МЕТОДИЧЕСКИЕ УКАЗАНИЯ СТУДЕНТАМ

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1 МЕТОДИЧЕСКИЕ УКАЗАНИЯ СТУДЕНТАМ CLASS 1. TPIC: Introduction in a subject. Safety regularions in a chemical laboratory. 1. Introduction in a subject. (0 min.) Control of a student s initial level in chemistry. (60 min.). 1.. Ways of expressing concentration of solutions (mass fraction, molarity) (30 min.).. Safety regularions in a chemical laboratory (0 min.). Safety regularions in a chemical laboratory. Working in a chemical laboratory requires following certain safety regulations. Using small amounts of chemicals reduces occurance of accidents but does not exclude them. That is why everyone working in a chemical laboratory must be aware of and keep to safety regulations. Be careful. Keep in mind that being careless and inattentive can cause an accident. Do not let metallic K, Na or Li touch water to avoid the inflammation and explosion of hydrogen that is formed in the traction. Alkaline metals, caustic alcalines, concentrated acids exposed to the skin and mucus cause severe burns. Injuring eyes is particularly hasardous. When a concentrated acid or alkaline gets onto the skin, wash the burn with the running water. Dilute H S 4 with water by adding the acid to the water drop by drop but not vice versa, stirring the solution all the time. Be careful when washing the glass with chromium mixture, so as the mixture does not get onto the skin, clothes or foot wear.

2 Wash off the concentrated acids and alkalies with a strong stream of the running water. Spilled concentrated acids and alcalies mist be cavered with sand and cleaned away. If a thermometre is broken up and mercury spills, it should be collected with a special trap or a bulb. The minute particles of mercury are collected with a crush made of white tin. The surface of the floor or the table, contaminated with mercury is thoroughly soaked with 0% of FeCl 3 solution or covered with sulphur. Heat the liquid in the test-tube gradually. Direct the outlet of the test tube sideway from youself and other students, working next to you as partial overheating can bring on a gush of the liquid. Don t bend over a test-tube with a boiling liquid. Do not breathe in chemical substances even if their amounts are very small. To smell the substance you should direct the vapour or gases towards yourself with the movement of your hand. You must not taste reagents. All the experiments with foul-smelling as well as poisonous substances (aniline, bromine) are done in the exhaust-hood. When working with a drain tube, you can take away the burner from under the test tube with the regent mixture only when the end of the pipe is removed from the liquid. ther wise the liquid can get sucked into the overheated test tube and cause the reagent mixture splashing about and getting on to the skin. Work with easily imflammable liquids (ether, benzene, acetone etc.) in the distance from the open fire. Their heating is done on the water-bath or sandbath. Taking liquids (acid and base solutions, salts of heavy metals, volatile substances) is done with a special pipette. The students should wear gouns and hats when in lab classes.

3 All experiments are done in clean glass. After finishing work, the glassware is washed and prepared for the next class. Containers with reagents are put in their place and closed with the same stoppers as they have been closed before. It is forbidden to move containers with reagents from one table to another. Books and bags as well as other students belongins are kept in the box of the laboratory table. A student on duty is appointed in every group. He/she is responsible for - bringing all the necessary things from the lab room; - keeping the laboratory clean and tidy; - checking every student s working place at the end of the class; - reporting to the lab assistant about the condition of the laboratory. CLASS. TPIC: Solutions. Ways of expressing concentration of solutions. 1. SEMINAR. (75 min.) Ways of expressing concentration of solutions (mass fraction, molarity, molality, molar concentration of the equivalent, titer). 1.. Preparation of a solution of a given concentration.. LABRATRY WRK. (35 min.)..1. Preparation of a solution from a fixanale... Preparation of a standard solution from an initial substance with a given weight..3. Preparation of a delute solution of a certain concentration from a concentrated solution of mineral acid. Literature 1. Lectures.

4 . Ebbing D.D. General Chemistry/ D.D.Ebbing, M.S.Wrighton;.- Boston. Third edition. Houghton Mifflin Company. Р , Lister T. Chemistry for Advanced Level / T. Lister, J. Renshaw; - Third edition. Stanley Thornes (Publishers) Ltd Р PRACTICAL PRBLEMS 1. What volume (in ml) of 40% H 3 P 4 solution (ρ=1,5 g/cm 3 ) is necessary to prepare 400mL of 0,5N of phosphoric acid solution (ρ=1g/cm 3 )? Calculate the mole fraction of H 3 P 4 in the obtained solution.. How many grams of Na C 3 10H are necessary to prepare 100mL of 10% Na C 3 solution (ρ=1,1 g/cm 3 )? What is the mole fraction and the molarity of the obtained Na C 3 solution? 3. How many grams of K S 4 are necessary to prepare 100mL of 0,1N K S 4 solution? Calculate the mass and the mole fraction of K S 4 in the obtained solution (density of the solution = 1,0 g/cm 3 ). 4. How many ml of 15% potassium chloride solution (ρ=1,09g/cm 3 ) are necessary to prepare 00mL of 8% solution (ρ=1,05g/cm 3 )? Calculate the mole fraction, molality, normality and titer of the obtained solution. 5. The mass of oxalic acid crystalline hydrate H C 4 H (3,1547g) was dissolved in a volumetric flask of 750mL. Calculate the molality, normality and titer of this solution.. 6. How many grams of crystal CuS 4 5H are necessary to prepare 500 ml of 5% CaCl solution with ρ = 1,049 g/cm 3. Describe how you would prepare this solution.

5 . LABRATRY WRK.1. Preparation of solutions of a given concentration from a fixed weight. In practice exactly weighed ready made amounts of chemically pure solid substances or exactly measured volumes of their solutions are used to prepare titrant of certain normality. These substances are placed in a special glass ampule and soldered. These ampules with strictly defined amounts of a substance are called fixed substances. They usually contain 0,1 mole substance equivalents; just this is the quantity required for preparing 1 liter of 0,1N solution. Normality of a given solution can be calculated by the equation: n( 1 X ) C ( 1 X ) = z z V ( solution) Where: n ( 1 X) is the number of moles of substance equivalent z (it must be shown on the container); V is a delivery flask volume, liter. To prepare a required titer solution the glass ampule is broken up over a funnel with a breaking rod and the substance is transferred into a delivery flask. The ampule and the funnel are washed with water and some water is added, up to the mark etched on the neck of the flask. Then it is carefully stirred... Preparation of a standard solution from an initial substance with a given weight. A standard solution is a solution, the concentration of which is known. The substances used for the preparation of a standard solution, are called initial or standard substances. They should meet the following requirements: be chemically pure and have a crystal structure;

6 be soluble well enough and not hydroscopic; the chemical composition of a substance must correspond to its formula; have a relatively big molar mass. the solutions of these substances must not change and they must be airresistant when stored; For preparation of a titrant from a given substance, it is weighed on a platform balance (the accuracy of ±0,0001g), then it is transferred carefully to a volumetric flask, and some water is added to bring the solution level to the mark on the neck of the volumetric flask. Calculate the normality and titer of the obtained solution. m( X ) T ( X ) =, V ( ) 1000 ( 1 T X C X ) = z M ( 1 X ) z Where: m(x) is the weight mass, g; V is the volume of the delivery flask, ml; M( 1 X) is the molar mass of the substance equivalent. z Problem 1. How many grams of oxalic acid should be added to a 100-mL volumetric flask to prepare 0,1 N oxalic acid solution when the flask is filled up to the mark with water? Describe, how you would prepare this solution. xalic acid, H C 4 H, is used in a neutralization method as a standard substance to determine normality and titer of alkali, as it meets all the requirements to initial substances. The prepared initial oxalic acid solution is used as a standard solution in acid base volumetric analysis. Solution Using the formula C( 1 X ) z m (H C 4 H ) = C( n( 1 X ) z m( X ) = V ( solution) M ( 1 X ) V ( solution) z =, we calculate 1 H C 4 H ) M( 1 H C 4 H ) V; m (H C 4 H ) = 0,1mol/L 63,04g/mol 0,1L = 0,6304 g.

7 . Preparation of a solution: g C H 4 H is weighed on a platform balance. The oxalic acid crystals are transferred carefully to a volumetric flask. Water is added to bring the solution level to the mark on the neck of the 100- ml volumetric flask. Problem. A sample of C H. 4 H weighing 0.68g is placed in a 100-mL volumetric flask. The flask is then filled with water up to the mark on the neck. What is the normality and titer of the resulting solution? Solution The concentration of the obtained solution is calculated by the formula: m( 1 H C 4 m( 1 H ) C4 H H ) = ( 1, M H ) ( ) C4 H V solution 0,68g C ( 1 H C H ) 0,108mol / L 4 = = 63,04 g 0,1L mol m( 1 H ) C4 H 0,68g T ( H C4 H ) = = = 0,0068( g / ml) V ( solution) 100mL.3. Preparation of a delute solution of a certain concentration from a concentrated solution of mineral acid. Problem 3 The given concentrated sulfuric acid is an aqueous solution, containing 40% of H S 4 by mass. The density is 1.30 g/ml. How many milliliters of concentrated H S 4 are required to make 00 ml of 0,1 N sulfuric acid solution? Solution We first calculate the mass of H S 4 that is contained in 50mL of 0,15N solution:

8 C( 1 H S ) = 4 m( H S4 ) M ( 1 H ) ( ) S4 V solution m (H S 4 ) = C( 1 H S 4 ) M( 1 H S 4 ) V = 0,1 mol 49 L mol g 0, L =0,98 g. Knowing the mass of H S 4 we can calculate the mass of the concentrated m( H S ) 100% 0, =, 40% 40 4 solution: m ( solution) = =,45( g) then the volume of concentrated solution will be: m,45 V = = = 1,88( ml) ρ 1,30 Preparation of a solution: use a measuring pipet to deliver 3.53 ml of 40% H S 4 solution into a 50-mL volumetric flask and then add water up to the mark on the neck of the flask and stir. Enter the results of the laboratory work in the record of proceedings. CLASS 3,4. TPIC : Acid base method of measure analysis. (Neutralization method). 1. SEMINAR. (60 min.) Classification of volumetric analysis methods.(neutralization, oxidation-reduction, precipitation, complex formation methods). 1.. Basics of the volumetric analysis. The Law of equivalent Theoretical grounds of neutralization method Titration curves and acid-base indicators The ways of calculation of normality and titer of solutions Measuring vessels. Volumetric analysis technique.. LABRATRY WRK. (45 min.)..1. Determination of normality and titer of alkali solution by a standard solution of oxalic acid... Determination of normality and titer of acid solution by the titrant of alkali.

9 Literature. 1. Lectures.. Ebbing D.D. General Chemistry/ D.D.Ebbing, M.S.Wrighton;.- Boston. Third edition. Houghton Mifflin Company. Р , Lister T. Chemistry for Advanced Level / T. Lister, J. Renshaw; - Third edition. Stanley Thornes (Publishers) Ltd Р PRACTICAL PRBLEMS 1. A 15,0 ml of NaH solution is titrated with N HCl, and 10,0 ml is required to reach the endpoint. a. What is the equivalent molar concentration and titer of NaH solution? b. If the density of the NaH solution is 1g/cm 3, what is the mass percent of NaH in the given solution?. Sodium hydroxide solution is usually standardized by titration of a pure sample of potassium hydrogen phthalate (KHP), an acid with one acidic hydrogen and a molar mass of 04. g. It takes 0.46 ml of a sodium hydroxide solution to titrate a 0,108 g sample of KHP. What is the molarity and normality of the sodium hydroxide? 3. A sample of CH 3 CH weighing.4 g is placed in a 100-mL volumetric flask. The flask is then filled with water to the mark on the neck. A 50,0 ml of the obtained solution is titrated with 14,4 ml of 0,15 N KH solution. Calculate the mass fraction of the formic acid in the original solution. 4. A 10-mL sample of vinegar, an aqueous solution of acetic acid (CH 3 CH), is titrated with 0,506 N NaH, and 16,58 ml is required to reach the endpoint.

10 a. What is the molarity of the acetic acid? b. If the density of the vinegar is 1,006 g/cm 3, what is the mass percent of acetic acid in the vinegar? EXAMPLE PRBLEMS Example 1. A 5,1-mL sample of NaH solution requires 5,0 ml of 0,1030 N oxalic acid for complete neutralization. What is the titer and molar concentration of KH equivalent. What indicator may be used in this case of titration? Solution The molar concentration of substance equivalent may be calculated using the law of equivalents. Two substances react completely in quantities, proportional to their equivalents, the volumes of the two reacting solutions inversely proportional to their normalities, that is: n( 1 X z 1 ) = n( from where C( 1 X z ); or C( 1 X z ) = 1 X z 1 ) V 1 = C( C( 1 X 1) V z V 1 1 X z ) V ; ; or C( 1 H ) ( ) C4 V H C4 Ñ( 1 KH ) = ( mol / L) ; 1 V ( KH ) C( 1 NaH ) M ( 1 NaH ) T ( NaH ) = 1 1 g / ml; 1000 Ñ 1 0,1030 5,0 0, ( NaH ) = = 0,1010mol / L ; T ( NaH ) = = 0,004040g / ml; 1 5, The titer of substance solution is determined sometimes by using such a value as the titer of standard solution per the determined substance. It shows

11 what mass of the substance that should be determined can be titrated by 1 ml of standard solution. The calculation is carried out by the equation: s tan d. C( 1 s tan d.) M ( 1 det erm.) T = z z ( g / ml) det erm ( 1 ( 1 H ) ) C4 H C H C4 H M NaH 1 0, T = = = 0,00410( g / ml) NaH It means that 0,00410g of KH is neutralized by 1 ml of 0,1030 N oxalic acid solution. The potassium hydroxide titer solution: s tan d. V ( s tan d.) T (det erm) = T ( g / ml) ; det erm. V (det erm.) H C T ( NaH ) = T NaH 4 V ( H C4 ) ( g / ml) V ( NaH ) 5,0 T ( NaH ) = 0,00410 = 0,004039( g / ml) 5,1 Knowing the titer of a solution, we may calculate its normality : ( det erm. ) 1 T 1000 C( (det erm.) = ( mol / L) z M ( 1 det erm.) z 1 T ( NaH ) , C ( NaH ) = 0,1010( / ) 1 ( 1 = = mol L M NaH ) 40 1 The mass of the determined substance in the volume to be titrated is found like s tan d. det erm. this: m(det erm.) = T V ( s tan d.)( g) H C4 m ( NaH ) = T ( ) V ( NaH ) = 0, ,0 = 0,006( g) NaH m( NaH ) 0,006 T ( NaH ) = = = 0,004039( g / ml) V 5,1 In this case we use phenolphthalein as an indicator. (See Appendix. Titration curves. Figure.)

12 Example. A 5,0-mL sample of Na C 3 solution is titrated with HCl solution, and 5,8 ml is required to reach the endpoint. Calculate the normality and the titer of HCl solution. The titer of Na C 3 solution equals 0,005936g/mL. Which of the indicators could be used for doing that titration? Solution A sodium carbonate, Na C 3, is the substance which may be used as a standard substance in neutralization method for the determination of mineral acids titer. In water Na C 3 is hydrolyzed forming alkali solution: Na C 3 + H = H + C + NaH This base, NaH, is titrated with HCl acid solution. To detect the equivalence point, we will add an indicator that changes color within the ph range 3-11 (dramatic change of ph on the titration curve of strong acid strong base. See Appendix. Figure 1). Calculation of normality and titer of HCl acid solution is the same as in the previous example. formula: The molar concentration of sodium carbonate equivalent is calculated by T ( Na C ) , C( 1 3 Na C ) 3 = M ( 1 = Na C ) 53 3 = 0,110mol / L The molar concentration of HCl substance equivalent is calculated by using the law of equivalents. C( 1 Na C ) V ( Na C ) C HCl 3 3 0,110 5,0 ( 1 ) = = = 0,0965mol / L 1 V ( HCl) 5,8 C( 1 HCl) M ( 1 HCl) 0, ,5 T HCl 1 1 ( ) = = = 0,0035g / ml The titer of HCl solution is calculated by using the such a value as titer of standard solution per detemined. NaC T HCl 3 C = 1 ( 3 Na C M ( 1 ) HCl) 1 0,110 36,5 = = 0,004088g / ml

13 NaC T ( HCl) = T HCl 3 V ( Na C V ( HCl) 3 ) 5,0 = 0, = 0,00354g / ml 5,8 T HCl C 1 ( ) , ( HCl) = 0,0965mol / L 1 M ( 1 = = HCl) 36,5 1. LABRATRY WRK.1. Determination of normality and titer of alkali solution by a standard solution of oxalic acid. Phenolphthalein is an indicator. Fill a buret with alkali solution. Put 5 ml of standard oxalic acid solution into a beaker by using a measuring pipette. Then add a small amount (-3 drops) of phenolphthalein indicator. The base is added drop by drop to the acidic solution in the beaker during the titration, the indicator changes color, but the color disappears on mixing. The stoichiometric (equivalence) point is marked by appearance of light pink color, that isn t vanishing for 30 seconds. Carry out titration to take three concurrent readings. Make a table of the titration readings. Calculate the average volume, and then determine normality and titer of alkali solution. Write the equation of the reaction... Determination of normality and titer of the mineral acid solution by a titrant of base. Fill a buret with a mineral acid solution. Put 5 ml of base solution into a beaker by using a measuring pipette. Then add a small amount (-3 drops) of methyl orange indicator. In basic solution the color of methyl orange is yellow. The acid is added drop by drop to the alkali solution in the beaker during the titration, the indicator changes color. Stoichiometric (equivalence) point is marked by appearance of yellow color.

14 Carry out titration to take three concurrent readings. Make a table of the titration readings. Calculate the average volume, and then determine normality and titer of alkali solution. Write the equation of the reaction. CLASS 5. TPIC: xidation-reduction measure analysis. (Permanganatometry). 1. SEMINAR. (60 min.) The characteristics of oxidation-reduction reactions. xidizing and reducing agents. The oxidation numbers. 1.. Balancing oxidation-reduction equations. xidizing and reducing agent equivalent Methods of oxidation-reduction titration xidation-reduction reactions in the body Volumetric analysis by using potassium permanganate as a standard solution.. LABRATRY WRK. (45 min.)..1. Determination of normality and titer of a potassium permanganate solution by a standard solution of oxalic acid... Determination of normality and titer of hydrogen peroxide solution by the titrant of potassium permanganate. Literature. 1. Lectures.. Ebbing D.D. General Chemistry/ D.D.Ebbing, M.S.Wrighton;.- Boston. Third edition. Houghton Mifflin Company. Р

15 3. Lister T. Chemistry for Advanced Level / T. Lister, J. Renshaw; - Third edition. Stanley Thornes (Publishers) Ltd Р PRACTICAL PRBLEMS 1. Define: oxidation, reduction; oxidizing agent, reducing agent.. During an oxidation process in an oxidation-reduction reaction the species oxidized electrons. 3. During an oxidation process in an oxidation-reduction reaction the species is the oxidizing agent. 4. During an oxidation process in an oxidation-reduction reaction the species is the reducing agent. 5. Assign oxidation states to all atoms in each compound. a) KMn 4, d) P 4 6 b) Ni e) Na C 4 c) K4[Fe(CN) 6 ] f) F 6. How to calculate a reducing and oxidizing agent equivalent? Is the equivalent of oxidizing and reducing agents a constant value? Give examples. 7. Balance the following oxidation-reduction reactions using the half-reaction method. Identify the oxidizing and reducing agents. a) KMn 4 + Na C 4 + H S 4 MnS 4 + K S 4 + Na S 4 + C + H b) KMn 4 + Na S 3 + H Mn + Na S 4 + KH c) KMn 4 + Na S 3 + KH K Mn 4 + Na S 4 + H d) HI + HN 3 I + N + H e) Cr(H) 3 + H + KH K Cr 4 + H f) MnCl + H + KH H Mn 3 + KCl + H g) K Cr 7 + H S 4 + Na S 3 Cr (S 4 ) 3 + Na S 4 + K S 4 + H h) P + HN 3 + H H 3 P 4 + N i) HCl 3 HCl 4 + Cl + H

16 j) K Cr 7 + KI + H S 4 Cr (S 4 ) 3 + I + K S 4 + H k) Ag + HN 3 AgN 3 + N + H l) CrCl 3 + KMn 4 + KH K Cr 4 + KCl + Mn + H 8. Balance the following skeleton equations. The reactions occur in acidic or basic solution, as indicated in parentheses. a) Mn 4 + S Mn + S (basic) b) I 3 + HS 3 I + S 4 (acidic) c) Cl Cl + Cl (basic) 9. What solutions are used as standard solutions in permanganatometry analysis? 10. What substance is used as an indicator in permanganatometry analysis? 11. A solution of permanganate is standardized by titration with oxalic acid (H C 4 ). It required 8.97 ml of the permanganate solution to react completely with g of oxalic acid. The unbalanced equation for the reaction is Mn 4 (aq) + H C 4(aq) Mn + (aq) + C (g) What is the molarity and normality of the permanganate solution? 1. The iron content of iron ore can be determined by titration with standard KMn 4 solution. The iron ore is dissolved in HCl, and all the iron is reduced to Fe + ions. This solution is then titrated with KMn 4 solution, producing Fe 3+ and Mn + ions in acidic solution, required 38,37 ml of 0,0198 N KMn 4 to titrate a solution made from 0,618 g of iron ore. What is the mass percent of iron in the iron ore? EXAMPLE PRBLEMS Example 1. Balancing oxidation-reduction equations occurring in acidic solution by the half-reaction method Write the equations for the oxidation and reduction half-reactions. For each half reaction: a. balance all the elements except hydrogen and oxygen; b. balance oxygen using H ; c. balance hydrogen using H + ;

17 d. balance the charge using electrons. If necessary, multiply one or both balanced half-reactions by integers to equalize the number of electrons transferred in the two half-reactions. Add the half-reactions, and cancel identical species. Check to be sure that the elements and charges balance. Balance the equation KMn 4 + FeS 4 + H S 4 MnS 4 + K S 4 + Fe (S 4 ) 3 + H using the half-reaction method. Solution The ionic equation is K + +Mn 4 +Fe + +S 4 +H + +S 4 Mn + +S 4 +K + +S 4 +Fe 3+ +3S 4 + H - Mn 4 + Fe + + H + Mn + + Fe 3+ + H First, carefully observe products and reactants to detect which ions, molecules, or atoms are oxidized or reduced. Examination of our unbalanced equation shows that Fe + changes to Fe 3+ and that Mn - 4 changes to Mn +. We therefore write our two half-reactions as Fe + Fe 3+ (incomplete half-reaction) - and Mn 4 Mn + (incomplete half-reaction) Balance each half-reaction. For the reduction reaction, we have Mn 4 (aq) Mn + (aq) a) The manganese is balanced. b) We balance oxygen by adding 4H to the right side of the equation: Mn 4 (aq) Mn + (aq) + 4H (l) c) Next, we balance hydrogen by adding 8H + to the left side: 8H + (aq) + Mn 4 (aq) Mn + (aq) + 4H (l)

18 d) All the elements have been balanced, but we need to balance the charge by using electrons. At this point we have the following charges for reactants and products in the reduction half-reaction: 8H + (aq) + Mn 4 (aq) Mn + (aq) + 4H (l) We can equalize the charges by adding five electrons to the left side: 8H + (aq) + Mn 4 (aq) + 5e - Mn + (aq) + 4H (l) Both the elements and the charges are now balanced, so this represents the balanced reduction half-reaction. For the oxidation reaction, Fe + (aq) Fe 3+ (aq) The half-reaction for iron is balanced in iron but not in charge. Thus, we add one electron to the right side. Fe + (aq) Fe 3+ (aq) + e - Because in Fe (S 4 ) 3 molecule there are two iron atoms, we may write oxidation reaction in such a way: oxidation. Fe + (aq) Fe 3+ (aq) + e - This half-reaction is balanced. Note that electrons are lost, as we expect for The last step is to equalize the number of electrons lost and gained in oxidation and reduction reactions. For that multiply the reduction reaction by, and oxidation reaction by 5 before the addition: 8H + (aq) + Mn 4 (aq) + 5e - Mn + (aq) + 4H (l) x Fe + (aq) Fe 3+ (aq) + e - x 5 16H + (aq) + Mn 4 (aq) + 10Fe + (aq) Mn + (aq) + 8H (l) +10Fe 3+ (aq) KMn FeS 4 + 8H S 4 MnS 4 + K S 4 + 5Fe (S 4 ) 3 + 8H

19 Example. Balancing oxidation-reduction equations occurring in basic solution by the half-reaction method Write the equations for the oxidation and reduction half-reactions. For each half reaction: a. balance all the elements except hydrogen and oxygen; b. balance oxygen and hydrogen atoms by adding H - or H ; c. balance the charge by using electrons. If necessary, multiply one or both balanced half-reactions by integers to equalize the number of electrons transferred in the two half-reactions. Add the half-reactions, and cancel identical species. Check to be sure that the elements and charges balance. Balance the equation MnCl + H + KH H Mn 3 + H + KCl by using the half-reaction method. Solution The ionic equation is: Mn + + Cl + H + K + + Cl H Mn 3 + H + K + + Cl Mn + + H H Mn 3 + H Mn + H Mn 3 (incomplete half-reaction) H H (incomplete half-reaction) Balance each half-reaction. For the oxidation reaction, we have Mn + (aq) H Mn 3 (s) a) The manganese is balanced. b) We balance oxygen and hydrogen atoms by adding H - to the left side and H molecules to the right side of the equation: Mn + (aq) + 4H (aq) H Mn 3 (s) + H (l)

20 c) All the elements have been balanced, but we need to balance the charge by using electrons. At this point we have the following charges for reactants and products in the reduction half-reaction: Mn + (aq) + 4H (aq) H Mn 3 (s) + H (l) + e We can equalize the charges by adding two electrons to the right side: Both the elements and the charges are now balanced, so this represents the balanced oxidation half-reaction. Note that electrons are lost, as we expect for oxidation. For the reduction reaction H (l) H (l) H (l) + H + (aq) + e - H (l) The last step is to equalize the number of electrons lost and gained in oxidation and reduction reactions. In our equations they are equal. Add the halfreactions, and cancel identical species: Mn + (aq) + 4H (aq) H Mn 3 (s) + H (l) + e - H (l) + H + (aq) + e - H (l) Mn + (aq) + H (l) + H + (aq) + 4H (aq) H Mn 3 (s) + 3H (l) H + H MnCl + H + KH H Mn 3 + H + KCl xidation-reduction reactions are commonly used as a basis for volumetric analytical procedures. For example, a reducing substance can be titrated with a solution of a strong oxidizing agent, or vice versa. ne of the most frequently used oxidizing agents is aqueous solution of potassium permanganate. The reaction that occurs in acidic solution is the one most commonly used: Mn 4 (aq) + 8H + (aq) + 5e - Mn + (aq) + 4H (l)

21 Permanganate has the advantage of being its own indicator the Mn - 4 ion is intensely purple, and the Mn + ion is almost colorless. As long as some reducing agent remains in the solution being titrated, the solution remains colorless (assuming all other species present are colorless), since the purple Mn - 4 ion being added is converted to the essentially colorless Mn + ion. However, when the whole of the reducing agent has been consumed, the next drop of permanganate titrant will turn the solution being titrated light purple. Thus the endpoint (where the color change indicates the titration should stop) occurs approximately one drop beyond the stoichiometric point (the actual point at which the whole of the reducing agent has been consumed completely).. LABRATRY WRK.1 Determination of a normality and titer of potassium permanganate solution by a standard sodium oxalate solution. Fill a buret with a potassium permanganate solution. Using a measuring pipette transfer 5 ml of standard sodium oxalate, Na C 4, solution, and ml of H S 4 (1:4) solution into a beaker. The mixture is heated to C, till strong vaporization and then the KMn 4 solution is added drop by drop to the hot mixture in the beaker. It should be stirred up to the endpoint of the reaction. The stoichiometric (equivalence) point is marked by appearance of light purple color of KMn 4 excess. Carry out titration to take three concurrent readings. Make a table of the titration readings. Calculate the average volume, and then determine normality and titer of potassium permanganate solution. The KMn 4 solution volume is estimated by the upper meniscus in the buret. Write the equation of the reaction. Calculate the KMn 4 normality and titer in two ways.

22 .. Determination of normality and titer of hydrogen peroxide solution by the titrant of potassium permanganate. Fill a buret with a potassium permanganate solution. Using a measuring pipet transfer 5 ml of H solution (the solution has already contained sulfuric acid) into a beaker. The KMn 4 solution is added drop by drop to the solution in the beaker constantly stirring up to the endpoint of the reaction. The stoichiometric (equivalence) point is marked by appearance of light purple color of KMn 4 excess. Carry out titration to take three concurrent readings. Make a table of the titration readings. Calculate the average volume, and then determine normality and titer of hydrogen peroxide solution. The KMn 4 solution volume is estimated by the upper meniscus in the buret. Write the equation of the reaction. Calculate the normality and titer of H solution by two ways. Calculate the equivalent of oxidizing and reducing agents for the given reaction. Calculate the mass of H in 100 ml of solution. CLASS 6. TPIC: xidation-reduction measure analysis. (Iodometry). 1. SEMINAR. 955 min.) Method of iodometry. 1.. Reactions lying in the base of the iodometry xidizing and reducing agents in iodometry Determination of the endpoint in iodometry. Indicators of the method Direction of oxidation - reduction reactions. Redox potential.. LABRATRY WRK. (40 min.).1. Determination of normality and titer of sodium thiosulfate solution by a standard potassium dichromate solution.

23 .. Determination of normality and iodine solution titer by a titrant of sodium thiosulfate. 3. CNTRL WRK. (10 min.). Literature. 1. Lectures.. Ebbing D.D. General Chemistry/ D.D.Ebbing, M.S.Wrighton;.- Boston. Third edition. Houghton Mifflin Company. Р Lister T. Chemistry for Advanced Level / T. Lister, J. Renshaw; - Third edition. Stanley Thornes (Publishers) Ltd Р PRACTICAL PRBLEMS 1. What major reactions are used in iodometry?. Continue the red-ox reactions, calculate coefficients by a half-reaction method and calculate the equivalent of reducing and oxidizing agents. a) NaBr 3 + KI + H S 4 b) KI 3 + Na S 3 + H S 4 c) NaN + KI + H S 4 N +. d) I + Cl + H HI 3 +. e) Ca(Cl) + KI + H S 4 CaCl.. 3. A sample of H 50-mL and excess of KI and HCl was titrated by 18. ml of 0.01 N Na S 3 solution. Calculate the mass of active chlorine in 1 liter of water. 4. What is the mass of iodine (I ) which is nessesary for obtaining of liters of iodine solution with I T = 0,001575g / ml NaS 3.

24 5. What is the volume of 5% I solution (ρ= 0,95 g/ml) may be obtained from 10 g of crystalline iodine? Calculate molarity and nomality of 5% solution. 6. Calculate the equivalence factor for reducing agents in the given red-ox reactions: a) NaN + KI + H S 4 I + N + Na S 4 + K S 4 + H b) KCl + H + KCl + H 7. Calculate the equivalence factor for oxidizing agents in the given red-ox reactions: a) NaBr 3 + KI + H S 4 NaBr + I + K S 4 + H b) KI 3 + Na S 3 + H S 4 KI + I + Na S 4 + K S 4 + H 8. Triiodide ions are generated in solution by the following (unbalanced) reaction in acidic solution: I 3 (aq) + I (aq) I 3 (aq) Triiodide ion is determined by titration with a sodium thiosulfate (Na S 3 ) solution. The products are iodide ion and tetrathionate ion (S ). a. Balance the equation for the reaction of I 3 - with I - ions. b. Write and balance the equation for the reaction of S 3 - with I 3 - in acidic solution.. LABRATRY WRK.1. Determination of normality and titer of sodium thiosulfate solution by a standard potassium dichromate solution. Fill a buret with a sodium thiosulfate solution. Put 5 ml: of standard potassium dichromate, K Cr 7, solution by using a measuring pipette, into the beaker, add 3 ml of 3% potassium iodide solution and ml of sulfuric acid solution (1:4). Shake the mixture, cover the beaker with glass and allow the solution to rest for 5 minutes. Then the educed iodine is titrated by sodium thiosulfate solution till a faint-yellow (straw) color appears. Then add 5 drops of 1% starch solution into the mixture (the solution will get blue) and continue titration, shaking the solution intensively till a light blue color, which is to

25 disappear from one drop of the sodium thiosulfate solution. At the equivalent point the solution has a greenish shade due to the presence of Cr 3+ ions. Carry out titration to take three concurrent readings. Make a table of the titration readings. Calculate the average volume, and then determine the normality and titer of the sodium thiosulfate solution. Write the equation of the reaction. Calculate the Na S. 3 5H normality and titer in two ways.. Determination of normality and titer of iodine solution by a titrant of sodium thiosulfate. Fill a buret with a sodium thiosulfate solution. Put 5 ml of iodine solution by using a measuring pipette into the beaker. Then the iodine solution is titrated by sodium thiosulfate solution till a faint-yellow (straw) color appears. Then add 5 drops of 1% starch solution into the mixture (the solution will get blue) and continue titration, shaking the solution intensively till a light blue color, which is to disappear from one drop of the sodium thiosulfate solution. CLASS 7. TPIC: Elements of chemical thermodynamics. 1. SEMINAR. (60 min.) The main notions of thermodynamics: a thermodynamic system; types of thermodynamic system (isolated, closed, opened); intensive and extensive parameters of a system; internal energy; work and heat are the forms of energy transmission; thermodynamic processes (isothermal, isobaric, isochoric); thermodynamically reversible and irreversible processes ; a standard condition. 1.. The first law of thermodynamics. Enthalpy. Exo-, and endothermic reactions. A standard enthalpy of substance formation; standard enthalpy of substance combustion.

26 1.3. A chemical reaction standard enthalpy. Hess s Law, consequences of Hess s law. Enthalpy of phase transitions and dissolving of substances The second Law of thermodynamics. Entropy. The standard entropy of substance. Spontaneous and non spontaneous processes. Gibb s energy. Prognosis of the direction of spontaneously proceeding processes in isolated and closed systems; entropy and enthalpy factors. Thermodynamics conditions of equilibrium The standard Gibbs energy of substance formation. The standard Gibb s energy of chemical reactions Examples of exothermic and endothermic processes in the body. The principle of energetic conjugation Chemical thermodynamics as a theoretical foundation of chemical and biochemical processes.. LABRATRY WRK. (50 min.)..1. Determination of neutralization reaction enthalpy... Determination of copper sulfate hydration enthalpy. Literature. 1. Lectures.. Ebbing D.D. General Chemistry/ D.D.Ebbing, M.S.Wrighton;.- Boston. Third edition. Houghton Mifflin Company. Р Lister T. Chemistry for Advanced Level / T. Lister, J. Renshaw; - Third edition. Stanley Thornes (Publishers) Ltd Р Given: PRACTICAL PRBLEMS

27 S(s) + 3 (g) S 3 (g) H = -395, kj S (g) + (g) S 3 (g) Calculate H for the reaction S(s) + ( (g) S (g) H = -198, kj. Hydrogen sulfide gas is a poisonous gas with the odor of rotten eggs. It occurs in natural gas and is produced during the decay of organic matter, which contains sulfur. The gas burns in oxygen as follows: H S(g) + 3 ( (g) H (l) + S (g) Calculate the standard enthalpy change for this reaction by using standard enthalpies of formation. (See Appendix, Table 3.) 3. Predict the sign of S for each of the following changes: a) AgCl(s) Ag + (aq) + Cl - (aq) b) H (g) + (g) H (l) c) 3 (g) 3 (g) 4. For the reaction Al(s) + 3Br (l) AlBr 3 (s) S 0 is equal to -144 J/K. Use this value and the data from Appendix to calculate the value of S 0 for solid aluminum bromide. 5. From the data in Appendix calculate H 0, S 0, and G 0 for each of the following reactions at 5 o C. a. CH 4 (g) + (g) C (g) + H (g) b. 6C (g) + 6H (l) C 6 H 1 6 (s) + 6 (g) Glucose 6. For the reaction at 98 K, N (g) N 4 (g) The values of H 0 and S 0 are kj/mol and J K -1 mol -1, respectively. What is the value of G 0 at 98 K? Assume that H 0 and S 0 do not depend on temperature. At what temperature is G 0 =? Is G 0 negative above, or below, this temperature?

28 LABRATRY WRK. We can determine the heat associated with a chemical reaction experimentally by using a device called a calorimeter. Calorimetry, the science of measuring heat, is based on observing the temperature change when a body absorbs or discharges energy as heat. Substances respond to heating differently. The heat required to raise the temperature of a substance is called its heat capacity, q C =. t The heat capacity (C) of a sample of substance is the quantity of heat to raise the temperature of the sample of substance by one degree Celsius (or one Kelvin). Changing the temperature of the sample from an initial temperature to a final temperature t f requires heat equal to q=c t, where t is the change of temperature and equals t f -t i. Heat capacity is directly proportional to the amount of substance. ften heat capacities are listed for molar amounts of substances. The molar heat capacity of a substance is the heat capacity for one mole of substance. It has the units J K -1 mol -1 or J o C -1 mol -1. Heat capacities are also compared for one-gram amounts of substances. The specific heat capacity (or simply specific heat) is the quantity of heat required to raise the temperature of one gram of a substance by one degree Celsius (or one Kelvin) at constant pressure, which has the units JK -1 g -1 or J o C -1 g -1. Although the calorimeters used for highly accurate work are precision instruments, a very simple calorimeter can be used to examine the fundamentals of calorimetry. All we need are two nested Styrofoam cups with a Styrofoam cover through which a stirrer and a thermometer can be inserted, as shown in Figure 1. This device is called a coffee cup calorimeter. The outer cup is used to provide extra insulation. The inner cup holds the solution in which the reaction occurs. The measurement of heat by using a simple calorimeter such as that shown in Figure 1. is an example of constant-pressure calorimetry. Constant-pressure calorimetry is used in determining changes in enthalpy occurring in solution. Under these conditions the enthalpy equals the heat.

29 Figure 1. A simple coffee-cup calorimeter. After reactants are added, the cup is covered to reduce heat loss by evaporation and convection. The heat of reaction is determined by noting the temperature rise or fall..1. Determination of neutralization reaction enthalpy. The method of carrying out the experiment. A group of students defines the heating affect of the neutralization reaction of a given variant: an acid a base. NaH + HCl = NaCl + H ; H 1 ; NaH + H S 4 = Na S 4 + H ; H ; NaH + CH 3 CH = CH 3 CNa + H ; H 3 ; NH 4 H + HCl = NH 4 Cl + H ; H 4 ; Put 100 ml of 1N acid solution in a calorimeter. In 5 minutes take the temperature of the solutions up to 0,1 C accuracy (t i ). Then add 100 ml of 1N base solution. Stir the solution with a stirring rod without stopping. When these reactants (both originally at the same temperature) are mixed, the temperature of the mixed solution is observed to increase (t f ). Thus the chemical reaction must be releasing energy as heat. This increases the random motions of the solutions components, which in turn increases the temperature. The quantity of energy released can be determined from the temperature increase, the mass of the solution, and the specific heat capacity of the solution. Working up of the experimental data. For an approximate result we will assume that the calorimeter does not absorb or leak any heat and that the solution can be treated as if it were pure water with a density of 1,0 g/ml.

30 We also need to know the heat required to raise the temperature of a given amount of water by 1 o C. The specific heat capacity (C) of water is 4,18 J 0 C -1 g From these assumptions and definitions we can calculate the heat (change in enthalpy) for the neutralization reaction using the formula: q = C m (t f t i ), where m the mass of the reaction mixture, the density of the reaction mixture is equals to 1g/mL.. Calculate the number of moles of the water, formed in the result of the neutralization reaction, then recalculate the heating effect of neutralization per 1 mole of water. 3. Record into the Table 1. results of the experiments carried out by other students and compare the data of heat of neutralization by various variants of reactions acid base: H 1 ; H ; H 3 ; H 4. Table 1. Results of the experiments Volume of acid solution, ml Volume of base solution, ml Initial temperature, Ti Final temperature, Tf Change of temperature, T Volume of mixture Mass of mixture Heat effect, q of experiment, J Heat,Q, of neutralization reaction, J H of neutralization reaction, kj.. Determination of copper sulfate (II) hydration enthalpy. The method of carrying out the experiment. a. Put 100mL of distilled water in a calorimeter. In 5 minutes take the temperature with 0,1 C accuracy (t i ). Transfer a 3,-g sample of a waterless copper

31 (II) sulfate, CuS 4, into the calorimeter, stir it quickly with a stirring rod and mark the maximum temperature. The pure water has a density of 1,0 g/ml. The specific heat capacity (C) of water is 4,18 J 0 C -1 g -1. Working up of the experimental data. 1. Calculation of the heat for dissolving of waterless CuS 4 according to the equation: q = c m 1 (t t 1 ), Where m 1 is the mass of CuS 4 solution, g.. Calculate the number of CuS 4 moles which were dissolved in water. 3. Recalculate the heat of waterless copper (II) sulfate dissolving per 1 mol of the substance ( H 1 ). Η 1 = b. Put 100mL of distilled water in a calorimeter. In 5 minutes take the temperature with 0,1 C accuracy (t i ). Transfer a 5.0-g sample of a copper (II) sulfate pentadydrate, CuS 4 5H, into the calorimeter, stir it quickly with a stirring rod and mark the maximum temperature. 4. Calculation of the heat for dissolving of CuS 4. 5H according to the equation: q = c m (t f t i ), water. Where m is the mass of CuS 4. 5H solution, g. 5. Calculate the number of CuS 4. 5H moles which were dissolved in 6. Recalculate the heat of copper (II) sulfate pentahydrate dissolving per 1 mol of the substance ( H ). Η 7. The enthalpy of CuS 4 hydration is calculated by the formula: = q n 1 q n Table. H hydration = H 1 - H Results of the experiments

32 Repeat the experiment with a 5-g sample of copper Solution mass, g (m1) ti initial temperature tf maximal temperature of dissolving CuS4 H1 heat of dissolving CuS4 Sample CuS4 5H, g Solution mass, g (m) tf final (maximum) temperature at dissolving H heat of dissolving CuS4. 5H Hhydration = H1 - H СLASS 8. TPIC: Elements of chemical and biochemical reaction kinetics. 1. SEMINAR. 960 min.) The subject and basic notions of chemical kinetics. 1.. Rate of reaction, the average rate and instantaneous rate. Classification of reactions: homogeneous, heterogeneous, simple and complex Molecularity of a reaction. The order of a reaction. Dependence of a reaction rate on concentration. Kinetic equations of zero order reactions, the first order and the second order reactions. A rate constant. A half - life of a reaction. The experimental determination of a reaction rate, rate constant and order of a reaction Dependence of reaction rate on temperature. Temperature coefficient of a reaction rate. Arrhenius equation. The activation energy. Energy profile of a reaction. Activated complex theory. Catalysis. Homogeneous and heterogeneous catalyst. Energy profile of a catalyzed reaction. Fermentative catalysis.. LABRATRY WRK. (50 min.)..1. Determination of rate and order of sodium thiosulfate decomposition reaction... Determination of order of iodide ions oxidation reaction with Fe 3+ ions

33 (the Clock Reaction). Literature. 1. Lectures.. Ebbing D.D. General Chemistry/ D.D.Ebbing, M.S.Wrighton;.- Boston. Third edition. Houghton Mifflin Company. Р Lister T. Chemistry for Advanced Level / T. Lister, J. Renshaw; - Third edition. Stanley Thornes (Publishers) Ltd Р , PRACTICAL PRBLEMS 1. Write the rate law for the following reactions: a) N (g) + (g) N(g) b) H (g) + (g) H (l) c) C (g) + C(s) C(g) d) Zn(s) + Cl (g) ZnCl (s). The reaction of nitrogen monooxide and chlorine is the third order. The reaction proceeds according to the equation: N(g) + Cl (g) NCl. How long would it take a) for the concentration of N to increase in double; b) for the concentration of Cl to decrease in double: c) for the concentration of both reactants to increase in double? 3. For the reaction of nitric oxide, N, with chlorine, N (g) + Cl (g) NCl the observed rate law is Rate = k[n] [Cl ]. What is the reaction order with respect to nitric oxide, with respect to Cl? What is the overall order?

34 4. The temperature of the reaction increases by How does the constant rate increase? (γ=3,). 5. Calculate the temperature coefficient (γ), if you know that the constant rate at 10 0 C equals 5, /s, and at C it equals 6, /s.? 6. Methyl isocyanide, CH 3 NC, isomerizes when heated to give acetonitrile, CH 3 CN: CH 3 NC(g) CH 3 CN(g). The reaction is the first order. At 30 0 C, the rate constant for the isomerization is 6, /s. What is the half-life? How long would it take for the concentration of CH 3 NC to decrease to 5% of its initial? 7. Sulfuryl chloride, S Cl, decomposes when heated. S Cl (g) S (g) + Cl (g) In an experiment, the initial concentration of sulfuryl chloride was 0,048 mol/l. If the rate constant is /s, what is the concentration of S Cl after 4,5 hr? The reaction is the first order. 8. The rate constant for the formation of hydrogen iodide from the elements H (g) + I (g) HI(g) is, l/mol. s at 600 K and 3, l/mol. s at 650 K. a. Find the activation energy E a. b. Then calculate the rate constant at 700 K. LABRATRY WRK..1. Determination of rate and order of sodium thiosulfate decomposition reaction. Sodium thiosulfate decomposes in sulfuric acid solution by the equation: Na S 3 + H S 4 = Na S 4 + S + S + H The reaction goes on in two stages: a. Na S 3 + H S 4 = Na S 4 + H S 3 ; b. H S 3 = S + S + H The first step proceeds very fast, the second one proceeds slowly, that s why the rate of the overall process is defined by the second stage.

35 Put the mixture of volume (A) of 0, 1M Na S 3 solution and volume (B) of distilled water into 5 test-tubes in ratio, given in the Table 3: Table 3. Results of the experiments Volume (number of drops) Concentration Na S 3 (A) H (B) H S 4 (С) of Na S , , , , ,05 Time of the solution dimness stativity τ (sec). R = τ 1 ( sec ) 1 Put 5 drops of 1M H S 4 solution into other 5 test-tubes. Then mix each sodium thiosulfate solution with H S 4 solution; switch on the stop watch (you can use a watch with a second hand) and fix the time (τ) when white sulfur appear. Carry out the same experiment with the other pairs of solutions. Enter the reading in the table. Calculate conditional rate of the reaction and make up a graph of conditional rate dependence on sodium thiosulfate concentration. Evaluate the order of the studied reaction. It is known that for the zero, the first and the second order reactions, the rate dependence on the concentration is expressed as follows: R R R Zero orger The first order The second order C C C.. Determining the order of the reaction from the rate law. xidation of iodide ions are by Fe 3+ ions (the Clock Reaction)

36 Iodide ions oxidation proceeds by the equation: KI + FeCl 3 = FeCl + I + KCl The iodide formation is fixed by the appearance of blue-colored complex in the presence of starch. Carry out experiments to study the reaction rate dependence on FeCl 3 concentration. table. f the testtube Place the components of reactional mixture in 4 test-tubes according to the Table 4. The reagents volume (number of drops) FeCl 3 HN 3 H 0,0 М 0,1 М starch (а) (b) (c) Results of the experiments Concentration of FeCl 3 solution, mol/l , ,5 Time of starchiodine complex appearance τ(sec) R = τ 1 ( sec ) 1 Then add 8 drops of 0,0M KI into the first test-tube and fix the time when the solution gets blue. Carry out the same experiment with the other reactants. Enter the results in the table; calculate conditional rate of the reaction and make up a graph of rate dependence on the concentration of FeCl 3 solution. Make a conclusion about a particular order of a reaction by Fe 3+ ion. The reaction order with respect to a given reactant species equals the exponent of the concentration of that species in the rate law, as determined experimentally. The initial method is a simple way to obtain reaction orders. It consists of doing a series of experiments in which the initial, or starting, concentration of reactants is varied. Then the initial rates are compared, from which the reaction

37 orders can be deduced. In experiments we calculate the rate of I formation with respect to each reactant species (concentration of one of the reactant is changed, and the second reactant concentration remains constant). The rate law for a given reaction is: Rate 1 = k[i ] n [Fe 3+ ] n1 1 = k observed. [Fe 3+ ] n1 1. Rate = k[i ] n [Fe 3+ ] n1 = k observed. [Fe 3+ ] n1. That is Rate = k. [Fe 3+ ] n 1 Applying logarithms, we get: ln Rate = lnk + n 1 ln[fe 3+ ] It is possible to determine the order of a reaction by graphcal plotting of the data for a particular experiment. The value of the reaction order (n 1 ) is the tangent of angleα in the Figure. A particular order by I - ions (n ) can be obtained in the same way. lnr α b a a tg α = = b lnc n 1 Figure. A plot of lnr versus lnc. СLASS 9. TPIC: Thermodynаmic and kinetic conditions of chemical equilibrium. 1. SEMINAR. (5 min.) Reversible and irreversible chemical reactions.