King Saud University College of Sciences Department of Chemistry Practical General Chemistry 101 chem & 104 chem Nabil Al-Sahly . Introduction: • Reference Book: Practical General Chemistry By : Dr. Ahmad Al-Owais Prof. Abdulaziz Al- wassil • Reports File :Present In building No. 5 Distribution of Marks: 1. 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) 6 Experiments then exam 2 of (10 Marks) c) Attendance on time & Reports (10 Marks) What must on the student? • • • • • Attendance at the time. Lab Coat. Reports File. Pin – Calculator Pencil & ruler in case Drawing (Graph) Exp(1):Determination of a liquid Density. • Density: is mass per volume unit. d= m(g) / V(ml) m: is mass V: is volume 1 ml = 1 cm3 Density is a physical property of a substance. In this experiment we will measure density of water Density of Water (H2O)= 1 g/ml ***************** Before starting the experiment...you should know What are tools(equipments) that we will use it for experiments?.. and How we use it? * Important of Tools(glassware) Used: Beakers Conical Flask(Erlenmeyer Flask) Volumetric Flask Funnel Pipette Burette Main Tools (glassware) used: Burette clamp Burette Valve (tap) Conical flasks Beakers Waste beaker Stand Indicator Volumetric pipette Bulb glass funnel Balance Distilled water Burette : 1 Pipette: 2 3 Very..Very important OK You can reading the volume ? . . . V = ………………ml Equipments (glassware) used: • 2 beakers (glass cups) • Burette • Graduated pipette • Plastic Funnel • Stand Instruments(apparatuses) used: • Digital (Electronic) Balance . Procedure: 1. Clean all of your glassware with water. 2. Find mass of the empty beaker in g = m1 3. Add the given volume of water(H2O) to the empty beaker and find mass of the beaker and H2O in g = m2 4. Calculate mass of H2O in g (m2 – m1 ) = m 5 . Calculate density of H2O in g / ml 6. Repeat the steps 3 to 5 = m/V= d * 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 Slope = m (g) y2 –y1 = x2 –x1 d y1 1 g/ml x1 x2 V (ml) Exp(2): * Introduction of Titrations(Reactions of Neutralization). and *Preparation of a standard solution of Sodium Carbonate (Na2CO3) (0.05 M). : is method of quantitative chemical analysis • that is used to determine the concentration of an unknown solution by a standard solution. OR : 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. Titration Terminology: :is the volume of titrant required to neutralize the sample (No. moles acid = No. moles base). :is the pH value at the equivalent point of a titration. :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) : is a solution of known concentration. Acid + Base • Example: HCl + NaOH Salt + Water NaCl + H2O Clamp Burette Titrant solution (acid or base) Valve Stand Conical flask Another solution Known of Volume 2 drops of indicator 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 pH Range 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Basic Acidic Neutral [H+]>[OH-] [H+] = [OH-] LecturePLUS Timberlake [OH-]>[H+] 19 pH Calculations (PH laws): pH pH = -log[H+] [H+] = 10 -pH [H+] [OH-] = 1 x10-14 pH + pOH = 14 pOH [H+] pOH = -log[OH-] [OH-] = 10 -pOH [OH-] 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 ? LecturePLUS Timberlake 21 It’s.................. Basic Laws: Where: n: No. moles of solute m: mass M.wt:molecular weight (molar mass) M: molarity V: volume of solution This is law used for Preparation of Solution from Solid Substance (material). To Preparation 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 gm of Na2CO3 # Equipments and Material used: Sodium Carbonate (Na2CO3). Volumetric flask (100 ml). Electronic Balance . … Procedure: 1-Weigh out about(0.53 gm)of Na2CO3 . 2- Put it in volumetric 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) 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 PH paper A PH meter Indicators Used : Indicator Phenolphthalein (ph.ph) Methyl Orange (M.O) Color in Color in Acidic middle Basic middle Range in PH Colorless Pink 8 - 10 Orange red yellow 3.1 - 4 The idea of the experiment: • To determine the suitable indicator we will make Titrations : a strong acid with strong base .. 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. A titration curve is a graph of the pH changes that occur during an acid-base titration versus the volume of acid or base added. The objective (purpose) of study of titration curve: Is to know the suitable Indicator for acid-base titration. But , The objective of titration: IS to know the concentration of an unknown solution. 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. 15_330 pH W eak acid Strong acid Generally: Vol NaOH This figure shows the titration curve for both cases. 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…………....... B)Titration between weak acid(CH3COOH)with a strong base(NaOH) CH3COOH + NaOH Ph.ph CH3COONa + H2O OK M.O *The suitable indicator for this titration is…………....... 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). Explanation for pH curve: This curve is similar to the previous curve, but the additive is acid and not the base 14 endpoint 7 X equivalent point equivalent point volume ( V NaOH (ml) ) 0 10 20 30 40 EXP(4): Determination of Sodium Hydroxide(NaOH) Concentration BY Titrations With A Standard Solution of Hydrochloric Acid (HCl) . 1) The titration by Ph.ph : *Reaction equation: 2) The titration by M.O : HCl + NaOH → NaCl + H2O *Results: ▪ Volume of NaOH: 123- *Calculations : ▪ Molarity (M) of NaOH : M × V /n = M’ × V’ /n’ ▪ Strength of Solution (S) of NaOH : S = M × M.wt NaOH M=? V = average HCl M’ = 0.1 M V’ = 10 ml + 2 drops of ph.ph or EXP(5):Determination of Acetic Acid (CH3COOH) Concentration BY Titrations With A Standard Solution of Sodium Hydroxide (NaOH) . * The titration by Ph.ph (only) : *Reaction equation: CH3COOH + NaOH CH3COONa + H2O * Results: … Volume of NaOH : 123- NaOH M = 0.1 M V = average *Calculations : .. Molarity ( M ) of CH3COOH : M × V /n = M’ × V’ /n’ .. Strength ( S )of CH3COOH : S = M × M.wt CH3COOH M’ = ? V = 10 ml + 2 drops of ph.ph EXP(6) : Determination of Hydrochloric Acid (HCl) Concentration By Titrations With A Standard Solution of Sodium Carbonate (Na2CO3). 1)The titration by M.O: 2 HCl + Na2CO3 2 NaCl +CO2 + H2O *Results: ..Volume of HCl 123*Calculations: . Molarity of HCl : M × V /n = M’ × V’ /n’ . Strength of HCl: S = M × M.wt 2) The titration by ph.ph: HCl + Na2CO3 HCl M =? V=average Na2CO3 M’ = 0.05M V’ = 10 ml + 2 drops of indicator NaHCO3 + NaCl *Results: ..Volume of HCl 123*Calculations: . Molarity of HCl : M × V /n = M’ × V’ /n’ .Strength of HCl : S = M × M.wt Is the molarity in two cases equal (same) or ?? 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 : ……………faster than ……………… Because : ........................................................... 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! 38 He N2 Which balloon will lose pressure sooner (quickly)? 39 N2 Big molecules He Small molecules NOW : Which balloon will lose pressure sooner ? The answer : .............. 40 Graham’s Law of diffusion: diffusion rate is inversely the proportional to square root of its molar mass or its density. In case 2 gases : r1 d1 2 r2 d 2 1 41 r1 2 r2 1 r = Diffusion rate of gas. M = Molar mass (Molecular Weight) of gas. the rates of diffusion of He and N2 . rHe 28.0 2.65 rN 2 4.00 He diffuses times as fast as N2 42 Dalton’s Law of Partial Pressures: He H2 N2 Ptotal = P1 + P2 + P3 +... 43 EXP(8): Determination of Critical Solution Temperature (C . S . T ) Principle of experiment: Warning: toxic H2O • 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). ………………………………………………………… C .S .T : Is temperature at which be all compositions of mixture or solution completely miscible. Above (C . S .T) be Homogeneous . mixture Under (C .S .T) be Heterogeneous . Calculation of percentage: % X in a sample (compound or mixture or solution) : = mass of X / mass of sample × 100 * Diagram between %phenol & miscibility temperature: * C.S.T =………OC T (O C) Homogeneous Heterogeneous %Phenol at C.S.T=…....% %Phenol % H2O at C.S.T = 100 ─ % phenol =….....OC EXP(9): Hess’s law • Hess’s law states that “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 L H4 W E D C H3 H1 H2 H3 H4 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. 48 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 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 endothermic Absorb heat H = + Liberate heat H = ─ that exothermic Procedure: Calorimeter: Thermometer Stirrer Cork stopper Cork cover(insulator) EXP(10): Effect of Concentration on Reaction Rate: (Determination of the order of the sodium thiosulphate and hydrochloric acid reaction) In this experiment We will determine the order of Na2S2O3 and HCl , and thus the order of total reaction . 2HCl + Na2S2O3 → 2NaCl + SO2 + H2O + S OR Rate = k [Na2S2O3]n [HCl ] m Precipitate yellow glutinous (slimy) Rate = [ S ] t Factors that effect on Reaction Rate : 1- Nature of reactants . 2- Presence of a catalyst . 3- Temperature . 4- Concentration of reactants . Log 1/t = log k + ? log M’ 1) Order of Na2S2O3 Slope= y2- y1/ x2-x1= Log 1/t n 2) Order of HCl n m Slope= y2- y1/ x2-x1= Log 1/t Log k Log k Log Na2S2O3 Order of total reaction = n+m Log HCl m EXP(11): Determination of the Molar Mass (M.wt) of An Organic Compound By The Depression of Its Freezing Point (ΔTf ) . If a nonvolatile solute is added to a liquid, a number of physical properties of the pure substance change, including vapor pressure depression, freezing point depression, and boiling point elevation. These alterations are collectively known as colligative properties of solutions. * Freezing point depression: Solution freeze at lower temperature than pure liquid(solvent). The amount that the freezing point is lowered called the freezing point depression (ΔTf) . molality (m) = moles of solute / mass of solvent(kg) * Freezing point ( T f ) : Is the temperature at which a liquid changes to a solid . The change in the freezing point (ΔTf) in °C for a nonvolatile organic solvent can be determined using the following equation, where kf is characteristic for the solvent used: ΔTf = kf . m is molality don’t mass ********* ΔTf = kf . m molality ΔTf = kf . n2 . 1000 / m1(g) Where: moles of solute mass of solvent m= n2 / m1(kg) m= n2 . 1000 / m1(g) ΔTf = kf . m2 . 1000 M2 . m 1 M Molar mass M.wt = kf × m2 × 1000 ΔTf × m1 Procedure: Split stopper Thermometer (4.8 ~ 5) gm of unknown 25 ml of H2O Stirrer Big beaker EXP(12): Determination of the Molar Mass(M.wt) of An Organic Compound By The Steam Distillation . When mixing two liquid do not mix (immiscible) the boiling point of the mix, will become less than boiling point of any of them . Boiling Point(Tb ): It is the temperature at which internal vapor pressure of the liquid is equal to the pressure exerted by its surroundings . Suppose that : H2O = A m A / mB = & Unknown liquid = B PA × ( M.wt)A / PB × (M.wt )B (M.wt)B = PA × ( M.wt )A × m B / mA × PB Raoult’s law: For a mixture of two miscible liquids (A and B), the total vapor pressure is the sum of the individual vapor pressures: PTotal = PA + PB Procedure: ح The steam distillation apparatus: Thermometer condenser 100 100 mlml of unknown compound (B) VA + 50 50 mlml of H2O (A) VB Graduated Cylinder **************************************************** Results: 1- Boiling point(Tb ) =………OC PA = …….. PB = PTOTAL ─ PA = ……. 2- Volume of H2O (A) after distillation = ……. ml 3- Volume of unknown solution (B) after distillation = …… ml Calculations: d B × VB dA × VA (M.wt)B = PA × ( M.wt )A × m B / mA × PB (M.wt)B = ....... gm / mol End of the experiments ***************************************