Ion exchange reactions going to completion. Ionic reaction equations

Problem 1. Calculate the concentration of hydrogen ions in the HCN solution (C m = 10 -3 M), if  = 4,2∙10 -3 .

Solution: The dissociation of hydrocyanic acid proceeds according to the equation HCN ↔ H + + CN - ; the concentrations of ions and in the solution are equal to each other (since  H+:  C N - = 1:1, where

 - stoichiometric coefficients) i.e. = =  C m, mol/l; Then = = 4.2∙10 -3 ∙ 10 -3 = 4.210 -7 mol/l.

Solution : Ammonium hydroxide dissociates as follows:

NH 4 OH ↔ NH 4 + + OH -, the dissociation constant has the form

K d =;

the concentrations of ammonium and hydroxide ions are the same (  (NH4+) :  (OH -) = 1:1), we denote them as X:

= = x mol/l , then the expression for K d will take the form

1,810 -5 = X 2 / 0,01-X. Considering that X<< С м, решаем уравнение

1.810 -5 =x 2 / 0.01, relative X: X=
=4.2∙10 -4 mol/l; = 4.2∙10 -4 mol/l.

The concentrations of hydrogen and hydroxide ions are related through the ionic product of water K w= =10 -14, let us express the concentration of hydrogen ions = K w/ and calculate its value:

110 -14 /4.210 -4 = 2.310 -11 mol/l.

Problem 3. Determine the pH of the HCl solution (  =1), if C m =2∙10 -3 M

Solution: The dissociation of hydrochloric acid proceeds according to the equation

HCl  H + + Cl - , concentration of hydrogen ions =  C m =1∙2∙10 -3 = =2∙10 -3 mol/l. Hydrogen indicator pH = - log = - log2∙10 -3 = 2.7.

Problem 4. Determine the molar concentration of ammonium hydroxide if pH=11 and Kd=1.8∙10 -5.

Solution: Concentration of hydrogen ions =10 - pH =10 -11 mol/l. From the ionic product of water we determine the concentration = K w / = 10 -14 /10 -11 =10 -3 mol/l. Ammonium hydroxide is a weak base and is characterized by the dissociation reaction equation

NH 4 OH ↔ NH 4 + + OH - . Expression for the dissociation constant

K d =.

From Ostwald's law it follows that = =  ∙C m, a TO d =  2 C m. Combining the equations, we get C m = 2 / K d = 10 -6 / 1.8∙10 -5 = 0.056 mol/l

Solubility product

Substances, depending on their nature, have different solubility in water, which ranges from fractions of a milligram to hundreds of grams per liter. Hardly soluble electrolytes form saturated solutions of very low concentrations, so we can assume that the degree of their dissociation reaches unity. Thus, a saturated solution of a sparingly soluble electrolyte is a system consisting of the solution itself, which is in equilibrium with a precipitate of the dissolved substance. Under constant external conditions, the rate of precipitate dissolution is equal to the rate of the crystallization process: K n A m ↔ n K+ m + m A- n (1)

precipitate solution

To describe this heterogeneous equilibrium process, an equilibrium constant is used, called the solubility product PR = n m, where and are the concentrations of ions in a saturated solution (mol/l). For example:

AgCl= Ag + +Cl - , PR = ; Here n=m=1.

PbI 2 = Pb 2+ +2I - , PR = 2; Here n=1, m=2.

PR depends on the nature of the dissolved substance and temperature. PR is a tabular value. Knowing PR , you can calculate the concentration of a saturated solution of a substance, and also estimate its solubility in g per 100 ml of water (value s, given in the reference literature) and determine the possibility of precipitation of the substance.

For equation (1), the relationship between the concentration of a saturated solution of a difficultly soluble substance (C m, mol/l) and the PR value is determined by the following equation:

,

Where n And m –stoichiometric coefficients in eq. 1.

Task 5. The concentration of the saturated solution (C m)Mg(OH) 2 is 1.1 10 -4 mol/l. Write down the expression for PR and calculate its value.

Solution: In a saturated solution of Mg(OH) 2, equilibrium is established between the precipitate and the solution Mg(OH) 2 ↔Mg 2+ + 2OH - , for which the PR expression has the form PR = 2 . Knowing the concentration of ions, you can find its numerical value. Given the complete dissociation

Mg(OH) 2, its concentration saturated solution C m = = 1.110 -4 mol/l, a = 2 = 2.210 -4 mol/l. Therefore, PR= 2 =1.1. 10 -4 (2.2 10 -4) 2 = 5.3. 10 -12.

Task 6. Calculate the concentration of the saturated solution and the PR of silver chromate if 0.011 g of salt is dissolved in 0.5 l of water.

Solution: To determine the molar concentration of a saturated solution Ag 2 CrO 4 we use the formula C M = , Where m- mass of solute (g), M - molar mass (g/mol), V- volume of solution (l). M (Ag 2 CrO 4 ) =332 g/mol. cm =9.48. 10 -5 mol/l. The dissolution of silver (I) chromate is accompanied by complete ( = 1) dissociation of the salt: Ag 2 CrO 4 ↔ 2Ag + +CrO 4 2-, PR = 2, where = C m = 9.48. 10 -5 mol/l, a = 2 =1.89610 -4.

Thus PR = (1.89610 -4) 2 (9.4810 -5) = 3.410 -12.

Problem 7. Is it possible to prepare solutions of CaCO 3 salt with concentrations of CaCO 3 C 1 = 10 -2 M and C 2 = 10 -6 M, if PR CaCO 3 = 3.810 -9.

Solution: Knowing the PR value, you can calculate the concentration

saturated salt solution and comparing it with the proposed

concentrations, draw a conclusion about the possibility or impossibility of preparing solutions. The dissolution of calcium carbonate proceeds according to the scheme CaCO 3 ↔Ca 2+ +CO 3 2- In this equation n = m = 1 then

=
≈ 6.2 10 -5 mol/l,

C 1 > C m – the solution cannot be prepared, since a precipitate will form;

C 2< С м – раствор приготовить можно.

Ion exchange reactions

Electrolyte solutions are characterized by ion exchange reactions. A prerequisite for such reactions to occur almost completely is the removal of certain ions from the solution due to:

1) sediment formation

FeSO 4 + 2 NaOH  Fe(OH) 2  + Na 2 SO 4 - molecular equation (MU)

Fe 2+ +SO 4 2- +2Na + +2OH - Fe(OH) 2 +2Na + +SO 4 2- ion-molecular equation (IMU).

Fe 2+ +2OH -  Fe(OH) 2  (PR Fe (OH) 2 = 4.810 -16) – a brief ion-molecular equation for the formation of precipitate;

2) gas release

Na 2 CO 3 + 2H 2 SO 4  H 2 CO 3 + 2NaHSO 4 (MU)

2Na + +CO 3 2- + 2H + + 2HSO 4 -  H 2 C0 3 + 2Na + + 2HSO 4 - (IMU)

2H + + CO 3 2-  H 2 C0 3  H 2 O + C0 2  - ion - molecular level

formation of a volatile compound.

3) formation of weak electrolytes

a) simple substances:

2KCN + H 2 SO 4 2HCN + K 2 SO 4 (MU)

2K + + 2CN - + 2H + +SO 4 2-  2HCN + 2K + +SO 4 2- (IMU)

CN - +H + HCN (K d HCN = 7.8 10 -10) – ion-molecular level of formation of the weak electrolyte HCN.

b) complex compounds:

ZnCl 2 + 4NH 3 Cl 2 (MU)

Zn 2+ + 2Cl - +4NH 3  2+ + 2Cl - -(IMU)

Zn 2+ +4NH 3  2+ - a brief ionic-molecular equation for the formation of a complex cation.

There are processes in which weak electrolytes or poorly soluble compounds are among the starting materials and reaction products. In this case, the equilibrium shifts towards the formation of substances that have the lowest dissociation constant or towards the formation of a less soluble substance:

A) NH 4 OH + HCl  NH 4 Cl + H 2 O (MU)

NH 4 OH + H + + Cl -  NH 4 + + Cl - + H 2 O

NH 4 OH + H +  NH 4 + + H 2 O (IMU)

K d ( NH 4 OH) =1.8 10 -5 > K d ( H 2 O) =1.810 -16.

The equilibrium is shifted towards the formation of water molecules.

B) AgCl + NaI AgI + NaCl (MU)

AgCl + Na + +I - AgI+ Na + +Cl -

AgCl + I - AgI + Cl - (IMU)

ETC AgCl =1.7810 -10 > ETC AgI =8.310 -17.

The equilibrium is shifted towards the formation of an AgI precipitate.

C) There may be processes in the equations of which there is both a poorly soluble compound and a weak electrolyte

MnS + 2HCl  MnCl 2 + H 2 S (MU)

MnS + 2H + +2Cl -  Mn 2+ + 2Cl - + H 2 S

MnS + 2 H +  Mn 2+ + H 2 S (IMU)

PR MnS =2.510 -10 ; =
=1.58.10 -5 mol/l

K d H 2 S = K 1 K 2 = 610 -22; =
=5.4.10 -8 mol/l

The binding of S 2- ions into H 2 S molecules occurs more completely than in MnS, therefore the reaction proceeds in the forward direction, towards the formation of H 2 S

Hydrolysis of salts

Hydrolysis is the result of polarization interaction of salt ions with their hydration shell. Hydrolysis is an exchange reaction in solution between water molecules and salt ions. As a result of hydrolysis, due to the formation of a weak electrolyte (weak acid or weak base), the ionic equilibrium H 2 O⇄H + + OH - changes due to the binding of H + or OH - and the pH environment changes. Salts that contain weak acid or weak base ions undergo hydrolysis. Salts formed by ions of a strong acid and a strong base do not undergo hydrolysis (NaCl, Na 2 SO 4). The products of hydrolysis can be weak electrolytes, poorly dissociating, sparingly soluble and volatile substances. Hydrolysis is a stepwise reaction; in the case of a multiply charged ion, the number of steps is equal to its charge. Hydrolysis by cation salts formed by strong acid anions and weak base cations are affected. For example, weak bases include hydroxides p- And d-metals (K d 10 -4), as well as ammonium hydroxide.

Zinc chloride is a salt formed by the weak base Zn(OH) 2 and the strong acid HCl. The zinc cation has a charge of 2+, so hydrolysis will occur in two stages:

Zn 2+ + HOH ↔ ZnOH + + H + I stage

ZnOH + +HOH↔ Zn(OH) 2 +H + II stage

As a result of this interaction, an excess of H + ions appears ([H + ]  [OH - ]), the solution is acidified (pH<7).

Hydrolysis by anion. This type of hydrolysis is characteristic of salts formed by anions of a weak acid (K d 10 -3) and cations of a strong base (K d >10 -3). Let's consider the hydrolysis of potassium carbonate - a salt formed by weak carbonic acidH 2 CO 3 (K d I = 4.5. 10 -7) and the strong base KOH, the carboxo anion has a charge (2-). Hydrolysis occurs in two stages:

CO 3 2- +H 2 O↔HCO 3 - +OH - Stage I

HCO 3 - +H 2 O↔H 2 CO 3 +OH - II stage

In this case, OH - ions are released ([H + ]  [OH - ]) - the solution becomes alkaline (pH > 7).

Irreversible hydrolysis. Salts formed by a weak base and a weak acid hydrolyze at the cation and anion. The result of hydrolysis will depend on the value To d bases and acids. Let us consider the hydrolysis of ammonium fluoride, a salt formed by weak

base NH 4 OH (K d = 1.8 . 10 -5) and weak acid HF (K d = 6.8 . 10 -4):

NH 4 F + HOH  NH 4 OH + HF

In this case K d ( NH 4 OH)  K d ( HF), therefore, hydrolysis (mainly) will proceed along the cation and the reaction of the medium will be slightly acidic.

Exchange reactions between electrolyte solutions
Reactions leading to the formation of a precipitate. Pour 3-4 ml of copper(I) sulfate solution into one test tube, the same amount of calcium chloride solution into the second, and aluminum sulfate into the third. Add a little sodium hydroxide solution to the first test tube, sodium orthophosphate solution to the second, and barium nitrate solution to the third. Precipitates form in all test tubes.
Exercise. Write reaction equations in molecular, ionic, and abbreviated ionic form. Explain why precipitation formed. Solutions of what other substances can be poured into test tubes to cause precipitation to form? Write equations for these reactions in molecular, ionic, and abbreviated ionic form.
Reactions that involve the release of gas. Pour 3-4 ml of sodium sulfite solution into one test tube, and the same volume of sodium carbonate solution into the second. Add the same amount of sulfuric acid to each of them. The first test tube releases a gas with a pungent odor, the second test tube releases an odorless gas.
Exercise. Write equations for the reactions occurring in molecular, ionic, and abbreviated ionic form. Think about what other acids could be applied to these solutions to obtain similar results. Write equations for these reactions in molecular, ionic, and abbreviated ionic form.
Reactions that occur with the formation of a slightly dissociating substance. Pour 3-4 ml of sodium hydroxide solution into one test tube and add two or three drops of phenolphthalein. The solution takes on a crimson color. Then add hydrochloric or sulfuric acid until the color becomes discolored.
Pour about 10 ml of copper(II) sulfate into another test tube and add some sodium hydroxide solution. A blue precipitate of copper(II) hydroxide is formed. Pour sulfuric acid into the test tube until the precipitate dissolves.
Exercise. Write equations for the reactions occurring in molecular, ionic, and abbreviated ionic form. Explain why discoloration occurred in the first test tube, and dissolution of the precipitate in the second. What common properties do soluble and insoluble bases have?
Qualitative reaction to chloride ion. Pour 1-2 ml of dilute hydrochloric acid into one test tube, the same amount of sodium chloride solution into the second, and calcium chloride solution into the third. Add a few drops of silver(I) nitrate solution AgNO3 to all test tubes. Check whether the precipitate dissolves in concentrated nitric acid.
Exercise. Write the equations for the corresponding chemical reactions in molecular, ionic, and abbreviated ionic form. Think about how you can distinguish: a) hydrochloric acid from other acids; b) chlorides from other salts; c) solutions of chlorides from hydrochloric acid. Why can you also use a lead(II) nitrate solution instead of a silver(I) nitrate solution?

The lesson will examine the conditions for ion exchange reactions to proceed to completion. In order to better understand what conditions must be observed for ion exchange reactions to proceed to completion, we will review what these reactions are and their essence. Examples are given to reinforce these concepts.

Topic: Chemical bond. Electrolytic dissociation

Lesson: Flow Conditionsion exchange reactions to completion

If you try to react sodium hydroxide with potassium chloride, the reaction will not occur. In a reaction, ions are exchanged, but no products are formed. Let's look at the reasons for this. Substances formed as a result of mutual attraction can dissociate.

1. Ion exchange reactions leading to the formation of a precipitate.

Previously, the reaction equations that resulted in the formation of a precipitate were considered.

All these reactions belonged to ion exchange reactions. It can be concluded that one of the conditions for the ion exchange reaction to proceed to completion is the formation of a precipitate.

BaCl 2 + Na 2 CO 3 → BaCO 3 ↓ + 2NaCl.

Ba 2+ +2Cl - + 2Na + + CO 3 2- →BaCO 3 ↓ + 2Na + +2Cl - complete ionic equation

Ba 2+ + CO 3 2- → BaCO 3 ↓ shortened ionic equation.

Let us write another reaction equation leading to the formation of a precipitate.

CuSO 4 + 2NaOH → Cu(OH) 2 ↓ + Na 2 SO 4

Cu 2+ + SO 4 2- +2Na + + 2OH - → Cu(OH) 2 ↓ + 2Na + + SO 4 2- complete ionic equation

Cu 2+ + 2OH - → Cu(OH) 2 ↓ abbreviated ionic equation.

Conclusion: Ion exchange reactions proceed to completion if the result is a precipitate.

Rice. 1. Neutralization reaction ()

Consider the reaction of neutralization of sodium hydroxide with hydrochloric acid.

NaOH + HCl → NaCl+ H 2 O

Na + + OH - + H + + Cl - →Na + + Cl - + H 2 O complete ionic equation

OH - + H + → H 2 O reduced ionic equation

This reaction proceeds to completion, because the result is a slightly dissociating substance - water.

Conclusion: Ion exchange reactions proceed to completion if the result is a slightly dissociating substance.

You know that calcium carbonate reacts well with hydrochloric acid.

CaCO 3 +2HCl → CaCl 2 + H 2 O + CO 2

CaCO 3 +2H + + 2Cl - → Ca 2+ +2Cl - + H 2 O + CO 2 complete ionic equation

2H + + CaCO 3 → Ca 2+ + H 2 O + CO 2 abbreviated ionic equation.

As a result of this reaction, carbon dioxide is produced, which is formed during the decomposition of weak carbonic acid. Please note that calcium carbonate is an insoluble substance and does not break down into ions. In the complete ionic equation, we write only hydrogen chloride and calcium chloride as ions. The remaining formulas remain unchanged, since these substances are not exposed.

Conclusion: Ion exchange reactions proceed to completion if they result in a gas.

In this lesson, you examined the conditions for ion exchange reactions to proceed to completion. Ion exchange reactions proceed to completion if the result is a precipitate, slightly dissociating substance or gas.

1. Rudzitis G.E. Inorganic and organic chemistry. 9th grade: textbook for general education institutions: basic level / G. E. Rudzitis, F.G. Feldman. M.: Enlightenment. 2009 119 p.: ill.

2. Popel P.P. Chemistry: 8th grade: textbook for general education institutions / P.P. Popel, L.S. Krivlya. -K.: IC “Academy”, 2008.-240 p.: ill.

3. Gabrielyan O.S. Chemistry. 9th grade. Textbook. Publisher: Bustard: 2001. 224s.

1. No. 3,4,5 (p.22) Rudzitis G.E. Inorganic and organic chemistry. 9th grade: textbook for general education institutions: basic level / G. E. Rudzitis, F.G. Feldman. M.: Enlightenment. 2009 119 p.: ill.

2. What do you observe when making dough when you add vinegar to soda? Write the reaction equation.

3. Why does scale form in the kettle? How to remove it? Write the reaction equations.

Goals:

• Students must acquire knowledge about ion exchange reactions and the conditions under which they occur.
• continue to develop the skills of writing equations of dissociations of substances;
• work with the solubility table;
• develop logical thinking when recognizing electrolytes and non-electrolytes, comparison, observation; develop practical skills and abilities, draw conclusions;
• write reaction equations in molecular, full ionic and reduced ionic forms.

Methods and methodological techniques: verbal-visual, heuristic, group frontal laboratory work.

Equipment:

• on students' desks: H 2 SO 4, BaCl 2, Na 2 CO 3, phenolphthalein, NaOH, 4 syringes, tablet, solubility table, table to fill out.
• to the teacher: H 2 SO 4, BaCl 2, Na 2 CO 3, phenolphthalein, NaOH, 3 test tubes, in 2 containers: soda and salt, water, acetic acid.

During the classes

1. Organizational moment.

2. Goal setting.

Teacher. Guys, imagine that in your kitchen there are salt and soda in 2 identical jars without labels. How can you recognize these two substances without tasting them?

Teacher. To find out, we need to get acquainted with ion exchange reactions, determine the conditions for their occurrence, and learn to write complete, abbreviated ionic equations. To better understand the mechanism of ion exchange reactions, let's remember what substances are called electrolytes.

Student. Electrolytes are substances that conduct electric current in melts and solutions.

Teacher. Why do electrolytes in solutions and melts conduct electric current?

Student. Electrolytes conduct electricity because ions are formed in solutions and melts.

Teacher. What is electrolytic dissociation?

Student. The process of electrolyte breaking down into ions is called electrolytic dissociation.

Teacher. Let's write the dissociation equations for various substances. (3 students work on the cards on the board):

• Card No. 1. Write summary dissociation equations for substances: H 2 SO 4, HCl.
• Map No. 2. Write the overall dissociation equations for the substances: Na 2 CO 3, BaCl 2.
• Card No. 3. Write the total dissociation equations for substances: NaOH, Ba(OH) 2

Teacher. Class assignment: select electrolytes and non-electrolytes from this list of substances.

KCl, CuO, CuSO 4, Cu(OH) 2, BaSO 4, K 2 SO 4. (from a sheet).

For electrolytes, write summary dissociation equations. (at the blackboard).

Teacher. Let's check the notes on the board.

Teacher. Guys, name from which ions the insoluble substance BaSO 4 is formed?

Student. Barium sulfate is formed from barium ions and sulfate ions.

Teacher. Name the substances, using the solubility table, whose solutions contain the Ba 2+ ion and SO 4 2- ?

Student. For example, barium chloride and sulfuric acid.

Teacher. Let's write down the equation for the reaction between H 2 SO 4 and BaCl 2 (student at the blackboard).

BaCl2 + H2SO4 = BaSO4 + 2HCl

Teacher. Reactions occurring in electrolyte solutions are called ion exchange reactions. To find out under what conditions ion exchange reactions occur, let’s do laboratory work:

Purpose: To become familiar with the conditions for such reactions. (write in notebook)

Experience No. 1. Obtaining BaSO 4. (instead of experience, you can use a fragment of a lesson from the “virtual school of Cyril and Methodius” 9th grade lesson No. 6)

The teacher does the same at the blackboard.

The teacher comments: add a solution of H 2 SO 4 to the BaCl 2 solution. What do we observe?

Student: A white precipitate fell.

Teacher: Let's write down the complete ionic equation; for this we write down what ions were in the solutions of the taken substances and what substances were formed.

2H 1+ + SO 4 2- + Ba 2+ +2Cl 1- - > BaSO 4v + 2H 1+ +2Cl 1-

This is the complete ionic equation.

If we reduce the right and left sides of the equation by identical ions, we get the abbreviated ionic equation.

SO 4 2- + Ba 2+ -> BaSO 4v

Discussion:

Questions for the class:

• What ions were contained in the solution before the reaction?
• What ions remained in solution after the reaction?
• What is the essence of these reactions?

Conversation with the class: we discuss that the essence of the reaction is that the binding of Ba 2+ and SO 4 2- ions occurred.

This equation shows the essence of this reaction.

Experience No. 2. Obtaining carbon dioxide.

The teacher comments: add a solution of H 2 SO 4 to the Na 2 CO 3 solution. (1 student writes the reaction on the board)

Na 2 CO 3 + H 2 SO 4 = Na 2 SO 4 + H 2 O + CO 2

What are we seeing?

Student: Release of gas bubbles.

The teacher writes down the full ionic equation and the abbreviated ionic equation.

2Na 1+ +CO 3 2- +2H 1+ +SO 4 2- - >2Na 1+ + SO 4 2- + H 2 O+ CO 2

CO 3 2- +2H 1+ -> H 2 O+ CO 2

Experience No. 3. Formation of H 2 O (low dissociating substance).

The teacher comments: add 1-2 drops of phthalein phenol to the NaOH solution, the solution turns crimson, add H 2 SO 4. (1 student writes the reaction on the board)

2 NaOH + H 2 SO 4 = Na 2 SO 4 + 2 H 2 O

What are we seeing?

Student. The solution became colorless.

Teacher. Let's write the full ionic equation and the abbreviated ionic equation on the board (1 student).

2Na 1+ +2OH 1- +2H 1+ + SO 4 2- ->2Na 1+ + SO 4 2- + 2H 2 O

2OH 1- +2H 1+ ->2H 2 O

Reaction conditions Between electrolyte solutions (to be completed by the student).	Examples of ion exchange reactions.
1.	Na 2 CO 3 +CaCl 2 =CaCO 3 +2NaCl 2Na + + CO 3 2- +Ca 2+ + 2Cl - = CaCO 3 +2Na + + 2Cl - Ca 2+ + CO 3 2- = CaCO 3
2.	K 2 CO 3 +2HCl =2KCl+H 2 O+CO 2 2K + + CO 3 2- +2H + +2Cl - =2K + +2Cl - +H 2 O+CO 2 CO 3 2- +2H + = CO 2 ^+H 2 O
3.	NaOH+HNO 3 = NaNO 3 +H 2 O Na + +OH - +H + +NO 3 =Na + +NO 3 - + H 2 O H + + OH - =H 2 O

Teacher: Guys, we carried out ion exchange reactions. Let's conclude: under what conditions do ion exchange reactions proceed to completion? (fill in the tables provided)

Student: Ion exchange reactions proceed to completion if, as a result, a precipitate is formed, a gas is released, and a poorly dissociating substance, such as water, is formed.

Teacher: Let's return to our problem. Suggest a method for recognizing salt (NaCl) and soda (Na 2 CO 3).

Student: You need to add acid to these substances. In which container the gas evolution will be observed, there will be soda.

Fixing the material:

Task at the board: 1 Select from this list the reactions that go to the end (one student)

NaOH+ NaCl -> NaCl+ H 2 O

AgNO 3 + NaCl ->NaNO 3 + AgCl

CuCl 2 +2NaOH ->Cu(OH) 2 +2NaCl

KNO 3 +LiCl ->KCl+LiNO 3

Given:

Complete ionic equation.

Fe 3+ +3Cl - +3Na + +3OH - = Fe(OH) 3 +3Na + +3Cl -

Write the molecular and abbreviated ionic equations accordingly.

Teacher. Let's summarize our lesson: What reactions did we get acquainted with in the lesson?

Student. We learned about ion exchange reactions.

Teacher. Under what conditions is it possible for these reactions to proceed to completion?

Student. Ion exchange reactions proceed to completion if a precipitate forms, a gas is released, and a slightly dissociating substance is formed.

Teacher. Homework: §37 exercise. 4, 5.

Literature.

1. Gabrielyan O.S. Chemistry. 8th grade: Bustard, 1999.
2. “Virtual School of Cyril and Methodius” Chemistry lessons 8-9 grades, 2004.

The proposed material presents methodological developments of practical work for the 9th grade: “Solving experimental problems on the topic “Nitrogen and Phosphorus”, “Determination of mineral fertilizers”, as well as laboratory experiments on the topic “Exchange reactions between electrolyte solutions”.

Exchange reactions between electrolyte solutions

Methodological development consists of three parts: theory, workshop, control. The theoretical part provides some examples of molecular, complete and abbreviated ionic equations of chemical reactions that occur with the formation of a precipitate, a slightly dissociating substance, and the release of gas. The practical part contains tasks and recommendations for students on how to perform laboratory experiments. The control consists of test tasks with the choice of the correct answer.

Theory

1. Reactions leading to the formation of a precipitate.

a) When copper(II) sulfate reacts with sodium hydroxide, a blue precipitate of copper(II) hydroxide is formed.

CuSO 4 + 2NaOH = Cu(OH) 2 + Na 2 SO 4.

Cu 2+ + + 2Na + + 2OH – = Cu(OH) 2 + 2Na + + ,

Cu 2+ + 2OH – = Cu(OH) 2.

b) When barium chloride reacts with sodium sulfate, a white milky precipitate of barium sulfate precipitates.

Molecular equation of a chemical reaction:

BaCl 2 + Na 2 SO 4 = 2NaCl + BaSO 4.

Full and abbreviated ionic reaction equations:

Ba 2+ + 2Cl – + 2Na + + = 2Na + + 2Cl – + BaSO 4 ,

Ba 2+ + = BaSO 4 .

2.

When sodium carbonate or bicarbonate (baking soda) interacts with hydrochloric or other soluble acid, boiling or intense release of gas bubbles is observed. This releases carbon dioxide CO 2, causing cloudiness in the clear solution of lime water (calcium hydroxide). Lime water becomes cloudy because... insoluble calcium carbonate is formed.

a) Na 2 CO 3 + 2HCl = 2NaCl + H 2 O + CO 2;

b) NaHCO 3 + HCl = NaCl + CO 2 + H 2 O;

Ca(OH) 2 + CO 2 = CaCO 3 + H 2 O.

a) 2Na + + + 2H + + 2Cl – = 2Na + + 2Cl – + CO 2 + H 2 O,

2H + = CO 2 + H 2 O;

b) Na + + + H + + Cl – = Na + + Cl – + CO 2 + H 2 O,

H + = CO 2 + H 2 O.

3. Reactions that occur with the formation of a slightly dissociating substance.

When sodium or potassium hydroxide reacts with hydrochloric acid or other soluble acids in the presence of the phenolphthalein indicator, the alkali solution becomes colorless, and as a result of the neutralization reaction, a low-dissociating substance H 2 O is formed.

Molecular equations of chemical reactions:

a) NaOH + HCl = NaCl + H 2 O;

c) 3KOH + H 3 PO 4 = K 3 PO 4 + 3H 2 O.

Full and abbreviated ionic reaction equations:

a) Na + + OH – + H + + Cl – = Na + + Cl – + H 2 O,

OH – + H + = H 2 O;

b) 2Na + + 2OH – + 2H + + = 2Na + + + 2H 2 O,

2OH – + 2H + = 2H 2 O;

c) 3K + + 3OH – +3H + + = 3K + + + 3H 2 O,

3OH – + 3H + = 3H 2 O.

Workshop

1. Exchange reactions between electrolyte solutions leading to the formation of a precipitate.

a) Carry out a reaction between solutions of copper(II) sulfate and sodium hydroxide. Write molecular, complete and abbreviated ionic equations of chemical reactions, note the signs of a chemical reaction.

b) Carry out a reaction between solutions of barium chloride and sodium sulfate. Write molecular, complete and abbreviated ionic equations of chemical reactions, note the signs of a chemical reaction.

2. Reactions that involve the release of gas.

Carry out reactions between solutions of sodium carbonate or sodium bicarbonate (baking soda) with hydrochloric or other soluble acid. Pass the released gas (using a gas outlet tube) through clear lime water poured into another test tube until it becomes cloudy. Write molecular, complete and abbreviated ionic equations of chemical reactions, note the signs of these reactions.

3. Reactions that occur with the formation of a slightly dissociating substance.

Carry out neutralization reactions between an alkali (NaOH or KOH) and an acid (HCl, HNO 3 or H 2 SO 4), after placing phenolphthalein in the alkali solution. Note observations and write molecular, complete and abbreviated ionic equations for chemical reactions.

Signs, accompanying these reactions, can be selected from the following list:

1) release of gas bubbles; 2) sedimentation; 3) the appearance of odor; 4) dissolution of the sediment; 5) heat release; 6) change in the color of the solution.

Control (test)

1. The ionic equation for the reaction that produces the blue precipitate is:

a) Cu 2+ + 2OH – = Cu(OH) 2;

c) Fe 3+ + 3OH – = Fe(OH) 3;

d) Al 3+ + 3OH – = Al(OH) 3.

2. The ionic equation for the reaction in which carbon dioxide is released is:

a) CaCO 3 + CO 2 + H 2 O = Ca 2+ +;

b) 2H + + SO 2- 3 = H 2 O + SO 2;

c) CO 2- 3 + 2H + = CO 2 + H 2 O;

d) 2H + + 2OH – = 2H 2 O.

3. The ionic equation for the reaction in which a low-dissociating substance is formed is:

a) Ag + + Cl – = AgCl;

b) OH – + H + = H 2 O;

c) Zn + 2H + = Zn 2+ + H 2;

d) Fe 3+ + 3OH – = Fe(OH) 3.

4. The ionic equation for the reaction that produces the white precipitate is:

a) Cu 2+ + 2OH – = Cu(OH) 2;

b) CuO + 2H + = Cu 2+ + H 2 O;

c) Fe 3+ + 3OH – = Fe(OH) 3;

d) Ba 2+ + SO 2- 4 = BaSO 4 .

5. The molecular equation that corresponds to the abbreviated ionic equation for the reaction 3OH – + 3H + = 3H 2 O is:

a) NaOH + HCl = NaCl + H 2 O;

b) 2NaOH + H 2 SO 4 = Na 2 SO 4 + 2H 2 O;

c) 3KOH + H 3 PO 4 = K 3 PO 4 + 3H 2 O;

d) Ba(OH) 2 + 2HCl = BaCl 2 + H 2 O.

6. Molecular equation that corresponds to the abbreviated ionic reaction equation

H + + = H 2 O + CO 2 , –

a) MgCO 3 + 2HCl = MgCl 2 + CO 2 + H 2 O;

b) Na 2 CO 3 + 2HCl = 2NaCl + CO 2 + H 2 O;

c) NaHCO 3 + HCl = NaCl + CO 2 + H 2 O;

d) Ca(OH) 2 + CO 2 = CaCO 3 + H 2 O.

Answers. 1 -A; 2 -V; 3 -b; 4 -G; 5 -V; 6 -V.

Solving experimental problems on the topic “Nitrogen and Phosphorus”

When studying new material on the topic “Nitrogen and Phosphorus,” students perform a series of experiments related to the production of ammonia, the determination of nitrates, phosphates, and ammonium salts, and acquire certain skills and abilities. This methodological development contains six tasks. To complete practical work, three tasks are enough: one on obtaining a substance, two on recognizing substances. When performing practical work, students can be offered tasks in a form that will make it easier for them to prepare a report (see tasks 1, 2). (Answers are given for the teacher.)

Exercise 1

Obtain ammonia and experimentally prove its presence.

a) Production of ammonia.

Heat a mixture of equal volume portions of solid ammonium chloride and calcium hydroxide powder in a test tube with a gas outlet tube. In this case, ammonia will be released, which must be collected in another dry test tube located with a hole ............ ( Why?).

Write the reaction equation for the production of ammonia.

…………………………………………………..

b) Determination of ammonia.

Can be identified by smell………… (name of substance), as well as by changes in the color of litmus or phenolphthalein. When ammonia is dissolved in water, ...... is formed. (name of base), so the litmus test...... (specify color), and colorless phenolphthalein becomes …………. (specify color).

Instead of dots, insert words according to their meaning. Write the reaction equation.

…………………………………………………..

* Ammonia, an aqueous solution of ammonia, smells like ammonia. – Note ed.

Task 2

Prepare copper nitrate in two different ways, having the following substances available: concentrated nitric acid, copper filings, copper(II) sulfate, sodium hydroxide. Write equations for chemical reactions in molecular form and note the changes. In method 1, for a redox reaction, write electron balance equations, determine the oxidizing agent and the reducing agent. In method 2, write abbreviated ionic reaction equations.

1st s p o s o b. Copper + nitric acid. Lightly heat the contents of the test tube. The colorless solution becomes….. (specify color), because is formed….. (name of substance); gas is released……..colored with an unpleasant odor, this is……. (name of substance).

2nd s p o s o b. When copper(II) sulfate reacts with sodium hydroxide, a precipitate of ..... color is obtained, this is ...... (name of substance). Add nitric acid to it until the precipitate is completely dissolved......... (name of sediment). A clear blue solution is formed...... (name of salt).

Task 3

Prove experimentally that ammonium sulfate contains NH 4 + and SO 2- 4 ions. Note the observations and write molecular and abbreviated ionic equations for the reactions.

Task 4

How to experimentally determine the presence of solutions of sodium orthophosphate, sodium chloride, sodium nitrate in test tubes No. 1, No. 2, No. 3? Note the observations and write molecular and abbreviated ionic equations for the reactions.

Task 5

Having the following substances: nitric acid, copper shavings or wire, universal indicator paper or methyl orange, prove experimentally the composition of nitric acid. Write the equation for the dissociation of nitric acid; molecular equation for the reaction of copper with concentrated nitric acid and electron balance equations, identify the oxidizing agent and the reducing agent.

Task 6

Prepare a solution of copper nitrate in different ways, using the substances: nitric acid, copper oxide, basic copper carbonate or hydroxycopper(II) carbonate. Write molecular, complete and abbreviated ionic equations for chemical reactions. Note signs of chemical reactions.

Benchmark tests

1. Give the reaction equation where a yellow precipitate forms.

2. The ionic equation of the reaction in which the white cheesy precipitate is formed is:

3. To prove the presence of nitrate ion in nitrates, you need to take:

a) hydrochloric acid and zinc;

b) sulfuric acid and sodium chloride;

c) sulfuric acid and copper.

4. The reagent for chloride ion is:

a) copper and sulfuric acid;

b) silver nitrate;

c) barium chloride.

5. In the reaction equation, the diagram of which

HNO 3 + Cu -> Cu(NO 3) 2 + NO 2 + H 2 O,

Before the oxidizing agent you need to put the coefficient:

a) 2; b) 4; at 6.

6. Basic and acid salts correspond to the pairs:

a) Cu(OH) 2, Mg(HCO 3) 2;

b) Cu(NO 3) 2, HNO 3;

c) 2 CO 3, Ca(HCO 3) 2.

Answers. 1 -A; 2 -b; 3 -V; 4 -b; 5 -b; 6 -V.

Determination of mineral fertilizers

The methodological development of this practical work consists of three parts: theory, workshop, control. The theoretical part provides general information on the qualitative determination of cations and anions included in mineral fertilizers. The workshop provides examples of seven mineral fertilizers with a description of their characteristic features, and also gives equations for qualitative reactions. In the text, instead of dots and question marks, you need to insert answers that are appropriate in meaning. To complete practical work at the teacher's discretion, it is enough to take four fertilizers. Testing students' knowledge consists of test tasks to determine fertilizer formulas, which are given in this practical work.

Theory

1. The reagent for chloride ion is silver nitrate. The reaction proceeds with the formation of a white cheesy precipitate:

Ag + + Cl – = AgCl.

2. Ammonium ion can be detected using alkali. When a solution of ammonium salt is heated with an alkali solution, ammonia is released, which has a sharp characteristic odor:

NH + 4 + OH – = NH 3 + H 2 O.

You can also use red litmus paper moistened with water, a universal indicator or phenolphthalein strip of paper to determine the ammonium ion. The piece of paper must be held over the vapors released from the test tube. Red litmus turns blue, the universal indicator turns purple, and phenolphthalein turns crimson.

3. To determine nitrate ions, copper shavings or pieces are added to the salt solution, then concentrated sulfuric acid is added and heated. After some time, a brown gas with an unpleasant odor begins to be released. The release of brown NO2 gas indicates the presence of ions.

For example:

NaNO 3 + H 2 SO 4 NaHSO 4 + HNO 3,

4HNO 3 + Cu = Cu(NO 3) 2 + 2NO 2 + 2H 2 O.

4. The reagent for phosphate ion is silver nitrate. When it is added to a phosphate solution, a yellow precipitate of silver phosphate precipitates:

3Ag + + PO 3- 4 = Ag 3 PO 4.

5. The reagent for sulfate ion is barium chloride. A white milky precipitate of barium sulfate, insoluble in acetic acid, precipitates:

Ba 2+ + SO 2- 4 = BaSO 4 .

Workshop

1. Sylvinite (NaCl KCl), pink crystals, good solubility in water. The flame turns yellow. When viewing the flame through blue glass, a violet color is noticeable. WITH …….. (name of reagent) gives a white precipitate...... (name of salt).

KCl + ? -> KNO 3 + AgCl.

2. Ammonium nitrate NH 4 NO 3, or…….. (name of fertilizer), white crystals, highly soluble in water. Brown gas is released with sulfuric acid and copper... (name of substance). With solution……. (name of reagent) When heated, the smell of ammonia is felt, its vapor turns red litmus into....... color.

NH 4 NO 3 + H 2 SO 4 NH 4 HSO 4 + HNO 3,

HNO 3 + Cu -> Cu(NO 3) 2 + ? + ? .

NH 4 NO 3 + ? -> NH 3 + H 2 O + NaNO 3.

3. Potassium nitrate (KNO 3), or…… (name of fertilizer), with H 2 SO 4 and ……… (name of substance) produces brown gas. The flame turns purple.

KNO 3 + H 2 SO 4 KHSO 4 + HNO 3,

4HNO 3 + ? -> Cu(NO 3) 2 + ? + 2H 2 O.

4. Ammonium chloride NH 4 Cl with solution……. (name of reagent) When heated, it forms ammonia, its vapors turn red litmus blue. WITH …… (name of the reagent anion) silver gives a white cheesy precipitate...... (name of sediment).

NH 4 Cl + ? = NH 4 NO 3 + AgCl,

NH 4 Cl + ? = NH 3 + H 2 O + NaCl.

5. When heated, ammonium sulfate (NH 4) 2 SO 4 with an alkali solution forms ammonia; its vapor turns red litmus blue. WITH …….. (name of reagent) gives a white milky sediment......... (name of sediment).

(NH 4) 2 SO 4 + 2NaOH = 2NH 3 + 2H 2 O + ? ,

(NH 4) 2 SO 4 + ? -> NH 4 Cl + ? .

6. Sodium nitrate NaNO 3, or…… (name of fertilizer), white crystals, good solubility in water, produces a brown gas with H 2 SO 4 and Cu. The flame turns yellow.

NaNO 3 + H 2 SO 4 NaHSO 4 + ? ,

Cu -> Cu(NO 3) 2 + ? + 2H 2 O.

7. Calcium dihydrogen phosphate Ca(H 2 PO 4) 2, or…… (name of fertilizer), gray fine-grained powder or granules, poorly soluble in water, with ..... (name of reagent) gives ….. (specify color) precipitate ……… (name of substance) AgH 2 PO 4.

Ca(H 2 PO 4) 2 + ? -> 2AgH 2 PO 4 + Ca(NO 3) 2.

Control (test)

1. Pink crystals, highly soluble in water, color the flame yellow; when interacting with AgNO 3, a white precipitate forms - this is:

a) Ca(H 2 PO 4) 2; b) NaCl KCl;

c) KNO 3; d) NH 4 Cl.

2. The crystals are highly soluble in water; in reaction with H 2 SO 4 and copper, a brown gas is released; with an alkali solution, when heated, it produces ammonia, the vapors of which turn red litmus blue - this is:

a) NaNO 3; b) (NH 4) 2 SO 4;

c) NH 4 NO 3; d) KNO 3.

3. Light crystals, highly soluble in water; upon interaction with H 2 SO 4 and Cu, brown gas is released; the flame turns purple - this is:

a) KNO 3; b) NH 4 H 2 PO 4;

c) Ca(H 2 PO 4) 2 CaSO 4; d) NH 4 NO 3.

4. The crystals are highly soluble in water; with silver nitrate it gives a white precipitate, with alkali when heated it gives ammonia, the vapors of which turn red litmus blue - this is:

a) (NH 4) 2 SO 4; b) NH 4 H 2 PO 4;

c) NaCl KCl; d) NH 4 Cl.

5. Light crystals, highly soluble in water; with BaCl 2 it gives a white milky precipitate, with alkali it gives ammonia, the vapors of which turn red litmus blue - this is:

c) NH 4 Cl; d) NH 4 H 2 PO 4.

6. Light crystals, highly soluble in water; when interacting with H 2 SO 4 and Cu, it produces a brown gas, the flame turns yellow - this is:

a) NH 4 NO 3; b) (NH 4) 2 SO 4;

c) KNO 3; d) NaNO 3.

7. Gray fine-grained powder or granules, solubility in water is poor, with a solution of silver nitrate it gives a yellow precipitate - this is:

a) (NH 4) 2 SO 4; b) NaCl KCl;

c) Ca(H 2 PO 4) 2; d) KNO 3.

Answers. 1 -b; 2 -V; 3 -A; 4 -G; 5 -b; 6 -G; 7 -V.