PHAR 270 Pharmaceutical Analytical Chemistry II Lab Manual PDF

**PHAR 270** **Chemistry II** **Lab Manual** ***Faculty of Pharmacy BAU, MISSION STATEMENT*** ***Faculty of Pharmacy, at BAU, is an academic institution founded in 1986 to provide high quality pharmacy education and scientific research. The faculty educational program was designed and developed...

**PHAR 270** **Chemistry II** **Lab Manual** ***Faculty of Pharmacy BAU, MISSION STATEMENT*** ***Faculty of Pharmacy, at BAU, is an academic institution founded in 1986 to provide high quality pharmacy education and scientific research. The faculty educational program was designed and developed to prepare competent pharmacists able to effectively participate in the advancement of pharmacy profession, nationally and internationally. The faculty supports the role of its graduates as health care providers in the frame of professional ethics. The faculty seeks to establish a well-built relation with peer institutions and the surrounding society. To accomplish its mission, the faculty relies on qualified staff members, laboratory facilities and educational tools.*** Contents ======== Fundamental Terms in Titrimetric Analysis 6 Standardization of Solutions 17 Acid-Base titration 20 Direct Titration of Bases 24 Back Titration 31 Back Titration with Blank Determination 33 Complex Formation Titration 46 Laboratory Regulations ---------------------- 1. Always bring your notebook with you. 2. Don't be late to come to the laboratory. 3. Put on your white coat while you are in the laboratory. 4. Switch off your mobile while you are in the laboratory. 5. Always read the label written on the reagent bottles before use. 6. Never use a chemical from an unlabeled bottle. 7. Don't return excess reagents to the stock bottle father reject them. 8. Follow the experimental procedures carefully. 9. Never remove a flask or a beaker or a test tube from the hot plate, but leave them for a while to cool. 10. All chemicals are hazardous and you must minimize your exposure time to them. 11. Don't talk with each other during work. 12. No chewing gum or wearing caps are allowed in the laboratory. 13. After finishing your work, you have to leave your place clean. 2- Deduction in your marks. Lab Policy and General Information ---------------------------------- ### I. Safety ### *Main Lab Safety Rules* 1. Dress appropriately for the lab: no sandals, no outfits you prefer over your safety, no contact lenses and long pants are preferable to shorts or short skirts. Tie your long hair back. Some clumsy people probably may set themselves on fire, spill acid on themselves, splash themselves in the eye, or be a bad example to others. 2. Uncontrolled long hair or clothing (loose sleeves, head covers, ties or jewellery) might touch a flame or become entangled in mechanical equipment. 3. Protective clothing, lab coat, is required, and highly recommended. You will not be permitted to work in the lab if you are dressed improperly. 4. Always conduct yourself in a professional manner. There will be NO jokes, running, or playing around or similar irresponsible activities while working in the laboratory. 5. No consumption of food (including gum) or beverages or application of cosmetics will be allowed. 6. Do not heat or mix anything near your face (or anybody's face). 7. Do not use any broken glassware. 8. You are not allowed to perform any unassigned experiments. 9. If something is spilled on you, wash it off immediately with lots of water, and then report the incident to the instructor. Clean up the spill later according to instructions. 10. Do not rub your eyes with your hands. Your hands are usually contaminated. 11. Be careful. Don't be a grabber and get burned by trying to pick up hot things with bare hands. 12. Do not dump reagents bottle near a sink or leave them near the balance or near the shelfedge where they could be in the way and get knocked over. Return all reagents to their proper location once you are finished with them. 13. Make sure, before leaving or in any emergency, to turn off all the utilities (gas, water, electricity) that have been used. 14. Never attempt to identify an unknown chemical by smelling or tasting it. 15. Read all chemical labels before use to be sure that you are using the exact reagent intended for the required step. 16. Do not place non-compatible chemicals close together (for example acids with bases or oxidizers with flammables). 17. Dispose used chemicals and shipped or broken glassware properly. If you are not sure how to do so, ask your instructor. 18. Do not operate electrical equipment with wet hands. 19. Do not block doors or fire exists. **II. Important Laboratory Rules & Guidelines** ### *Lab code* ### *All of the following rules will be closely watched at any times you are in the Laboratory* **[Calculations with Molarity ]** **Mole:** is the formula weight of a substance expressed in grams m moles = mg/form. Wt **Molar Concentration** = Molarity = is the number of moles of solute present in one liter of solution. or moles = (M) (Liters) mmoles = (M) (ml) g = (moles) (from, wt.) mg = (m moles) (form. wt) g = (liters) (M) (form. Wt) mg = (MI) (ml) (form. wt) **[Calculating the results of a titration:]** To calculate the results of a titration it is necessary to know the volume and molarity of titrant: used; it is also necessary to know the reacting ration of substance titrated to titrant in the chemical reaction. The reacting ratio can be obtained easily from the balanced chemical equation: a A + b B products where a moles of titrant A react with b moles of titrated substance B. A clear understanding, calculations of the results with be broken down into several steps. 1- The mmoles of A may be calculated from the volume and molarity of the titrant used for the titration. m moles~A~ = ml~A~ M~A~ 2- The mmoles of B may be calculated from the mmoles of A using the combining ration R, of B to a, which is given by the coefficients b/a in the chemical equation. mmoles ~B~ = mmoles ~A~ R = ml~A~ M~A~ R 3- The weight of B in milligrams is obtained by multiplying both sides of this equation by the formula weight of B. mg~B~ = ml~A~ M~A~ R (form. wt.) 4- Finally the percentage of B is calculated by dividing the weight of B by the weight of sample and multiplying by 100. \%B = **Example (1):** What weight of acetic acid is in a 5.00ml sample of vinegar which requires 35.00ml of 0.100M NaOH for titration? The reacting ratio is R = 1 (ml~OH~^-^) (M ~OH~^-^) (form. wt. CH~3~COOH) = mg CH~3~COOH (35.00) (0.10) (60.03) = 210.1mg of acetic acid. **[Calculating the molarity of a solution from standardizing titration.]** If an accurately weighed sample of a primary standard, B, is dissolved and titrated with A, the molarity of A is calculated as follows: a A + b B products R = b/a (ml~A~) (M~A~) (R) (form. wt. B) = mg~B~ **Example (2):** Exactly 410.4mg of primary standard potassium acid phthalate (KHP), formula weight 204.2, is weighed out and dissolved in water. If titration of the (KHP) requires 36.70ml of a NaOH solution, what is the molarity of the sodium hydroxide? (NaOH) (KHP) (36.70) (M) (204.2) = 410.4 **[Calculation With Normality]** In instances in which the reacting ratio of reactants A and B is not unity the calculations are sometimes made easier if the concentrations of solutions are expressed in terms of normality instead of the molarity and the equivalent weight of the substance titrated is used instead of the formula weight. In the called normal system the concentration of solution is adjusted, or normalized, to account for differences in reacting ratios in chemical reactions. A 1N (1 normal) solution contains one equivalent per liter or one milliequivalent per milliliter: An equivalent is so defined that one equivalent of titrant will react with one equivalent of substance titrated and 1 milliequivalent of titrant will react with one milliequivalent of substance titrated. Keep in mind that 1 mole = 1 Avogadro number (6.023×10^23^) of atoms or molecules. The definition of an equivalent depends on the reaction a substance undergoes. In acid\--base reactions one equivalent is the number of grams of a substance that supplies, or combines with, one Avogadro number of hydrogen ions (one H^+^). 1 eq of HCI = 1 mole, or 36.5g of HCI 1 eq of H~2~ SO~4~ = 1/2mole, or 49 g of H~2~ SO~4~ and accordingly. eq. wt. HCI = form. wt eq. wt. H~2~SO~4~ = 1/2 form. wt. So in acid \-- base titrations, eq. **[Calculating the results of a titration:]** Calculations with normality are similar to those with molarity. The difference is that at the stoichiometric point (E.P.) in a titration the milliequivalents of titrant A always equal the milliequivalents of titrated substance B. the reacting ratio of B to A used in the molar system has been taken into account in preparing normal solution. It follows that a given volume of a normal solution will react with an equal volume of another solution having the same normality. ml~A~ N~A~ = ml~B~ N~B~ **Example:** A HCI solution is standardized by titration with standard NaOH. If 25.00ml of HCI requires 32.20ml of 0.0950N NaOH for titration, what is the normality of HCI solution? (32.20) (0.0950) = (25.00) (N~HCl~) N~HCI~ = 0.1224 **\ Example:** What volume of 12.1N HCL must be diluted to 1 liter to give a 0.1N solution? (ml~HCl~) (12.1) = (1000) (0.1) ml ~HCl~ = 8.26 ml In calculating the results of a titration of substance B with titrant A, one of the following equations is used: (ml~A~) (N~A~) (eq. wt.~B~) = mg~B~ **Example:** A 1.0000g H~3~PO~4~ sample requires 28.16ml of 0.1000 N NaOH for titration to a phenolphthalein indicator. Calculate the percentage of H~3~PO~4~ in the sample. eq. wt H~3~ PO~4~ (to ph. Ph) = If an accurately weighed primary standard B is dissolved and titrated with A, the normality of A is calculated as follows: (ml~A~) (N~A~) (eq. wt.~B~) = mg~B~ **Example:** A 150.0mg sample of pure Na~2~ CO~3~ requires 30.06ml of a HCl solution for titration: (2 HCl) + (Na~2~CO~3~) **[Calculate the normality of HCI]** eq. wt. (30.06)(N~HCI~)(52.99) = 150.0 N~HCI~ = 0.09416 **[Standardization factor:]** It is more often convenient to prepare standard solutions which although of known strength, are only approximately N, 0.5N, 0.1N etc the relationship of the exact strength of such a solution is then indicted by a \"Standardization Factor\". It is the number by which the actual volume of the solution of approximate strength must be multiplied to obtain the equivalent volume of a standard solution of exact normality. In the above example, the standardization factor is calculated as follows: Volume of standard solution Factor = = Volume of solution of unknown strength That is the strength of hydrochloric acid solution is 0.1N (1.02). The factor of the solution is 1.02 and all volumes of this solution when multiplied by 1.02 will give the equivalent volume of an exactly 0.1N solution. ml hydrochloric acid solution titrated = 25.5ml ml 0.1 N sodium hydroxide used in the titration = 25.5ml ∴ Normality of hydrochloric acid solution = **[Weighing devices ]** **[For solids]**  **[For liquids]** [ **Measuring devices:**]  [ ] Volumetric flask Bulb pipettes Graduated pipettes \ Washing bottle Burette  **Dessicator** **[Titration Technique]** **[Method for quantitative transfer of precipitate]**  **[Technique for using pipettes]**:  **Lab1** **Standardization of Solutions** **1- [Preparation of 0.1N sodium carbonate]:** \- Weigh accurately about x gm Na~2~CO~3~ in a small dry beaker. \- Add dist. Water about 30 ml to dissolve it. \- Transfer quantitatively into 100ml measuring flask. \- Complete to the mark with dist. Water, stopper, and mix well. **2- [Preparation of 0.1 N HCI from 1N HCl:]** \- Pipette 10ml IN HCI into 100ml measuring flask. \- Complete to the mark with water, stopper, and mix well. **3- [Standardization of the prepared HCl with standard Na~2~CO~3~ solution]:** **[Principle: ]** Hydrochloric acid is a strong acid, but is not a primary standard. Sodium carbonate is a primary standard that can be used to standardize the acid solutions like HCl. Na~2~CO~3~ + HCl \-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\--\> CO~2~ + 2NaCl +H~2~O **[Procedure:]** \- Pipette10ml of 0.1N Na~2~CO~3~ in conical flask. \- Add (1-2) drops M.O. indicator. \- Titrate against the prepared HCl. \- Determine the exact normality of HCl and the standardization factor. **[Calculation:]** X~(g)~ of Na~2~CO~3~ = E.P= N ~HCl~= **Lab2** **Standardization of sodium hydroxide** **1- [Preparation of 0.1N NaOH from 1N NaOH:]** \- Pipette 10ml 1N NaOH into 100ml measuring flask. \- Complete to the mark with dist. water, stopper and mix well. **2- [Using Secondary Standard HCl]:** \- Pipette 10ml HCl in a conical flask. \- Add 10 drops ph. Ph. Indicator, or M.O. indicator \- Titrate with standard NaOH. \- Determine standardization factor and exact normality of NaOH solution. **[Calculation:]** E.P.= N~NaOH~= F~NaOH~= **3-[Determination of CH~3~COOH]** **[Principle:]** Acetic acid is a weak acid that can be determine by direct titration with standard sodium hydroxide. CH~3~COOH + NaOH \-\-\-\-\-\-\-\-\-\-\--\> H~2~O + CH~3~COOH **[Procedure:]** \- Pipette 10ml of CH~3~COOH in a conical flask. \- Add 10 drops ph. Ph. Indicator. \- Titrate with standard NaOH. **[Calculations:]** 1ml 0.1N NaOH = 0.006 gm CH~3~COOH E.P. = \%CH~3~COOH= **Lab3** **Determination of Boric Acid** **(Percentage of H~3~BO~3~)** **[Principle:]** Boric acid is a weak acid, that it cannot be titrated with alkali hydroxides in an aqueous solution using a visual indicator, because the metaborate ion is strongly basic and forces the above reaction to the left, so that the end point appears early. If glycerol is added, it forms a slightly dissociated complex with the metaborate ion thus prevents reversal of the reaction. Thus in the presence of glycerol, boric acid behaves like a sufficiently strong monobasic acid. **[Procedure:]** 1- Pipette 10 ml of boric acid sample in a conical flask. 2- Add 10 ml neutral glycerol + 10 drops ph. Ph. 3- Titrate with 0.1 N NaOH. **[Calculations:]** 1ml 0.1N NaOH = 0.006184gm H~3~ BO~3~ E.P. = \% H~3~BO~3~= **\ Determination of dilute Phosphoric acid** **(Percentage of H~3~ PO~4~)** **[Principle:]** \- Phosphoric acid when titrated against st. NaOH, using M.O. as indicator, only one proton is liberated with the formation of Mono-sodium dihydrogen phosphate. \- But when phosphoric acid titrated against st. NaOH using ph. ph. as indicator and adding NaCl to suppress hydrolysis 2 protons are liberated with the formation of disodium monohydrogen phosphate. **[Procedure:]** **[Calculations:]** E.P. (1) = E.P. (2) = \% W/V = % W/V = **Lab4** **Analysis of mixture of Phosphoric acid** **and Hydrochloric acid** **[Principle:]** 1- Total Acidity: By the Titrating the mixture with 0.1N NaOH and using M.O. as indicator will give total HCl contents and the 1^st^. ionisable proton of H~3~PO~4~. 2- And by titrating the mixture with standard NaOH in presence of ph.ph. as indicator and NaCl will give total HCl contents and the two ionisable protons of H~3~PO~4~. **[Procedure:]** **[Calculations:]** E.P. (1) = E.P. (2) = \% W/V HCl = \% W/V H~3~PO~4~ = **Analysis of mixture of HCl and H~3~BO~3~** **[Principle:]** The presence of the strong acid (HCl) in the medium suppresses the ionization of the weak acid (H~3~BO~3~). Thus neutralization of the strong acid will occur first then the solution is considered one of a weak monoprotic acid (H~3~BO~3~), to which glycerol is added to increase its acidic strength that it can be titrated with standard NaOH. **[Procedure:]** **[Calculations:]** E.P. (1) = E.P. (2) = \% W/V HCI = \% W/V H~3~BO~3~ = **Lab 5** **Direct Titration of Bases** **Determination of Borax** **[Principle:]** Borax in solution may be regarded as boric acid which is half neutralized. The overall reaction is **[Procedure:]** **[Calculations:]** 1ml 0.1 N HCI = 0.01907gm Na~2~B~4~O~7~.10 H~2~O E.P. (I) = V~1~ = E.P. (2) = V~2~ = V~2~ = 2V~1~ \% W/V Borax= **Analysis of a mixture of Borax and Boric acid** **[Principle:]** The analysis depends on first determining the borax content in the mixture by titration with standard acid, using methyl orange as indicator. The titrated solution contains the boric acid formed from neutralization of borax, together with the boric acid originally present. Thus the total boric acid content is determined by titration with standard alkali hydroxide solution after adding glycerol (previously neutralized to ph. ph.) **[Procedure:]** For Borax 1. Pipette 10 ml sample mixture in a conical flask+ 2 drops M.O., + 10 H~2~O titrate with 0.1 N HCI. For total Boric acid **[Calculations:]** 1ml 0.1 N HCI = 0.01907gm Na~2~B~4~O~7~.10 H~2~O 1ml 0.1 N NaOH = 0.006184 gm H~3~BO~3~ E.P. (I) = V~1~ = E.P. (2) = V~2~ = \% Borax W/V = \% Boric W/V = **Analysis of a mixture of Borax and Sodium Carbonate** **[Principle:]** \(1) For total alkalinity of the mixture: add known excess of standard acid, boil off CO~2~, then back titrate the residual unused acid with a standard alkali hydroxide solution using M.O. as indicator. \(2) For Borax Content To the titrated solution, add neutral glycerol and titrate against standard NaOH solution using ph. ph. as indicator. **[Procedure:]** **[Calculations:]** 1ml 0.1NaOH = 0.009535 gm Na~2~B~4~O~7~ 10 H~2~O 1ml 0.1N HCI = 0.0053 gm Na~2~CO~3~ E.P. (1) = V~1~ = E.P. (2) = V~2~ = \% W/V Borax = \% W/V Na~2~CO~3~ = **Analysis of a mixture of Alkali Hydroxide and Carbonate** **(NaOH + Na~2~CO~3~)** **[Method I:] [Winkler\'s Method]** \(1) Total Alkalinity: () is determined in one portion of the mixture by titration with standard acid using M.O. as indicator. \(2) Hydroxide Content: Is determined in another portion of the mixture, by first precipitating the carbonate with neutral barium chloride solution, and titrating the hydroxide with standard acid using ph. Ph. as indicator. There is no need to filter off the precipitated BaCO~3~, because it will not be affected by the acid titrant as long as the pH does not fall below pH 9 (pH at which ph. ph. Change its color). **[Procedure:]** **[Calculations:]** 1ml 0.1 N HCl = 0.0053 gm Na~2~ CO~3~ 1ml 0.1 N HCl = 0.00401 gm NaOH \% OH^-^ W/V = E.P. (1) = V~1~ = \% W/V = E.P. (2) = V~2~ = **[Method II] [(Two Indicators Method)]** \(2) In this method all hydroxide and half carbonate, are determined by titrating with standard acid using ph.ph. as indicator. \(1) Total Alkalinity: Make as Winkler's method. **[Procedure:]** \(1) For total alkalinity: Make as Winkler's method (yellow --orange). (2) Pipette 10 ml sample add 10 drops ph.ph. titrate with 0.1N HCI. **[Calculations:]** 1ml 0.1N HCl = 0.00401 gm NaOH 1ml 0.1N HCl = 0.0053 gm Na~2~ CO~3~ E.P. (1) = V~1~ = E.P. (2) = V~2~ = \% W/V = \% W/V = **Analysis of a mixture of Alkali Carbonate** **And Bicarbonate ()** **[Method] I ([Winkler's Method)]** The carbonate formed together with the carbonate originally present, are precipitated by neutral BaCl~2~ solution. Then the excess unused alkali hydroxide is titrated (without filtration) with standard acid using ph.ph. as indicator. **[Procedure:]** 2- For Bicarbonate Content: **[Calculations:]** 1ml 0.1 N NaOH = 0.0084 gm NaHCO~3~ 1ml 0.1 N HCl = 0.0053 gm Na~2~ CO~3~ E.P. (I) = V~1~ = E.P. (2) = V~2~ = \% CO~3~ W/V = \% HCO~3~ W/V = **\ [Method] II [(Two Indicator's Method)]** **(1) Total Alkalinity:** It determined by titration with standard acid using M.O. as indicator. **(2) Half Carbonate:** In other portion of the mixture, ½ is determined by titration with standard acid using ph. ph. as indicator. **[Procedure:]** \(1) Total Alkalinity: \(2) 1/2 Contents: **[Calculations:]** 1ml 0.1 N HCl = 0.0084 gm NaHCO~3~ 1ml 0.1 N HCl = 0.0053 gm Na~2~CO~3~ E.P. (I) = V~1~ = E.P. (2) = V~2~ = \% W/V = \% HW/V = **Lab 6** **[Back Titration]** Procedures employing back titrations consist in the addition of a known volume in excess of a standard volumetric solution to a weighed amount of sample, and the determination of the excess not used of the standard solution. Hence the amount of volumetric solution used by the substance is determined. **[In general, this procedure is used for:]** **Determination of Zinc Oxide** **[Principle:]** The determination depends on decomposing the zinc oxide by heating with a measured volume in excess of standard acid, when an equivalent amount of acid is used up. The excess unused acid is then back titrated with standard alkali solution. Ammonium chloride must be added before the back -- titration, in order to prevent the precipitation of zinc as hydroxide during the back -- titration with the alkali solution. **[Procedure:]** \- Weigh accurately about 0.2 g of ZnO in a conical flask dissolved by adding 2N HCl. \- Add 20 ml NH~4~CI solution. \- Add 2 drops M.R. indicator. \- Titrate with 1N NaOH. **[Calculations:]** 1ml 1N HCl = 0.04069 gm ZnO E.P. = \% W/W ZnO = **[\ Back Titration With Blank Determination]** If the volumetric solution is unstable or if it alters in strength during a determination a blank experiment must be made. A notable example is the heating of a liquid containing excess of standard alkali, cooling, and back titrating the excess. The heating and cooling results in an apparent change in strength of alkali liquid if certain indicators are used. This may be attributed to interaction of the reagent with the glass or the absorption of CO~2~ from the atmosphere. Therefore, the alkali must be standardized under the conditions to be used in the determination; this is called Blank Determination. **[Determination of Ammonium Sal]ts** **[Principle:]** Most Ammonium salts can be determined by heating with a measured excess of standard sodium hydroxide solution, when ammonia is evolved and an equivalent amount of sodium hydroxide is used up. The ammonia is boiled off completely, and the excess unused sodium hydroxide is back titrated with standard acid, performing a blank determination at the same time. **[Procedure:]** 1- Pipette 10 ml sample 2- Dilute with 20 ml water. 3- Add 20 ml 1N NaOH, boil gently for 5 min, till no NH~3~ odor. 4- Cool, back titrate with 1N HCl using 2 drops of M.O. as indicator. **[Calculations:]** **[\ ] [Determination of Acetylsalicylic acid (Aspirin)]** **[Principle:]**  **[Procedure:]** Transfer one tablet of Aspirin in a conical flask, add 20 ml 1N NaOH, dilute with 20 ml water, heat for 5 min, cool, titrate the excess unused NaOH with 1N HCl using 10 drops of ph. ph. as indicator. **[Calculations:]** 1ml 1N NaOH = 0.09008 gm CH~3~ -- COOC~6~H~4~. COOH E.P. = The content of aspirin per one tablet = **Lab 7** **[Precipitate Formation Titration]** **[Determination of Chloride by Mohr Method]** **[% NaCl]** **[Principle:]** Depends on the precipitation of chloride ions with an equivalent amount of silver ions to form silver chloride then any excess silver ions will precipitate chromate ion as silver chromate which is reddish in color. **[Procedure:]** 1- Pipette 10 ml sample into a stopper conical flask. 2- Add 10 ml distilled water. 3- Add 1 ml potassium chromate indicator. **[Calculation:]** 1 ml 0.05 N Ag NO~3~ = 0.002925g NaCl \% NaCl = = **[Determination of Bromide by Mohr Method]** **[% KBr]** **[Principle:]** **[Procedure:]** 1- Pipette 10 ml sample into a stopper conical flask. 2- Add 10 ml distilled water. 3- Add 1 ml potassium chromate indicator. **[Calculation:]** 1 ml 0.05 N Ag NO~3~ = 0.00595g KBr \% KBr = = **[\ Determination of Chloride by ]** **[Fajan Method % NaCl]** **[Principle:]** Depends on the precipitation of halides (Cl^-^, , I^-^) with silver nitrate to form colloidal silver halide at the end point this will acquire a positive charge and a secondary layer of negatively charged indicator ions will be absorbed from solution causing immediate change in the color of the precipitate. **[Procedure:]** 1- Pipette 10 ml sample into a stopper conical flask. 2- Add 10 ml distilled water. 3- Add 2 drops of fluorescein indicator. **[Calculation:]** 1 ml N/20 Ag NO~3~ = 0.002925g NaCl \% NaCl = = **[Determination of Bromide by ]** **[Fajan Method % KBr]** **[Principle:]** As under determination of chloride. **[Procedure:]** 1- Pipette 10 ml sample into a stoppered conical flask. 2- Add 10 ml distilled water. 3- Add 2 drops eosin indicator. **[Calculation:]** 1 ml N/20 Ag NO~3~ = 0.00595g KBr \% KBr = = **[Determination of Iodide by ]** **[Fajan Method % KI]** **[Principle:]** As under determination of chloride. **[Procedure:]** 1- Pipette 10 ml sample into stoppered conical flask. 2- Add 10 ml distilled water. 3- Add 1-2 drops eosin indicator. **[Calculation:]** 1 ml N/20 Ag NO~3~ = 0.0083g KI \% KI = = **Lab 8** **[Determination of Chloride by Volhard Method]** **[% NaCl]** **[Principle:]** The chloride is determined by precipitating it quantitatively with a measure excess of standard silver nitrate solution. The silver chloride is then removed by filtration or by coating with nitrobenzene, finally the excess unused of silver nitrate is determined by titration with standard thiocyanate. **[Procedure:]** 1- Pipette 10 ml sample into stopper conical flask. 2- Add 20 ml Silver nitrate. **[Calculation:]** 1 ml 0.05N Ag NO~3~ = 0.002925g NaCl \% NaCl = = **[Determination of Bromide by Volhard Method]** **[% KBr]** **[Principle:]** **[Procedure:]** 1- Pipette 10 ml sample into stopper conical flask. 2- Add 20 ml Standard 0.05N silver nitrate. 3- Add 3 ml conc nitric acid. **[Calculation:]** 1 ml 0.05N silver nitrate = 0.00595 g KBr \% KBr = = **Lab 9** **[Determination of Iodide by Volhard Method]** **[% KI]** **[Principle:]** **[Procedure:]** 1- Pipette 10 ml sample into stoppered conical flask. 2- Add 20 ml Standard 0.05N AgNO~­3~. 3- Add 2 ml conc nitric acid. **[Calculation:]** 1 ml 0.05N AgNO~3~ = 0.0083 g KI \% KI = = **[Determination of mixture HCl and HNO~3~]** **[Principle:]** The total acidity is determined by direct titration with standard alkali like NaOH using ph.ph. as indicator. OH^-^ + H^+^ \-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\--\> H~2~O The hydrochloric acid is determined by Volhard method. **[Procedure:]** [1-For total acidity:] Pipette 10 ml of sample in a conical flask, add 10 ml dist. water and 10 drops ph.ph. indicator.Titrate with 0.1 N NaOH [ ] [2-For HCl:] \- Pipette 10 ml of sample in a beaker 100 ml. \- Add 2 ml HNO~3~ (conc). \- Add by pipette 25 ml 0.05N AgNO~3.~ \- Stir the solution with glass rod for about 3 minutes to coagulate the precipitate. \- Filter the precipitate by decantation in a stopper conical flask. \- Wash the precipitate by dist. water and transfer to the filter paper. \- Titrate the filtrate with 0.05N ammonium thiocyanate using 3ml of ferric alum as indicator till faint red color appears in the supernatant layer. **[Calculation:]** 1ml 0.1N NaOH = 0.0063g HNO~3~ 1ml 0.05N AgNO~3~ = 0.001835g HCl E.P.(1)= E.P.(2)= \% HNO3= \%HCl = **Lab 10** [ **Mercurimetric Titration**] **[Determination of Chloride]** **[% NaCl]** **[Principle:]** The determination depends on titrating the chloride with standard solution of mercuric nitrate in presence of nitric acid. The indicator is a solution of nitroprusside at the end of the titration when all the chloride has been converted into the slightly ionized mercuric chloride. The first excess of mercuric ions (from the titrant), produces a white turbidity of mercuric nitroprusside. **[Procedure:]** 1- Pipette 10 ml sample. 2- Add 2 ml nitric acid. 3- Add 1 ml sodium nitroprusside indicator. **[Calculation:]** 1 ml 0.1N Hg (NO~3~)~2~ = 0.00585 g NaCl. \% NaCl = = [ **Complexometric Titration**] **[Determination of Magnesium]** **[% MgSO~4~]** **[Principle:]** The determination depends on titrating the magnesium salt at pH 10 with M/20 sodium edentate solution using Eriochrome black T as indicator. **[Procedure:]** 1- Pipette 10 ml sample. 2- Add 50 ml H~2~O. 3- Add 10 ml buffer solution pH~10~. 4- Add 3 drops EBT. **[Calculation:]** 1 ml M/20 Na~2~ H~2~Y = 0.0060185 g MgSO~4~. \% MgSO~4~ = = **[\ Determination of Calcium]** [ **% CaCl~2~. 6H~2~O**] **[Principle:]** Calcium salts are determined in much the same way as magnesium salts. However the color of the calcium indicator complex is less distinct than the color of magnesium indicator complex. Therefore a certain volume of standard magnesium sulphate solution is added to the calcium solution in order to sharpen the end point of the titration. The volume of magnesium sulphate is to be subtracted from the volume of the edetate solution used in the titration. **[Procedure:]** 1- Pipette 5ml sample calcium chloride. 2- Add 5 ml M/20 magnesium sulphate. 3- Add 5 ml distilled water. 4- Add 5 ml buffer solution. 5- Add 5 drops E.B.T. **[Calculation:]** 1 ml M/20 Na~2~ H~2~Y = 0.01095g CaCl~2~ 6H~2~O. \% CaCl~2~ 6H~2~O = = **Lab11** **[Determination of Calcium using Murexide Indicator]** **[% CaCl~2~. 6H~2~O]** **[Principle:]** At pH 12 murexide indicator is specific for calcium ion, so that calcium salt can be determined directly using murexide indicator at this specific pH. **[Procedure:]** 1- Pipette 10 ml sample. 2- Add 150 ml distilled water. 3- Add 5 ml 2N sodium hydroxide. **[Calculation:]** 1 ml M/20 EDTA = 0.01095 g CaCl~2~.6H~2~O. \% CaCl~2~.6H~2~O = = **[Determination of Calcium -- Magnesium mixture]** **[Principle:]** **[Procedure:]** **A) Total content:** 1- Pipette 10 ml sample. 2- Add 50 ml H~2~O. 3- Add 10 ml buffer solution. 4- Add 3 ml drops EBT indicator. **b) Calcium content:** 1- Pipette 10 ml sample. 2- Add 150 ml distilled water. 3- Add 5 ml 2N sodium hydroxide. 4- Add 0.05g murexide indicator. **[Calculation:]** 1 ml M/20 EDTA = 0.01095 g CaCl~2~.6H~2~O. 1 ml M/20 EDTA = 0.0060185 g MgSO~4~. \% CaCl~2~.6H~2~O = = \% MgSO~4~ = = ##### Performance Evaluation *Use the following indicators to evaluate your student during laboratory work.* Scoring Rubric: +-----------------------+-----------------------+-----------------------+ | **0- Below | **1- Meets | **2- Above** | | expectations** | expectations** | | +-----------------------+-----------------------+-----------------------+ **Performance checklist** --------------------------------------------------------------------- -- j- Behave in a manner that does not lead to fill a misconduct form.