerates acid salt
Alkali excess (small amount of CO2): CO22OH ===CO32- H2O;
Insufficient alkali (excess CO2): CO2 OH ===HCO3-.
2.A salt solution of multiple weak acids (or their anhydrides) with weaker acids. If CO2 is introduced into Naalo2 solution, it becomes carbonate and then acid salt
Naalo2 excess (small amount of CO2): 2alo2-CO2 3H2O===2Al(OH)3CO32-;
Naalo2 insufficiency (CO2 excess): AlO2-CO2 2H2O===Al(OH)3 HCO3-.
3.Reaction of multiple weak salts with strong acids. For example, Na2CO3 solution and dilute hydrochloric acid react first to form acid salts, and then to generate carbon dioxide
Hydrochloric acid insufficiency: CO32- h ===HCO3-;
Hydrochloric acid excess: CO32- 2H ===CO2 H2O.
4.Aluminum salt solution and strong alkali solution, such as dropping strong alkali into aluminum salt, Mr. aluminum hydroxide precipitation, and then dissolved to form metaaluminate:
Excess aluminium salts (small amounts of NaOH): Al3OH===AL(OH)3;
Strong base overdose (NaOH overdose): Al3 4OH ===alo2- 2H2O.
5.Naalo2 solution with a strong acid solution, drop the acid in metaaluminate, store aluminum hydroxide, and then dissolve to generate aluminum ions: Naalo2 excess: AlO2- H2O===Al(OH)3;
Excess of strong acid: AlO2-4H ===Al3 2H2O.
6.Fe with dilute Hno3 solution, gradually add iron to nitric acid, Mr. save trivalent iron, iron excess, generate divalent iron:
Fe excess: 3Fe 2NO3- 8H ===3Fe2 2NO4H2O;
HNO3 excess: Fe NO3- 4H ===Fe3 NO2H2O.
Benchmarking case analysis and tracking drill].
Basic example] Quantity-related ion equations.
Example: 1Sodium phosphate solution with hydrochloric acid.
1) Press 1:1 to join:
2) Press 1:2 to join:
3) Press 1:3 to join:
2.Aluminum ammonium sulfate solution with sodium hydroxide solution.
1) Press 1:3 to join:
2) Press 1:4 to join:
3) Press 1:5 to join:
3.Barium hydroxide solution with sodium bisulfite solution.
1) Barium hydroxide excess:
2) A small amount of barium hydroxide:
4.Ammonium iron sulfate solution with. Barium hydroxide solution.
1) Press 2:3 to join:
2) Press 1:2 to join:
Follow-up drill] 1. Continuous response.
1.SO2 is passed into NaOH solution:
SO2 in small amounts:;
SO2 overdose:.
2.Na2CO3 solution with dilute sulfuric acid:
Sulfuric acid insufficiency:;
Sulfuric acid overdose:.
3.The Al2(SO4)3 solution reacts with the NaOH solution.
Al2(SO4)3 overdose:;
NAOH overdose:.
4.Fe reacts with dilute Hno3.
Fe overdose:;
Hno3 overdose:.
Second, the response sequence.
1.NH4HSO4 solution reacts with NaOH solution.
NAOH Deficiencies:;
NAOH overdose:.
2.Mix Fe(NO3)3 solution with Hi solution.
Hi insufficient:;
Hi overdose:.
3.Cl2 is introduced into the Febr2 solution.
Cl2 in small amounts:;
Amount of Febr2 and Cl2: ;
Cl2 sufficient:.
4.To the solution containing OH, CO32-, ALO2-, dilute hydrochloric acid is added drop by drop to excess, and the ion equation of the reaction is as follows:
5.To the solution containing H, Al3 and NH4+, NaOH solution was added drop by drop to the excess, and the ion equation of the reaction was as follows:
3. Substance proportioning type.
1.Ca(HCO3)2 solution reacts with NaOH solution.
NAOH Deficiencies:;
NAOH overdose:.
2.The NaHCO3 solution reacts with the Ca(OH)2 solution.
NaHCO3 insufficiency:
NaHCO3 overdose:
3.The Ba(OH)2 solution reacts with the NaHSO4 solution.
n[ba(oh)2]∶n(nahso4)＝1∶1
At this point, the solution is sexual;__
n[ba(oh)2]∶n(nahso4)＝1∶2
At this time, the solution is sexual, if the Ba(OH)2 solution is added to the solution, the ion equation is
Prerequisite knowledge point 4: Quantity-related ionic reactions.
Benchmarking case analysis and tracking drill].
Basic example] Quantity-related ion equations.
Example: Answer] 1（1）h++po43-=hpo42-
2）2h++po43-===h2po4-
3）3h++po43-===h3po4
2.（1）3oh－＋al3+===al(oh)3↓
2）al3＋nh4+＋4oh－===al(oh)3↓+nh3.h2o
3）al3＋nh4+＋5oh－===alo2-+nh3.h2o+2h2o
3.（1）ba2＋＋oh－＋hco3-===baco3↓＋h2o
2）ba2＋＋2oh－＋2hco3-===baco3↓＋h2o+co32-
4.（1）2fe3+＋3so42-＋6oh－＋3ba2＋===2fe(oh)3↓＋3baso4↓
2）fe3+＋2so42-＋4oh－＋2ba2＋===fe(oh)3↓＋2baso4↓+nh3.h2o
Follow-up drill] 1. Continuous response.
Answer: 1 ①so2＋2oh－===so32-＋h2o
so2＋oh－===hso3-
2.①co32-＋h＋===hco3-
co32-＋2h＋===co2↑＋h2o
3.①al3＋＋3oh－===al(oh)3↓
al3＋＋4oh－===alo2-＋2h2o
4. ①3fe＋8h＋＋2no3-===3fe2＋＋2no↑＋4h2o
fe＋4h＋＋no3-===fe3＋＋no↑＋2h2o
2. Answer]: 1①h＋＋oh－===h2o
nh4+＋h＋＋2oh－===nh3·h2o＋h2o
2.①8h＋＋2no3-＋6i－===4h2o＋3i2＋2no↑
fe3＋＋12h＋＋3no3-＋10i－===fe2＋＋5i2＋3no↑＋6h2o
3.①2fe2＋＋cl2===2fe3＋＋2cl－
2fe2＋＋2br－＋2cl2===2fe3＋＋br2＋4cl－
2fe2＋＋4br－＋3cl2===2fe3＋＋2br2＋6cl－
4.oh－＋h＋===h2o、
alo2-＋h＋＋h2o===al(oh)3↓、
co32-＋h＋===hco3-、
hco3-＋h＋===h2o＋co2↑、
al(oh)3＋3h＋===al3＋＋3h2o
5.h＋＋oh－===h2o、
al3＋＋3oh－===al(oh)3↓、
nh4+＋oh－===nh3·h2o、
al(oh)3＋oh－===alo2-＋2h2o
3. Answer]: 1①ca2＋＋hco3-＋oh－===caco3↓＋h2o
ca2＋＋2hco3-＋2oh－===caco3↓＋2h2o＋co32-
2.①hco3-＋oh－＋ca2＋===caco3↓＋h2o
ca2＋＋2oh－＋2hco3-===caco3↓＋co32-＋2h2o
3.Ba2 oh H SO42 -===Baso4 H2O base.
Ba2 2OH 2H SO42 -===BaSO4 2H2O.
so42-＋ba2＋===baso4↓