General Chemistry Experiments with Dr. Nabil Al-Sahly

1. King Saud UniversityCollege of SciencesDepartmentofChemistry Practical General Chemistry 101 chem & 104 chem Nabil Al-Sahly .

2. Introduction: • Reference Book:Practical General Chemistry By : Dr. Ahmad Al-Owais Prof. Abdulaziz Al- wassil • Reports File :Present In building No. 5

3. Distribution of Marks: • Theoretical Part: In Class of (70 Mark) 2. Practical Part: In Laboratory of (30 Mark) a) 6 or 5 Experiments then exam 1 of (10 Marks) b) 6Experiments thenexam 2 of (10 Marks) c) Attendance on time & Reports (10 Marks)

4. What must on the student? • Attendance at the time. • Lab Coat. • Reports File. • Pin – Calculator • Pencil & rulerin case Drawing (Graph)

5. Exp(1):Determination of a liquid Density. • Density: is mass per volumeunit. d=m(g)/V(ml) m: is mass V: is volume • 1 ml = 1cm3 • Density is a physical property of asubstance. • In this experiment we will measure density of water • Density of Water (H2O)= 1 g/ml

6. ***************** Before starting the experiment...you should know What are tools(equipments) that we will use it for experiments?.. and How we use it?

7. * Important of Tools(glassware) Used: Beakers Conical Flask(Erlenmeyer Flask) Funnel Burette Pipette Volumetric Flask

8. Main Tools (glassware) used: Burette clamp Burette Valve (tap) Conicalflasks Beakers Waste beaker Stand Indicator Distilled water Volumetric pipette glass funnel Balance Bulb

9. How we use glassware ??? Burette : 1 2 3 Pipette:

10. Very..Very important OK

11. You can reading the volume ? . . . V = ………………ml

12. Equipments (glassware) used: • 2beakers (glass cups) • Burette • Graduated pipette • Plastic Funnel • Stand • Instruments(apparatuses) used: • Digital (Electronic)Balance .

13. Procedure: • Clean all of your glassware with water. • Find mass of the empty beaker in g =m1 • Add the given volume of water(H2O) to the empty beaker and find mass of the beaker and H2O in g =m2 • Calculate mass of H2O in g (m2 – m1) = m 5 . Calculate density of H2O in g / ml = m / V = d 6. Repeat the steps 3 to 5

14. * Calculation of the density(d) by graph: • Plot the relationship between the mass of H2O(m) on the Y-axis versus its volume of H2O(V) on the X-axis , and find density(d) of H2O from the slope = y2 – y1 / x2 – x1 = d y2 –y1 x2 –x1 d y2 Slope = = m(g) y1 1 g/ml x1 x2 V(ml)

15. Exp(2): *Introduction of Titrations(Reactions of Neutralization). and *Preparation of a standard solution of Sodium Carbonate (Na2CO3) (0.05 M). • Titration: is method of quantitative chemical analysis • that is used to determine the concentration of an unknown solution by a standard solution. OR • Titration: is a process transfer of a solution from a burette (called the titrant) into a measured volume of another solution for determining an unknown solution concentration.

16. Titration Terminology: .Equivalent(equivalence) point:is the volume of titrant required to neutralize the sample (No. moles acid = No. moles base). • End point:is the pH value at the equivalent point of a titration. • Indicator:is a chemical which is added to the sample that changes color at the equivalent point of a titration. (Indicator is used to show when neutralization is occurs) • A standard Solution: is a solution of known concentration. • Remember that: Acid + Base Salt + Water • Example: HCl + NaOH  NaCl + H2O

17. Summary of Titration: Burette Clamp Titrantsolution (acidorbase) Valve Stand Anothersolution Conical flask KnownofVolume 2 drops ofindicator

18. The Concentration:There are several ways of expressing concentration and but we will focus on : • Molarity(M): Is number of moles of solute dissolved in one liter of solution. M= n(mol) / V(L) • In case of the titration : M × V/n = M’ × V’ /n’ This is law very important. • Strength of Solution(S): Is number of grams of solute dissolved in one liter of solution. Relation between M & S : S = M ×M.wt S=………. g/l S=m(gm) of solute / V(L)of Solution

19. pH Range 0 1 2 3 4 5 6 7 8 910 11 12 13 14 Neutral [H+]>[OH-][H+] = [OH-][OH-]>[H+] Acidic Basic LecturePLUS Timberlake

20. pH Calculations (PH laws): pH [H+] pH = -log[H+] [H+] = 10 -pH pH + pOH = 14 [H+] [OH-] = 1 x10-14 pOH [OH-] pOH = -log[OH-] [OH-] = 10 -pOH

21. Example: The [H+] of lemon juice is 1.0 x 10-3 M . What is the [OH-] , PH and POH? [OH-] = 1.0 x 10 -14 = 1.0 x 10-11 M 1.0 x 10 -3 PH= - log [H+] = -log(0.001) = 3 POH = - log [OH- ] = -log(1.0 x 10-11 ) = 11 OR POH + PH =14   POH = 14 – 3 = 11 IS the solution acidic or alkaline(basic) or neutral ?It’s.................. LecturePLUS Timberlake

22. Basic Laws: n = m / M.wt & M = n / V(L) Where: n:No. moles of solute m: mass M.wt:molecular weight (molar mass) M:molarity V: volume of solution m = M × V(ml) ×M.wt / 1000 This is law used for Preparation of Solution from Solid Substance (material).

23. ToPreparation of Solution of Sodium Carbonate (Na2CO3) in( 100 ml) & (0.05 M ) We apply the previous law: m= 0.05 ×100 ×106 / 1000 =0.53 gmof Na2CO3 # Equipments and Material used: Sodium Carbonate (Na2CO3). Volumetric flask (100ml). ElectronicBalance . …

24. Procedure: 1-Weigh out about(0.53 gm)of Na2CO3 . 2- Put it involumetric flask(100 ml) . 3- Dissolve it with little of water(H2O). 4-Complete with water until the mark. 100 ml Solution of Na2CO3(0.05M)

25. Exp(3)Determination of Organic Indicators for Acid base Titrations At first we will know on methods of measuring pH, where there are two ways …. A) Using pH meter.B) Using pH paper, a special type of paper.. B A PH paper PH meter

26. Indicators Used :

27. Theidea of ​​the experiment: • To determine the suitable indicator we will make 2 Titrations : A) a strong acid with strong base .. B) a weak acid with a strong base .. and draw a graphic relationship between the volume of base added and the pH value .. where we get Curve called (Titration Curve) ... through which we can know on the suitable indicator.

28. Titration Curves:  A titration curve is a graph of the pHchanges that occur during an acid-base titration versus the volumeof acid or base added. The objective (purpose) of study of titration curve:Is to know the suitable Indicator for acid-base titration. The objective of titration: IS to know the concentration of an unknown solution. But ,

29. Procedure: • Where we will be adding different volumes of the burette filled with a strong base(NaOH) to beaker by the presence of known amount of strong acid(HCl) or weak acid(CH3COOH), then follow the change in pH value. Generally: This figure shows the titration curve for both cases.

30. A)Titration between a strong acid(HCl) with a strong base(NaOH)HCl + NaOH  NaCl + H2O ph.ph OK M.o OK *The suitable indicator for this titration is………….......

31. B)Titration between weak acid(CH3COOH)with a strong base(NaOH)CH3COOH + NaOH  CH3COONa + H2O Ph.ph OK M.O *The suitable indicator for this titration is………….......

32. Results from the drawing (graph): • After that draw a relationship between the volume of base added and the change in pH value. • What determine from the drawing (graph)? 1- PH range at the equivalent point (….. to…..). 2- Draw two parallel lines for each indicator based on range of each indicator, of which we can know the suitable indicator (……………….) . 3- Find the volume of base at the equivalent point, through dropping line from the region that occur then the sudden change of the curve on the axis of the volume of base(NaOH) added ( VNaOH = …… ml).

33. Explanation for pH curve: This curve is similar to the previous curve, but the additive is acid and not the base 14 pH endpoint equivalent point 7X equivalent point volume ( V NaOH (ml) ) 0 10 20 30 40 Volume of titrant added (mL)

34. EXP(4): Determination of Sodium Hydroxide(NaOH) Concentration BY Titrations With A Standard Solution of Hydrochloric Acid (HCl) . 1) The titration by : 2) The titration by : Ph.ph M.O HCl + NaOH → NaCl + H2O *Reaction equation: *Results: NaOH M =? V = average ▪ Volume of NaOH: 1- 2- 3- *Calculations : HCl M’ = 0.1 M V’ = 10 ml + 2drops of ph.ph or M.O ▪ Molarity (M) of NaOH : • M × V /n = M’ × V’/n’ ▪ Strength of Solution (S) of NaOH : S = M × M.wt

35. EXP(5):Determination of Acetic Acid (CH3COOH) Concentration BY Titrations With A Standard Solution of Sodium Hydroxide (NaOH) . * The titration by(only) : Ph.ph *Reaction equation: CH3COOH + NaOH CH3COONa + H2O * Results: … Volume of NaOH : 1- 2- 3- *Calculations : .. Molarity ( M ) of CH3COOH : M × V /n = M’ × V’ /n’ .. Strength ( S )of CH3COOH : S = M × M.wt NaOH M = 0.1 M V = average CH3COOH M’ = ? V = 10 ml + 2drops of ph.ph

36. EXP(6) : Determination of Hydrochloric Acid (HCl) Concentration By Titrations With A Standard Solution of Sodium Carbonate (Na2CO3). 2 HCl + Na2CO3 2 NaCl +CO2 + H2O HCl + Na2CO3 NaHCO3 + NaCl 1)The titration by M.O: 2) The titration by ph.ph: HCl M =? V=average *Results: ..Volume of HCl 1- 2- 3- *Calculations: . Molarity of HCl : M × V/n = M’ × V’/n’ .Strength of HCl : S = M × M.wt *Results: ..Volume of HCl 1- 2- 3- *Calculations: . Molarity of HCl : M × V/n = M’ × V’/n’ . Strength of HCl: S = M × M.wt Na2CO3 M’ = 0.05M V’ = 10 ml + 2 drops of indicator Is the molarity in two cases equal (same) or ??

37. EXP(7): Measurement of Gas Diffusion (Graham’s Law of Diffusion). • In this experiment we will measure diffusion rate to 2 gases : NH3(g) & HCl(g) Which is faster diffusion you think ? And Why? The answer :……………fasterthan ……………… Because:...........................................................

38. Molecular diffusion : Diffusion: “gas molecules spreading out to fill a room are diffusion.” Its not easy since an average gas molecule at room temperature and pressure will experience about 10 billion collisions per second! It only travels about 60 nm between collisions!

39. He N2 Which balloon will lose pressure sooner (quickly)?

40. N2 He Small molecules Big molecules NOW: Which balloon will lose pressure sooner ? The answer : ..............

41. Graham’s Law of diffusion: diffusion rate is inversely the proportional to square root of its molar mass or its density. In case 2gases:

42. r = Diffusion rate of gas. M =Molarmass (Molecular Weight) of gas. Example: Comparethe rates of diffusion of HeandN2 . Hediffuses2.65 times as fast as N2

43. Dalton’s Law of Partial Pressures: He H2N2 Ptotal = P1 + P2 + P3 +...

44. EXP(8): Determination of Critical Solution Temperature(C . S . T ) Principle of experiment: • Liquid water and phenol show limited(partially) miscibility below( C.S.T). In this experiment, miscibility temperatures of several water-phenol mixtures of known composition will be measured. Then determination of the critical solution temperature(C.S.T). Warning: toxic H2O

45. …………………………………………………………………………………………………………………… C .S .T : Is temperature at which be all compositions of mixture or solution completely miscible. Above (C . S .T) be Homogeneous . Under (C .S .T) be Heterogeneous . Calculationof percentage: % X in asample (compound or mixture or solution) : = mass of X/ mass of sample × 100 mixture

46. * Diagram between %phenol & miscibility temperature: * C.S.T =………OC Homogeneous T(O C) Heterogeneous %Phenol at C.S.T=…....% %Phenol % H2O at C.S.T = 100 ─ % phenol =….....OC

47. EXP(9): Hess’s law • Hess’s law statesthat “the change in enthalpy H for any chemical reaction is constant, whether the reaction occurs in one step or in several steps” . H1 A B F H2 H4 L W E D C H3 H4 H1 H2 H3

48. Explanation of Hess’s Law: Start Finish A State Function: Does not depend on path . Both lines accomplished the same result, they went from start to finish. Net result = same.

49. In this experiment you will measure: the change in enthalpy(H ) of through knowing amount of heat absorbed(q), where: # q = m . P . t q: amount of the heat m: mass P: Specific heat (constant) t :The difference in temperature(t2 – t1) # The Change in Enthalpy (H) = Q / n Q: Total (q1 + q2) n: No. moles of NaOH

50. Specific heat & Heat capacity: .Specific heat : Is the amount of heat (energy) required to raise the temperature of (1 g ) of a substance by (1C). .Heat capacity: Is the amount of heat (energy) required to raise the temperature of something of a substance by (1 C). H H = + endothermic Absorb heat that H = ─ exothermic Liberate heat