1.00 Measurements. Chemistry 251. Dr. Fred Omega Garces

Transcription

1 1.00 SI -Units Mass and Weights The Mole Millimole Solutions and Concentration Analytical Molarity Equilibrium Molarity Composition by Parts % Composition ppm, ppb Density and specific gravity Titration p-function Chemical Stoichiometry Dr. Fred Omega Garces Chemistry 251 Miramar College 1

2 1.01 System of What units are used? America The Rest of the World Scientific Community English System 1 ft = 12 in 1 yd.= 3 ft 1 mi. = 1,760 Yd. 1 mi = 5280 ft Metric System 1 km = 1000 m 1 m = 100 cm 1 m = nm Le System International d Unites time g sec length g m mass g kg temperature g Kelvin 2

3 SI units; International System of Units There are seven quantities in which all units can be derived; these are the base units and the SI defines these as: Quantity Unit Symbol length meter m mass kilogram kg time seconds s temperature kelvin K amount mole mol electric current ampere A luminous intensity candela cd All others units are derived units 3

4 SI units; International System of Units The definition of units: All others units are derived units 4

5 SI Derived Units Some units that are defined in terms of the fundamental SI units: These units are derived from 7 basic units 5

6 Prefixes used instead of exponents Some units that are defined in terms of the fundamental SI units: Prefix are spaced in multiple of e e e -9 - e -6 - e e 3 - e 6 Femto - pico - naano - micro - milli kilo - mega 6

7 Chemical Concentrations 7

8 Defining the MOLE What is a mole? SI base unit for amount anything. a magnifier between the atomic scale and macroscopic scale. a conversion factor. a chemist dozen. the sum of the atomic mass of all elements in a substance when the mass is expressed in grams. (Molar Mass) the ratio of elements in a substance as indicated by its chemical formula. the ratio of elements combining in a chemical reaction as indicated by the balance equation. The milli-mole useful when working with ml. 2.0 ml of 0.50 M solution = 1.0 mmol 8

9 1.02 Solution and Concentration Analytical Conc.: total moles of solute in 1 L solution (F) Equilibrium Conc.: the concentration of species at equilibrium 6 ways of expressing concentration- Molarity(M) - moles solute / Liter solution Molality * (m) - moles solute / Kg solvent Mass percent (% m)- (grams solute/total grams of solution) 100 Normality (N) - Number of equivalent / Liter solution Formality (F) - moles strong electrolyte / Liter solution mole fraction * ( c A) - moles solute / Total moles solution * Note that molality is the only concentration unit in which denominator contains only solvent information rather than solution. 9

10 Concentration Relationship Molecular Weight Moles mass } Solute Molc Wt Density* Moles mass Volume } Solvent c m** %m M * Density and Specific Gravity **Volume used must be volume of solvent. N (analytical conc) F (analytical conc) 10 Formality (F): Molarity of strong electrolyte to emphasize the dissociation process of the chemical, ii.e, chemical has change to other species in solution.

11 Percent % Concentration % Concentration w/w = Wt Solute 100 g 100 g % (pph) (pph) Wt Soln g w/v = Wt Solute 100 g 100 g % (pph) Vol Soln ml v/v = Vol Solute 100 ml 100 g % (pph) Vol Soln ml ppm & ppb (For dilute solution) v/v = Vol Solute 10 6 g 10 6 g ppm (ppm) Vol Soln g v/v = Vol Solute 10 9 g 10 9 g ppb (ppb) Vol Soln g 11 1 ppm = 1mg/L or 1ug/mL 1 ppb = 1ug/L or 1ng/mL

12 Solution Preparation 12

13 1. Suppose 1.00 g of K2Cr2O7 is dissolved in enough water to give 100. ml of solution. What is the molar concentration of K2Cr2O7? MM= Concentration M K2Cr2O Solution Prep 2. From Previous problem: Suppose 10 ml of the M K2Cr2O7 solution is diluted to 50.0mL. What is the new concentration? 100 ml of M K2Cr2O7 Before Dilution: After Dilution: 10-mL Aliquot: 50-mL Volume: 10-ml of M 50-mL vol. contains mol Moles in aliquot- Moles of new solution- Moles = mol L mol L = mol = Mnew Vol new = Mnew L M = Mnew M1 V1 = M2 V2 13

14 Concentration Example Calculate the molar (Formal) concentration of sulfuric acid (98 g/mol) with specific gravity of 1.84 and a concentration of 96.5% w/w. 96.5g H 2 SO 4 100g Solution 1mol 98g 1.84g ml 1000mL 1L = 18.1 F i) Describe how to prepare 500-mL of 5.0 M solution from this stock. ii) What is the density of this new solution? ii) Describe how to prepare 1.0L of 25ppm (m/v) solution from part i. 14

15 Concentration Example Calculate the molar (Formal) concentration of sulfuric acid (98 g/mol) with specific gravity of 1.84 and a concentration of 96.5% w/w. 96.5g H 2 SO 4 100g Solution 1mol 98g 1.84g ml 1000mL 1L = 18.1 F i) Describe how to prepare 500-mL of 5.0 M solution from this stock. ii) Describe how to prepare 1.0L of 25ppm (m/v) solution from part i. 15

16 Gravimetric Analysis 16

17 Gravimetric Analysis Chemical analysis based on the mass of final product. This is a classical stoichiometry type of calculations. In an experiment in which the calcium concentration is to be determined from an unknown solution, exactly ml of unknown solution is transferred to a 250-mL beakers, and diluted with ~75 ml of 0.1 M HCl. Ammonium oxalate is added and stirred into the solution after which urea is added and the solution is heated slowly. After cooling the solution, the precipitate is filtered, dried in an oven and cooled. The mass of calcium oxalate monohydrate product was determine to be g. (MM CaC2O4 H2O is ± g/mol). Calculate the molarity of calcium in the original sample. Ca +2 + C2O4 2- g CaC2O4 H2O (ppt) 1 mol CaC g CaC 2 O 4 H 2 O 2 O 4 H 2 O g CaC 2 O 4 H 2 O 1 mol Ca 1 mol CaC 2 O 4 H 2 O L = M Ca+2 17

18 Gravimetric Analysis Chemical analysis based on the mass of final product. This is a classical stoichiometry type of calculations. In an experiment in which the calcium concentration is to be determined from an unknown solution, exactly ml of unknown solution is transferred to a 250-mL beakers, and diluted with ~75 ml of 0.1 M HCl. Ammonium oxalate is added and stirred into the solution after which urea is added and the solution is heated slowly. After cooling the solution, the precipitate is filtered, dried in an oven and cooled. The mass of calcium oxalate monohydrate product was determine to be g. (MM CaC2O4 H2O is ± g/mol). Calculate the molarity of calcium in the original sample. Ca +2 + C2O4 2- g CaC2O4 H2O (ppt) 1 mol CaC g CaC 2 O 4 H 2 O 2 O 4 H 2 O g CaC 2 O 4 H 2 O 1 mol Ca 1 mol CaC 2 O 4 H 2 O L = M Ca+2 18

19 Volumetric Analysis 19

20 Volumetric Analysis Chemical analysis based in which volumes of reagent are used to react with analyte. This too is a classical stoichiometry type of calculations. A 1-mL sample of unknown was pipetted into a 250-mL flask and fill to the mark with deionized water. A 50mL aliquot samples was placed in a 250mL Erlenmeyer flasks and then 3 ml of ph 10 buffer and 6 drops of Eriochrome black T indicator was added and fill to the mark with water. This solution was titrate with ml of M EDTA. What is the Ca +2 concentration (M) of the original unknown? Ca +2 + EDTA -4 -(Acidic)g CaH2EDTA (to end pt) L EDTA mol EDTA 1 L 1 mol Ca +2 1 mol EDTA L aliquot = M mol Ca +2 L L = mol Ca mol Ca L = M 20

21 Volumetric Analysis Chemical analysis based in which volumes of reagent are used to react with analyte. This too is a classical stoichiometry type of calculations. A 1-mL sample of unknown was pipetted into a 250-mL flask and fill to the mark with deionized water. A 50mL aliquot samples was placed in a 250mL Erlenmeyer flasks and then 3 ml of ph 10 buffer and 6 drops of Eriochrome black T indicator was added. This solution was titrate with ml of M EDTA. What is the Ca +2 concentration (M) of the original unknown? Ca +2 + EDTA -4 -(Acidic)g CaH2EDTA (to end pt) L EDTA mol EDTA 1 L 1 mol Ca +2 1 mol EDTA L aliquot = M 4 Ca mol Ca +2 L L = mol Ca mol Ca L = M 21

22 1.04 Mapping out Stoichiometry Vol ( L) Liquid phase Density (g / cc) # of molecules / atoms N Av ( ) particle (atomic) phase # of molecules / atoms N Av ( ) particle (atomic) phase Vol (L) Liquid (l) phase Density (g / cc) Mass (g) Conc. (mol / L) Vol (L) Pressure (atm) Temperature (K) Volume (L) Molar Mass (g / mol) Solid phase Aqueous phase Gas phase Balance equation Moles A ## ## ## ## Moles Stoic. coefficient. B R (.0821 atm L ) mol K Molar Mass (g / mol) Solid phase (g) Gas phase R (.0821 atm L ) mol K Aqueous phase Mass (g) Conc. (mol / L) Vol (L) Pressure ( atm ) Temperature ( K) Volume ( L) 22

23 The same Problem with different faces Type of Stoichiometry Questions Consider: aa + bb g cc + dd Questions: Products:Given X grams of A, how much C or D is produced Given X grams of B, how much C or D is produced Given X grams of C, how much D is produced Given X grams of D, how much C is produced Reactants: Producing X grams of C, how much A or B is required Producing X grams of D, how much A or B is required Co-product: Producing X grams of C, how much D is also produced Producing X grams of D, how much C is also produced Also problem can ask what is the limiting and excess reagents? 23

24 Let the Titrations Begin (Volumetric Analysis) 24

25 Reaction Between Acid - Base What is the ph of a solution when an acid is mixed with a base? Stoichiometry Problem: HA g H 3 O + + A- MOH g OH - + M + MOH + HA g H 2 O + MA Stoichiometry Problem: The amount of H 3 O + or OH - remaining after a portion is neutralize determines the ph of the solution. In an acid - base reaction, H + & OH - always combine together to form water and an ionic compound (a salt): Neutralization Reaction. HCl (aq) + NaOH (aq) g H 2 O (l) + NaCl (aq) Analysis is a Stoichiometry problem only if a strong acid is combined with a strong base. 25

26 Acid/Base Titration Titration A technique of chemical analysis to determine the amount of a substance in a sample. i.e., What is the acidic content of Lake Ellsinore? A sample can be tested by titration. In a titration experiment, a known volume of a standard concentrated solution (the titrant) is used to analyze a sample (the analyte). One is usually an acid, the other a base. An indicator is added to the analyte to signal when the titration is complete. This is called the endpoint. When the moles of acid(h 3 O + ) and moles of base (OH - ) are equal in a titration experiment, the stoichiometric equivalent point is reached. This is called the equivalent point. Indicator changes endpoint moles titrant = moles equivalent point. 26

27 Setting up a titration experiment. Titration Setup Titration is a common laboratory method of quantitative/chemical analysis which can be used to determine the concentration of a known reactant. Because volume measurements play a key role in titration, it is also known as volumetric analysis. A reagent, called the titrant, of known concentration (a standard solution) and volume is used to react with a measured quantity of reactant (Analyte). Using a calibrated burette to add the titrant, it is possible to determine the exact amount that has been consumed when the endpoint is reached. The endpoint is the point at which the titration is stopped. This is classically a point at which the number of moles of titrant is equal to the number of moles of analyte, or some multiple thereof (as in di- or tri- protic acids). In the classic strong acid-strong base titration the endpoint of a titration is when the ph of the reactant is just about equal to 7, and often when the solution permanently changes color due to an indicator. There are however many different types of titrations (see below). Many methods can be used to indicate the endpoint of a reaction; titrations often use visual indicators (the reactant mixture changes color). In simple acid-base titrations a ph indicator may be used, such as phenolphthalein, which turns (and stays) pink when a certain ph (about 8.2) is reached or exceeded. Methyl orange can also be used, which is red in acids and yellow in alkalis. Not every titration requires an indicator. In some cases, either the reactants or the products are strongly colored and can serve as the "indicator". For example, an oxidation-reduction titration using potassium permanganate (pink/purple) as the titrant does not require an indicator. When the titrant is reduced, it turns colorless. After the equivalence point, there is excess titrant present. The equivalence point is identified from the first faint pink color that persists in the solution being titrated. Due to the logarithmic nature of the ph curve, the transitions are generally extremely sharp, and thus a single drop of titrant just before the endpoint can change the ph significantly leading to an immediate color change in the indicator. That said, there is a slight difference between the change in indicator color and the actual equivalence point of the titration. This error is referred to as an indicator error, and it is indeterminate. (wikipedia.org) 27

28 Acid-Base Indicators Some common acidbase indicators. The color changes occur over a range of ph values. Notice that a few indicators have two color changes over two different ph ranges. Acid-Base Indicator Mechanism for phenolphthalein indicator. At Low ph phenolphthalein is colorless and has a structure in which there is a five membered ring. In the presence of excess base the five membered ring is broken and the resulting change in conformation gives rise to a compound which is pink. 28

29 Acid-Base Calculations Indicator moles H 3 O + = moles OH - M acid V acid = M base V base M acid V acid M base V base For monoprotic A & B For polyprotic A & B but N acid V acid = N base V base for monoprotic and polyprotic A & B eq H 3 O + L H 3 O + = eq OH - L OH - L Soln L Soln For Acid base calculation at Equivalence pt: Sometimes: M acid V acid = M base V base but always: N acid V acid = N base V base changes endpoint. Indicator is chosen so that endpoint occurs at equivalent pt. The following is true at the equivalence point. The difference between the two is the titration error. This error can be corrected by titration of a blank 29

30 The Precipitation Titration Curve Shape of titration curve First derivative- DpH / DV vs Vol Base Second derivative- D 2 ph / DV vs Vol Base 30

31 Misc Topics: Titration of a mixture Calculating Titration Curves with Excel End Point Detection (blank titration) Experiment design efficiency 31

32 Titration: Notes to Consider Indicator is chosen so that endpoint occurs at equivalent pt. At the true equivalent point, all the analyte is neutralize H + + OH - è H 2 O, no excess OH-, but excess OH- is needed. Once there is an excess titrant the indicator is deprotonated and the indicator changes changes form and a color change is observed. How much changed indicator form is necessary to show the endpoint? The difference between the two is the titration error. This error can be corrected by titration of a blank. Carried out without analyte. 32

33 Summary In this chapter we reviewed the basic fundamentals that you will use in analytical chemistry. The basic topic covered here are quantitative calculations in chemistry. Using the moles as the fundamental method of counting atoms. 33

34 Exercise Try These Exercises 1.11 Pb accumulates at rate of 0.03mg m -2 day -1. How many metric tons (1MT = 1000kg) of Pb fall on the 535 km 2 of Chicago in 1 year? 6 Tons 1.23 Find molarity of glucose (180.2 g/mol) in 120 mg / 100mL blood. 6.7 e-3 M 1.26 How many grams of NaF ( g/mol) must be added to a reservoir with diameter of m and depth of 10.0 m, so the Fluoride (F -, g/mol) concentration is 1.6 ppm? 5.6e-6 NaF 1.32 Conc sulfuric acid is 18.0 M, with a 98.0% m/m. Calculate the density of this solution. How would you prepare 1.00 L of with a concentration of 1.00 M solution from this stock? 1.80 g soln. / ml 1.35 How many grams of wt. % aq HF (20.01 g/mol) are required to provide a 50% excess to react with 25.0 ml of M Th4+ by the reaction: 14.4 g Th 4+ (aq) + 4F - (aq) ThF 4 (s) 34

35 Answer to Lecture Notes Exercise Slide g H 2 SO 4 100g Solution 1mol 98g 1.84g 1000mL ml 1L = 18.1 F Slide 14 1 mol CaC g CaC 2 O 4 H 2 O 2 O 4 H 2 O g CaC 2 O 4 H 2 O 1 mol Ca 1 mol CaC 2 O 4 H 2 O L = M Ca+2 Slide L EDTA mol EDTA 1 L 1 mol Ca +2 1 mol EDTA L aliquot = M mol Ca +2 L L = mol Ca mol Ca L = M 35