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Phph 1 Phph 1 Document Transcript

  • Lab .1 Concentration expression The concentration of a solution may be expressed either in terms of thequantity of solute in a definite volume of solution or as the quantity ofsolute in definite mass of solvent or solution. The expressions used are Morality, Normality, Mole fraction, Molepercent, percent by weight, percent by volume, percent weight in volume,milligram percent Morality, Normality and percent expressions are the most commonlyused in analytical work.Experimental work:Prepare the following solution using volumetric flasks & pipettes:A. 50 ml of 0.5 M NaCL.B . 50 ml of 2N NaCL.C. 50 ml of 0.1N Na2CO3.D. 50 ml of 0.1 M Na2CO3.E . 50 gm of 2% w/w NaCL solution.F. 50 ml of 10% w/v NaOH or NaCL.G. 50 ml of 10% v/v alcohol.
  • Note: Morality is gram molecular weight per one liter solution.Normality: is gram equivalent weight per one liter solution.Dilution-From stock solution 1% NaCL prepares 50 ml 0.5% solution usingequation:V1 * C1= V2* C2Materials and equipment-NaCL, Na2CO3, NaOH, Alcohol, H20.-Volumetric flasks (50cc), pipettes.
  • Lab.2TWO COMPONENT SYSTEM CONTAINING LIQUIDPHASES We may define a phase as a homogenous, physically distinct portionsystem, which is separated from other portions of system by boundingsurfaces e.g. a system containing water and its vapor is a two phasesystem. An equilibrium mixture of ice, water and water vapor is a threephase system. We know ethyl alcohol and water are miscible in all proportions, whilewater and mercury are completely immiscible.Between these two extremes lie whole ranges of system which exhibitpartial miscibility. Such a system is phenol and water.Their miscibility is affected by 2 factors Conc. And temp. Bimodal curve: is the curve that separate two phase area from one phasearea. Tie line: is the line drawn across the region containing two phases, is All system prepared along tie line at equilibrium separated into twophases of constant composition. These phases are termed conjugatephases. Mass ratio: is a relative amount by weight of conjugate phases.It depends on the position of the original system along the tie line.Procedure: 1. Prepare the following percent W/W phenol/water(10 gm total) 2%,7%,9%,11% ,24%,40%,55 %,63%,70%,75%. 2. Put test tube in afixed temperature in water bath (25 C0) or (left test tube at room temp.) and keep it for 10 minutes at that temp. 3. Take the test tubes out and before their temp has changed record which one has 2 phases and which has one phase.
  • 4. Repeat the work at higher temp using the following temp.40C0, 50C0, 70C0. 5. Draw a curve temp verses concentrations showing your 2 phases area and one phase area in the curve. 6. Draw tie line for each temp. 7. Take 40% W/W for example to find the mass ratio and the composition of each phase at different temp. 8. Mention the upper consulate temp.Material: phenol and water.Instruments and equipment: water bath, test tubes, glass stirrer
  • Lab.3 Three component systemThe rules that relate to use of triangular co-ordinates graph paper 1- Each of the three corners or apexes of triangular represent 100% byweight of one component and zero% of other two components 2- The three lines joining the corner points represent two componentmixtures &each line is divided into one hundred equal units .In goingalong a line e bounding the triangles so as to represent concentrations in atwo component system , it does not matter whether we proceed in a clockwise or counter clock wise direction around the triangle , provided we areconsistent.3- The area within the triangle represents all possible combinations ofthree components to give three components systems.4- If a line is drawn through any apex to appoint on opposite side, then allsystem represented by points on such aline have a constant ratio of twocomponents.5-Any line drawn parallel to one side of tiangle, represent ternary systemsin which the proportion (or percent by weight) of one component isconstant.Procedure: 1. Prepare 10 gm of the following combination of HAc & CHCl3:5%, 10%, 20%, 30%, 40%, 50%, 60%, 70%, 80%, and 90% w/w HAc: HCCl3 in a small clean &dry flask which form one single phase. 2. To these mixtures slowly add water from a burette until a turbidity just appears. Check the weight of water (which is equal to its volume). 3. Obtain a miscibility curve by calculating the percent w/w of eachcomponent in the turbid mixture and plot this triangular diagram.
  • Note:To prepare samples in step no.1, the required amount of HAc &CHCl3from burettes by converting the weight in to volume according to the law:Specific gravity (sp.gr) = weight/volumeSp.gr of HAc =1.009 and for CHCl3 =1.3Materials and equipment:-Acetic acid (HAc), CHCl3 distilled water.-Conical flasks, burette.
  • Lab.4 Tie Line for three component system In the three component system , the direction of tie lines are related tothe shape of the bimodal which, in turn ,depend on the relative solubilityof the third component in the other two components. Only when added component equally on the other two components tobring them into solution will the bimodal perfectly symmetrical and thetie lines run parallel to the base line.Procedure: 1. In a small reparatory funnel prepare 50 gm of a mixture having composition giving rise to a two phase system (e.g. 4gm HAc +16 gm CHCl3 +30gm H2O). 2. Separate each layer in two conical flasks. 3. Weigh 10 gm for each layer. 4. Titrate each layer with standard 1N NaOH solution using phenolphthalein as indicator. The end point from colorless to pink. 5. Obtain tie line, calculate the percent W/W of HAc in each layer and locate the values on the miscibility curve. The straight line joining these points should pass through compositions of the two phase system.Calculation: HAc + NaOH NaAc + H2O1 M.Wt. of HAc = 1 M.Wt. of NaOH1 eq.wt of HAc = 1eq.wt of NaOH 60 = 1000 ml 1N NaOH 60/1000 = 1 ml 1N NaOH So, each 1 ml of 1N NaOH is equivalent to 0.06 gm, this is the chemicalfactor (it is the no. of gms of substance which is equivalent to 1 ml ofstandard solution).
  • E.P 1 x o.o6 =gm HAc in 10 gm aqueous layer (upper layer).E.P 2 x o.o6 =gm HAc in 10 gm CHCl3 layer (lower layer).Change these values to percent.Note: - For the upper layer (100% HAc) represent mostly water with little chloroform. This layer represents aqueous layer. - For the lower layer (100% HAc) represent mostly chloroform with little water. - This layer represents chloroformic layer.Materials & equipments: 1. H2O, HAc, CHCl3, 1N NaOH solution, phenolphthalein indicator. 2. Burettes, separatory funnel, conical flasks, balance.
  • Lab.5 Partition coefficientDistribution of solute between immiscible solvent: If an excess of liquid or solid is added to mixture of two immiscibleliquids, it will distribute itself between the phases so that each becomessaturated. If the substance is added to the immiscible solvent in an amountinsufficient to saturate the solutions, it will still become distributedbetween the two layers in a definite conc. ratio. If C1&C2 are the equilibrium concentrations of the substance insolvent (1) and solvent (2) ,the equilibrium expression becomes: C1/C2 =K The equilibrium constant K is known as distribution ratio, distributioncoefficient, or partition coefficient, which is defined as solubility orconcentration ratio at constant temp.Experimental work : Determination the partition coefficient of iodine between water andchloroform.Procedure: 1. in dry Stoppered conical flask (iodine flask) put 20 ml of 1% iodine in chloroform (use burette). 2. Add 50 ml D.W. to it. 3. The flask is the thoroughly shaken from time to time half hour after equilibrium is established ,allow to stand for complete phase separation, this need another half an hour.
  • 4. 10 ml of the sample are taken from the upper aqueous layer, care is taken to avoid touching the chloroforming layer. Then titrate against 0.02 N sodium thiosulphate.the end point is the disappearance of light brownish color. 5. 5 ml are taken from the organic layer (lower layer) .the inside wall of the pipette must be kept dry as it passes through aqueous phase by placing the finger tightly over the upper end of the pipette. Then titrate against 0.1 N sodium thiosulphate. Before titration, add 5 ml of 10% pot. Iodide to facillate extraction of I2 from the organic layer and its titration with aqueous sodium thiosulphate. The end point is the disappearance of the brownish color. Calculation:Iodine distributed between the aqueous phase and chloroformic phase.Aqueous phase:-the no. of ml of sodium thiosulphate (0.02N) consumedin the titration is equivalent to the amount of Iodine present. (Na2S2O3) V1 X C1 = V2 X C2 (iodine) E.P X 0.02 N =10 X N2 N2 =conc. Of iodine in waterChloroformic phase:-the no. of ml of sodium thiosulphate(0. 1N) consumed in the titration is equivalent to the amount of Iodinepresent. (Na2S2O3) V1 X C1 = V2 X C2 (iodine) E.P2 X 0.1N = 5 X N2 N2 =conc. Of iodine in chloroform
  • Conc. Of iodine in CHCl3 Partition coefficient = --------------------------------------- Conc. Of iodine in water Material &equipment : Iodine, potassium iodide, chloroform, water, sodium thiosulphate. Solutions: 10%w/v KI , 0.02N&0.1N sodium thiosulphate , 1% I2 / CHCL3 Conical flask (iodine flask) , pipette , burette . Home work :-1. What idea will the p.c value give you about the solubility?2. What is the importance of p.c to the pharmacist?
  • Lab .6 Solubility Methods of solubility Solubility: is defined in quantitative term as the concentration of solute in a saturated solution at certain temperature. In a quantitative way it may be stated as spontaneous interaction of two or more substances to form a homogenous molecular dispersion. The solubility of a compound depend upon the physical and chemical properties of solute and solvent , as well as upon such factors as temperature , pressure , pH of solution and to lesser extent the state of subdivision of solute . Methods of solubility :1-Solvent combinations: The solubility of the solute is qualitatively related to dielectric constant of the solvent system. For example given solute will have qualitatively similar solubility profile with respect to dielectric constant for various co- solvent combinations. The objective of this experiment is to increase solubility of salicylic acid (weak organic acid, slightly soluble in water) by solvent combination, changing the dielectric constant of the solvent system used. Procedure : 1- Put 0.1 gm salicylic acid (S.A)in conical flask 2-Add 10 ml distilled water and shake the flask to see its solubility. 3-Add from burette drop by drop absolute alcohol with continuous shaking until all crystals of S.A dissolve. 4- Measure the amount of alcohol used.
  • Calculate the percent of alcohol in the final mixture.Express the solubility, as1part of S.A is soluble in x parts of y% ofalcohol.2- Salt formation (pH control): Most weak electrolytes can be retained in solution by adjusting pHso as to keep drug in ionized form. The aim of this experiment is to increase the solubility of S.A by saltformation using Na2CO3.Procedure :1- Put 0.1 gm salicylic acid (S.A) in conical flask.2-Add 10 ml distilled water and shake the flask to see its solubility.3-add 0.1 gm Na2CO3 to the flask with shaking and observe the result.4-add 5 ml dilute HCL (10%) slowly. Observe the result and develop an equation to account for theobservation in step 3 and 4.3- solubilization by complexation : It has been found that insoluble drug can form a soluble complex withsome compound. In organic and organic materials which do not themselves ionize maybe rendered soluble in polar solvent by complexation with electrolytes.The aim of this experiment is to solubilize I2 in water with KI.Procedure1-put 0.1 gm I2 in conical flask.2-add 10 ml water, shake and observe.3-add 0.2 gm KI.Observe the result and write an equation for it.
  • Materials &equipments : Salicylic acid ,water , alcohol , iodide , potassium iodide , sodiumcarbonate , 10% HCL solution , 2conical flask , pipette , burette .Home work1- How solubility express?2-what is the saturated solution?3-what is the difference between saturated and unsaturated solution?4-what is relation between the dielectric constant and polarity of solvent?5-what is the importance of solubility to the pharmacist?
  • Lab.7 Buffer solution Buffer solutions are solutions that tend to resist changes in pH whenacids or bases are added. Buffer solutions usually contain a salt of weak acid or weak base andcorresponding acid or base. Buffer equation: the pH of Buffer solution and the change in pH uponthe addition of an acid or base may be calculated by use of Bufferequation .this expression is developed by considering the effect of a salton the ionization of a weak acid when the salt and acid have an ion incommon. The buffer equation or Henderson- Hassel balks equation, for a weakacid and its salt: [Salt]PH= pka +log ------------- [Acid]For weak base and its salt: [Base]PH= pkw- pkb+ log ------------- [Salt]Buffer capacity :- is the ability of a buffer solution to resist pH change . The smaller PH change caused by the addition of a given amount ofacid or alkaline, the greater the buffer capacity of a solution. The buffer has its greatest capacity before any base is added where[salt]/ [acid] =1 (when equimolar amount of acid and salt are used)therefore, according to buffer equation, pH = pKa. The buffer capacity is also influenced by an increase in the totalconcentration of the buffer constituents since obviously a greaterconcentration of salt and acid provides a greater alkaline acid reserve.
  • Approximate formula to calculate buffer capacity was introduced byVan Slyke is: ∆Bβ=------------------ PH May be used, in which delta (∆) has its usual meaning, a finite change,and B is the small increment in gm equivalent/liter of strong base. Various buffer systems have been suggested for differentpharmaceutical solutions:1-Sorensen phosphate: buffer which consists of two stock solutionsM /15 sodium acid phosphate and M/15 disodium phosphate, which aremixed in various volumes to obtain a desired PH2-Acetate buffer: consist of 0.2 M solution of acetic acid (A) and0.2Msolution of sodium acetate (B) by mixing x ml of A and Y ml of B andenough water to produce 100 ml.Experimental work:Procedure:1-prepares acetate buffer solution according to the following table thenmeasure the PH in each case:Solution (A) ml + Solution (B) ml D.W to 100ml pH 46.3 3.7 30.5 19.5 25.5 24.5 20 30 14.8 35.2 4.8 45.2
  • The PH of each solution is measured by using:1. Paper indicator: by immersing a strip of wide range PH paper intosmall quantity of buffer solution and observing the color changes of thepaper which changes according to the PH value of the buffer.2. Liquid universal indicator: which consists of several indicators e.g. amixture of methyl yellow, methyl red, bromothymol blue,thymol blue and phenolphthalein which covers PH range (1-11). The PH of buffer solution is measured by the addition of 2 drops ofuniversal indicators to 10 ml buffer solution ,then compare the colorresult with color found on the bottle of liquid universal indicator .eachcolor represents certain value of PH . Method (1) and (2) are called colorimetric methods for thedetermination of PH. they are probably less accurate and less convenientbut also less expensive than electrometric method (by using PH meter) .They are used in the determination of PH of aqueous solutions which arenot colored or turbid. PH indicators: indicators may be considered as weak acids or weakbases which exhibit color changes as their degree of dissociation varieswith PH. Indicators therefore offer a convenient alternative method toelectrometric techniques for the determination of PH of solution. Thecolor of an indicator is a function of the PH of the solution.3-PH meter :1. Put the electrode of the PH meter in the buffer solution & read the PHNote: a- the PH meter should be opened before reading the PH about 30min.b- The electrode should always be immersed in D.W when its not usedc- Standardization for instrument should be done using standard buffersolution PH 7 &PH 4 when our samples are acidic, or PH 7 & PH 10when our samples are basic.
  • 2-a. Take a certain volume of acetate buffer solution, add to it 0.0004 Msodium hydroxide portions (0.1 ml of 0.1 M).B- Measure the PH. C- Calculate the buffer capacity.Materials ,equipment , and instruments- acetic acid, sodium acetate, water, liquid universal indicator, paperindicator.-solution: 0.2 M HAc, 0.2 M NaAc, 0.1 M NaOH.- conical flask, beaker, pipette, volumetric flasks.-PH meter.Home work :1-explain how buffer resist the changes in PH by using equations?2-derive buffer equation for acidic buffer?3-How can you differentiate between buffer and non buffer system inlab?
  • Lab .8Determination of solubility product constant ofslightly soluble salts When slightly soluble electrolytes are dissolved to form saturatedsolution, the solubility is described by a special constant, known assolubility product Ksp .Theory: in case of inorganic salt, when solute goes into solution such as AB, wethen write the following equation for solution product:AB (solid) A− (solution) + B+(solution) ………(1)There is an equilibrium between the solute and the saturated solutionphases, the law of mass action defines as equilibrium constant, Keq as: a A− (solution) × a B+(solution)Keq = ---------------------------------------- …………….(2) a AB(solid) Where a A− (solution) , a B+(solution) and a AB(solid) are the activities of Aand B in solution and AB in solid phase . Since the activity of a solid isdefined as unity, and that in dilute solution (e.g. where we have slightlysoluble salt) , concentration may be substituted for activities ; equation(2) then becomes : Keq = CA . CB
  • Where CA and CB are the concentration of A and B in solution .Keqin this situation has a special name , the solubility product , Keq , thus : Ksp = CA . CB ……….(4)This equation will hold true theoretically only for slightly soluble salt. As an example of this type of solution, consider the solubility of silverchloride, as we can write:Ksp = Ag + Cl - Where the brackets [ ] designate concentration in molar /litter(molar concentration). So, the Ksp is the multiplication of molar conc. of two ions. If thesilver ion concentration is increased by the addition of soluble silver salt,the chloride ion concentration must decrease until the product of theconcentration is again numerically equal to the solubility product. In order to affect the decrease in chloride ion concentration, silverchloride is precipitated, and hence its solubility is decreased .in a similarmanner an increase chloride ion concentration by the addition of solublechloride salt, a decreased in the silver ion conc. until the numerical valueof the solubility product is attained. Again this decrease in silver ionconc. Is brought about by the precipitation of silver chloride. The solubility of AgCl in saturated aqueous solution of the salt may becalculated by assuming that the concentration of silver ion is the same asthe conc. Of chloride ion , both expressed in gm mol/l and that the conc.of dissolve silver chloride is numerically the same since each silverchloride molecule gives rise to one silver ion and one chloride ion , since:AgCl Ag+ + CL-Procedure :-1-into five clean dry conical flasks, add 1gm of KHT (potassium acidtartar ate) +40 ml of different molarities of KCL.a- In the first flask add 40 ml D.W.b- In the second flask add 40 ml of 0.01 M KCl.c- In the third flask add 40 ml of 0.02M KCl.
  • d- In the fourth flask add 40 ml of 0.03 M KCl.e- In the fifth flask add 40 ml of 0.05 M KCl.Note : you are provided with 0.1 M KCl e.g. in order to prepare 40 mlof 0.01 M KCl and complete the volume with water to 40 ml dependingon the dilution equation .C1 ×V1 = C2 ×V2 and so on.2- Shake for 10 min, leave 15 min for equilibration.3-filter, rinse the flask with first portion of the filtrate (e.g. 1ml) ,complete the filtrate .4-take 10 ml of the filtrate, titrate against M/50 NaOH usingphenolphthalein as indicator.5-calculate the solubility product of potassium acid tartar ate.Calculation: O O C OH C OH H C OH H C OH H C OH H C OH C OK C OH O O KHT HTFlask no.(1) :Ksp= [HT-] [K+]in other flasks : Ksp = [HT-] [K+ + K+ from KCl] KCl K+ + CL-
  • in titration : HT + NaOH NaHT + H2O1 eq.wt of HT- ≡ 1 eq.wt of NaOH1M.wt of HT- ≡ 1 M.wt of NaOH1L 1M NaOH ≡ 1M.wt of HT- =149 gm1ml 1M NaOH ≡ 149 / 1000 gm 1 149 11ml ------- M NaOH ≡ ------------ × ----------- gm 50 1000 50 149 1For flask no .1 E.P1×-----------×---------------- = gm HT/10ml 1000 50 149 1 100E.P1×-----------×---------------- ×------------- = mole /liter 1000 50 149So E.P 1 /500 = molar concentration of HT = molar conc. of K+ E.P1 E.P1 KSP for flask (1) = [-------] [--------] 500 500 E.P2 E.P2 KSP for flask (2) = [-------] [-------- + 0.01] 500 500
  • E.P3 E.P3 KSP for flask (3) = [-------] [-------- + 0.02] 500 500 E.P4 E.P4 KSP for flask (4) = [-------] [-------- + 0.03] 500 500 E.P5 E.P5 KSP for flask (5) = [-------] [-------- + 0.04] 500 500Tabulate your result as follows:Flask no. E.P M KCL [HT-] [K+] KspMaterials and equipments:-potassium acid tartarate , potassium chloride , sodium hydroxide .–solution: 0.1 M KCl , M/50 NaOH, phenolphthalein indicator. –Conical flask, pipette , burette , filter paper , measuring cylinder .