CONCENTRATION EXPRESSION

Transcription

1 CONCENTRATION EXPRESSION 1

2 Moles = grams formula weight(g/mol) Where formula weight represents the atomic or molecular weight of the substance. Thus, Moles Na2SO4 = g f wt = g g/mol + Moles Ag = g f wt = g g/mol 2

3 Calculate the number of grams in one mole of CaSO 4 7H 2 O Solution One mole is the formula weight expressed in grams. The formula weight is Ca S 11 O 14 H g/mol 3

4 Mi lim oles = milligrams formula weight(mg/mmol) g Na 2 SO 4 = moles X f wt = moles X g/mol g Ag = moles X f wt = moles X g/mol We usually work with millimole quantities, so Miligrams = millimoles X formula weight (mg/mmol) Note that g/mol is the same as mg/mmol, g/l the same as mg/ml, and mol/l the same as mmol/ml. 4

5 Calculate the number of moles in 500 mg Na 2 WO 4 (sodium tungstate). Solution mg mg/mmol x mol/mmol = mol How many milligrams are in mmol Fe 2 O 3 (ferric oxide)? Solution mmol x mg/mmol = 39.9 mg 5

6 Millimoles = molarity X milliliters (or mmol = M X ml) A solution is prepared by dissolving 1.26 g AgNO 3 in a 250 ml volumetric flask and diluting to volume. Calculate the molarity of the silver nitrate solution. How many millimoles AgNO 3 were dissolved? Solution 1.26g/169.9g/mol M = 0.250L = mol/L(or mmol/mL) Then, Millimoles = ( mmol/ml)(250 ml) = 7.42 mmol 6

7 How many grams per milliliter of NaCI are contained In a M solution? Solution mol/l = mmol/ml mmol/ml X 58.4 mg/mmol X g/mg = g/ml How many grams Na 2 SO 4 should be weighed out to prepare 500 ml of a M solution? Solution 500 ml X mmol/ml = 50.0 mmol 50.0 mmol X 142 mg/mmol X g/mg = 7.10 g 7

8 Expressing concentrations of solution Molarity = moles of solute (mol) volume of solution (L) = mmoles of solute (mmol) volume of solution (ml) Normality (N) = no. of equivalents (eq) volume of solution (L) = no. of equivalents (meq) volume of solution (ml) 8

9 No. equivalents (eq) = mass (g) equivalent mass (g/eq) = normality (eq/l) x vol (L) No. equivalents (meq) = mass (mg) equivalent mass (mg/meq) = normality (meq/ml) x vol (ml) Equivalent mass = Relative Molec Mass (g/mol) No. of reacting units (eq/mol) Reacting units: H + (acid/base) or e - (redox) 9

10 Calculate the normality of a solution prepared by dissolving g K 2 Cr 2 O 7 in 1 L water, which is used to oxidise FeCl 2 in the unbalanced equation. RMM K 2 Cr 2 O 7 is K 2 Cr 2 O 7 + FeCl 2 + HCl CrCl 3 + FeCl 3 + KCl + H 2 O N = mass K 2 Cr 2 O 7 (g) equiv. mass K 2 Cr 2 O 7 (g/equiv) x ml Cr 2 O H + + 6e 2Cr3+ + 7H 2 O Equiv. mass = RMM K 2 Cr 2 O 7 = (mg/mmol) no. e 6 (meq/mmol) = mg/meq N = mg = N mg/meq x 1000 ml 10

11 Calculate the volume of a conc solution required to prepare 1 L M HCl solution that was taken from a bottle of conc. HCl. The label on the bottle indicates 37.0% HCl, density of 1.18 g/ml. RMM for HCl is

12 37.0% HCl = 37.0 g HCl 100 g conc. solution Mass HCl = 1.18 g conc soln x 37.0 g HCl Vol HCl ml conc soln 100 g conc. = g/ml Mass HCl required = 1000 ml x mmol x g ml mmol = 3.65 g Vol HCl required = 3.65 g = 8.36 ml g/ml 12

13 Solid Samples Express as weight % (% w/w) % = mass solute (g) x 100 mass of sample (g) For trace concentrations, expressed as ppt, ppm, ppb Ppt = mass solute (g) x 10 3 mass of sample (g) Ppm = mass solute (g) x 10 6 mass of sample (g) Ppb = mass solute (g) x 10 9 mass of sample (g) 13

14 Mass Units mg 10-3 g µg 10-6 g ng 10-9 g Volume Units L ml 10-3 L µl 10-6 L 14

15 A sample weighing g contains g. Calculate the % Fe in the sample. What is the Fe content in ppt and ppm? % Fe = g x 100 = 34.73% g Ppt Fe = g x 10 3 = ppt g Ppm Fe = g x 10 6 = ppm g 15

16 Liquid Samples Normally expressed as % mass/volume (% w/v) % w/v = mass solute (g) x 10 2 vol of sample (ml) Ppt = mass solute (g) x 10 3 vol of sample (ml) Ppm = mass solute (g) x 10 6 vol of sample (ml) Ppb = mass solute (g) x 10 9 vol of sample (ml) 16

17 % volume/volume (% v/v) % v/v = volume of solute (ml) x 10 2 vol of sample (ml) Ppt = volume of solute (ml) x 10 3 vol of sample (ml) Ppm = volume of solute (ml) x 10 6 vol of sample (ml) Ppb = volume of solute (ml) x 10 9 vol of sample (ml) 17

18 Common Units For Expressing Trace Concentrations Unit Abbreviation wt/wt wt/vol vol/vol Parts per million Ppm mg/kg mg/l µl/l (1ppm = 10-4 %) µg/g µg/ml nl/ml Parts per billion Ppb µg/kg µg//l nl/l (1ppb = 10-7 % = ng/g ng/ml pl/ml a 10-3 ppm) Milligram percent mg% mg/100g mg/100ml A pl = picoliter = L 18

19 DILUTIONS PREPARING THE RIGHT CONCENTRATION We often must prepare dilute solutions from more concentrated stock solutions. The millimoles taken for dilution will be the same as the millimoles in the diluted solution, i.e., M stock X ml stock = M Diluted X ml Diluted 19

20 You wish to prepare a calibration curve for the spectrophotometric determination of permanganate.you have a stock M solution of KMnO 4 and a series of 100 ml volumetric flasks. What volumes of the stock solution wll you have to pipet into flasks to prepare standards of 1.00, 2.00, 5.00, and X 10-3 M KMnO 4 solutions? Solutions A 100 ml solution of 1.00 X 10-3 M KMnO 4 will contain 100 ml X 1.00 X 10-3 mmol/ml = mmol KMnO 4 We must pipet this amount from the stock solution mmol / ml X χ ml = mmol χ = 1.00 ml stock solution For other solutions, will need 2.00, 5.00 and 10.0 ml of stock solution. 20

21 A 2.6 g sample of plant tissue was analyzed and found to contain 3.6 µg zinc. What is the concentration of zinc in the Plant in ppm? In ppb? Solution 3.6µg 2.6g = 1.4µg/g ß 1.4ppm 3.6X10 3 ng 2.6g =1.4X10 3 ng/g ß 1400ppb 21

22 (a) Calculate the molar concentrations of 1.00 ppm solution each of Li + and Pb 2+ Solution (a) Li concentration = 1.00 ppm = 1.00 mg/l Pb concentration = 1.00 ppm = 1.00 mg/l M 1.00mg Li /LX10 3 g/mg Li= =1.44X g Li /mol 4 mol/l Li M 1.00mg Pb/LX10 3 g/mg Pb= = 4.83X10 207g Pb/mol 6 mol/l Pb 22

23 What weight of Pb(NO 3 ) 2 will have to be dissolved in 1 liter of water to prepare a 100 ppm Pb 2+ solution? (b) 100 ppm Pb 2+ = 100 mg/l = g/l g Pb 207g/mol = 4.83X 10 4 mol Pb Therefore, we need 4.83 X 10-4 mol Pb (NO 3 ) X 10-4 mol X 283.3g Pb (NO 3 ) 2 /mol = g Pb (NO 3 ) 2 23

24 KAT 141nm2 24

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26 A selective reaction or test is one that can occur with other substances but exhibits a degree of preference for the substance of interest. A specific reaction or test is one that occurs only with the substance of interest. Unfortunately, few reactions are specific but many exhibit selectivity. Selectivity may be achieved by a number of strategies. Some examples are: Sample preparation (e.g., extractions, precipitation) Instrumentations (selective detectors) Target analyte derivatization (e.g. derivatize specific functional groups with detecting reagents) Chromatography, which provides powerful separation 26

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28 Material Max. Working Temperature (C) Sensitivity to thermal Shock Chemical Inertness Notes Borosilicate glass C change OK Attacked somewhat by alkali solutions on heating Trademarks : Pyrex, Kimax, Soft glass Poor Attacked by alkali solutions Boron free. Trademark : Corning Alkaliresistant glass More sensitive Than Borosilicate Fused quartz 1050 Excellent Resistant to most acids, halogens Quartz crucibles used for fusions 28

29 Material Max. Working Temperature (C) Sensitivity to thermal Shock Chemical Inertness Notes High silica glass 1000 Excellent More resistant to alkalis than borosilicate Similar to fused quartz Trademark : Vycor (Corning) Porcelain 1100 (glazed) Good Excellent 1400 (unglazed) Platinum ca.1500 Resistant to most acids, molten salts. Attacks by aqua regia, fused nitrates, cyanides, chlorides at > C. Alloys with gold, silver, and other metals Usually alloyed with iridium or rhodium to increase hardness. Platinum crucibles for fusions and treatment with HF. 29

30 Material Max. Working Temperature (C) Sensitivity to thermal Shock Chemical Inertness Notes Nickel and iron Fused samples contaminated with the metal Ni and Fe crucibles used for peroxide fusions Stainless steel Excellent Not attacked by alkalis and acids expect conc. HCI, dil H 2 SO 4 and boiling conc. HNO 3 Polyethylene 115 Not attacked by alkali solutions or HF. Attacked by many organic solvents (acetone, ethanol OK) Flexible plastic 30

31 Material Max. Working Temperature (C) Sensitivity to thermal Shock Chemical Inertness Notes Polystyrene 70 Not attacked by HF. Attacked by many organic solvents Somewhat brittle Teflon 250 Inert to most chemicals Useful for storage of solutions and reagents for trace metal analysis 31

32 Some Common Drying Agents Agent Capacity Deliquescent a Trade Name CaCI 2 (anhydrous) High Yes CaSO 4 Moderate No Drierite (W.A. Hammond Drierite Co). CaO Moderate No MgCIO 4 (anhydrous) High Yes Anhydrone (J.T. Baker Chemical Co.); Silica gel Low No AI 2 O 3 Low No P 2 O 5 Low Yes Q Becomes liquid by absorbing moisture 32