1. SOLUTION PREP. AND BEER’S LAW Experiment #1

2. What is this experiment about? This experiment has 2 parts to it. They are as follows: 1.How to make solutions of required concentration? 2. Using Beer’s law to determine concentration of unknown solutions?

3. Solution Preparation and Beer’s Law Experiment 1

4. What is the game plan? First: Briefly, review the concepts underlying making solutions. Second: Dilution of solutions. Third: Deal with the BEER’S law……..

5. Solution Preparation How do we make a solution? solution Solvent (it dissolves the solute) + Solute (it dissolves in the solvent)

6. Solution Preparation The ratio of the amount of solute dissolved in a certain volume of solution gives us the concentration of the solute in that solution. Units of concentration: g/mL, g/L, g/g, mL/mL Molarity, Normality g/g: concentration expressed as % by mass mL/mL: concentration expressed as % by volume Normality: Number of gram equivalents per liter of solution

7. Molarity Defined as the number of moles of solute per liter of solution A Mole refers to a collection of 6.022 x 10 23 items. 6.022 x 10 23 is also called Avogadro’s number

8. Always remember! Moles refers to a collection of particles. Moles is a number Moles is not a weight but Moles can be calculated from the weight of the substance So our goal is to know the moles of the solute that is to be dissolved in a particular volume of solvent, to make up a solution of required molarity (a unit of concentration).

9. Why is preparing a solution correctly so important? How about these simple life situations? 1.Excess salt in your soup 2.Insufficient amount of sugar in the coffee 3.More or less alcohol in an alcoholic beverage How about these life threatening situations? 1.Excess or insufficient amount of the drug in a medication. 2.Excess chlorine in a swimming pool

10. A hypothetical example A box containing tennis balls tennis ball outside the box Find the number of tennis balls inside the box without opening the box, if each and every ball inside the box weighs the same as the one outside the box? The weight of empty box is 5.0 g. The Weight of the box with the balls is 55.0 g. The weight of the tennis ball outside the box is 2.0 g.

11. Solution: Mass of Balls + box = 55.0 g Mass of the box with no balls in it = 5.0 g Mass of just the balls = 55.0 -5.0 = 50.0 g Mass of the ball outside the box = 2.0 g Since each ball inside the box weighs the same as the ball outside the box, If we divide the mass of all the balls by the weight of the single ball that is outside the box, we can know how many balls are inside the box (without opening the box). Proceeding, 50.0 ÷ 2.0 = 25 balls.

12. Lessons learned from the example: If we know the total mass of all the balls inside the box and the mass of one ball, we can determine the number of balls inside the box, without opening it. Important condition: Mass of the ball outside the box = mass of each and every ball inside the box.

13. How about we apply this example to our problem of finding the number of solute particles that have to be dissolved in our solution? What are these particles that make up our solutions? They are chemical substances. They are made up of atoms. The atoms are combined in a certain way to form molecules. For example: salt solution: salt + water Salt = Sodium Chloride, NaCl So now we can call these particles as molecules. Think that all the particles that we are adding to make our solution are inside the box that had the tennis balls. Let us say just one of those particles is outside the box.

14. New lesson: If we know the total mass of all the molecules inside the box and the mass of a single molecule outside the box, we can find the number of molecules that we have inside the box. Important condition: Mass of the molecule outside the box = mass of each and every molecule inside the box

15. How do we find the following? 1. Total mass of all the molecules inside the box? Weigh the substance on a balance. 2. The mass of a single molecule? Mass of a single molecule is called its molecular weight. Since a molecule is made up of atoms. We can find the molecular weight by adding the mass of the individual atoms (atomic mass) that make up the molecule. 3. Mass of a single atom? Mass of a single atom can be obtained for each and every atom from the atomic mass values given in a periodic table.

16. Atomic mass is given below the symbol of the element

17. Units of Atomic Mass Most common unit of mass: g, kg, lb Mass of an atom or atomic mass has units: amu 1 a.m.u or atomic mass unit = 1.66 x 10 -24 g For example mass of a single atom of sodium or the atomic mass of sodium = 22.99 amu = 22.99 x 1.66 x 10 -24 g = 3.816 x 10 -23 g = 0.00000000000000000000003816 g Since it is impossible to measure a single atom on a common lab balance, we always measure a collection of atoms or molecules.

18. How many are atoms/molecules are in this collection? the atomic mass of sodium = 22.99 amu = 22.99 x 1.66 x 10 -24 g = 3.816 x 10 -23 g = 0.00000000000000000000003816 g 6.022 x 10 23 Na atoms will be required make up 22.99 g of Na = 22.99 x 1.66 x 10 -24 g x 6.022 x 10 23 Na atoms = 22.98 g This collection of 6.022 x 10 23 Na atoms = 1 mole of Na atoms So we can say that 1 mole of Na atoms weighs 22.99 g. Therefore, the weight off 1 Na atom = 22.99 g/mol atomic mass has unit: amu or g/mol

19. Example Problem 1 What is the molarity of a 500 mL solution that contains 10 g of sodium chloride (NaCl)? Given: Volume of the solution: 500 mL or 0.5 L (Remember! 1000 mL = 1.0 L) Mass or weight of the solute: 10 g Name and chemical formula of the compound: Sodium Chloride (NaCl)? To be found: Concentration or Molarity of the solution:? Methodology:

20. Example Problem 1 Calculation: Molecular Weight of NaCl= Atomic Weight of Na + Atomic Weight of Cl From Periodic Table

21. Example Problem1

22. Example Problem 2 How will you prepare 500 mL of 0.84 M solution of glucose(C 6 H 12 O 6 )? Given: Volume of the solution to be prepared: 500 mL Concentration or Molarity of the solution: 0.84 Molar Name and chemical formula of the compound: glucose(C 6 H 12 O 6 )? Information that needs to be found: The mass or weight of glucose that is required to make up this 500 mL 0.84 M solution. Methodology:

23. Example Problem 2 Calculation:

24. Example Problem 2

25. When making solutions 500 mL mark 500mL Volumetric flask Erlenmeyer flask Beaker Very accurate less accurate

26. Summary I 1. 2. Moles refers to a collection of particles. Moles is a number Moles is not a weight but moles can be calculated from the weight of the substance 3. Atomic mass has unit: amu or g/mol 4. Molecular mass or Molecular weight, which is sum of the atomic mass of atoms in that molecule, also has units amu or g/mol

27. Summary I Contd. 5. 6. Mass of one mole of atoms or molecule is called its molar mass. Unit of molar mass is g. Ex. Molar mass of sodium = 22.99 g Molar mass of NaCl= 58.44 g.

28. Dilution of solutions Dilution means making a solution of lower concentration from a solution of higher concentration. More concentrated (stock solution) Less concentrated M1M1 M2M2 Need to take a certain volume from the more concentrated solution And make it up to a certain volume of diluted solution

29. How will you prepare 500 mL of 0.16 M solution of glucose(C 6 H 12 O 6 ) from a 500mL 0.84 M glucose solution? Use water as the solvent. What makes a solution more or less concentrated is the number of solute molecules per liter of solution. Example problem Given: Concentration of the stock solution: 0.84 M Total volume of the stock solution: 500 mL or 0.5 L Concentration of the dilute solution: 0.16 M Volume of the dilute solution: 500 mL= 0.5 L To be found: The volume of the stock solution that needs to be taken out of 500 mL stock so that The dilution can be made

30. Number of moles of glucose in the 500 mL stock solution: Example problem Methodology and calculation:

31. Number of moles of glucose in the 500 mL of 0.16 M solution If we want to make a dilute (0.16 M, 500 mL) solution of glucose from the stock Solution, the dilute solution dictates that we need to have only 0.08 moles of glucose. If we use all the 500 mL of 0.84 M stock to make the dilute solution, we will end up With 0.42 moles of glucose which is much more than the 0.08 moles that we want. So we will have to figure out what volume of stock will give us the required 0.08 moles of glucose. Example problem

32. Molarity of Stock (M) Volume of stock (L) Moles of stock (M  L) 0.840.005 (5 mL)0.0042 0.84 0.010 (10 mL)0.0084 0.84 0.015 (15 mL)0.0126 0.840.020 (20 mL)0.0168 0.840.025 (25 mL)0.0210 We can do this by trial and error by changing the volume of stock and figuring Out which value of volume would give the required 0.08 moles. Or we can do this by solving a simple equation. Example problem

33. Molarity of stock ( )  Vol. of stock (L)  Vol. of the dilute soln. (L) Molarity of dilute soln. ( ) = 0.84 M  V stock L = 0.16 M  0.5 L To make the solution: Take 95 mL stock solution and water and fill a 500 mL volumetric flask to the mark. Stopper the flask and shake the flask few times so as to enable uniform mixing. Example problem

34. Lessons learned So far!! 1.From Example 1. We learnt how to make a solution of specific concentration, given the molarity, volume and the of course, the name of the compound 2. From Example 2. We learnt how to make a dilution from a concentrated stock solution. 3. If solution need to be made accurately to a certain concentration, they need to be made in a volumetric flask and not in a beaker or a erlenmeyer flask.

35. Properties of light, Beer’s Law and its application

36. Properties of light Light exhibits both wave-like and particle like properties wave-like properties c (not a greek Symbol) Speed nm or Å 1nm=10 -9 m 1 Å = 10 -10 m (lambda) Wave length (nu) Frequency Value/unitGreek symbolProperty 3 x 10 8 m/s

37. 1 second    wave-like properties Wavelength is inversely proportional to frequency When wavelength increases, the frequency decreases

38. Particle-nature of light  n Energy is directly proportional to number of photons and their frequency  = n h photon

39.  = nh n – number of photons. It is not number of moles of photons. h – planck’s constant= 6.626 x 10 -34 Js – frequency of light, units  - Change in energy Units of  :

40.  = nh  = Substituting the value of from the above equation into the equation for  E (the equation below), we get

41. human eye can see only these colors

42. Interaction of light with matter I in I out Definition of absorbance & transmittance Absorbance Transmittance Absorbance (A) and transmittance (T) are unitless s

43. Graph #1 Graph #2 max =530 nm max =740 nm Wavelength of maximum absorption, max

44. Beer-Lambert’s law I in I out [2] I out [1] Absorbance Vs. Concentration (c ) I in I out [2] I out [1] Absorbance Vs. Path length (b) Path length- the distance that light travels in the sample Unit of path length = cm Unit of concentration = M s s s s This is not true at high concentrations

45. Beer-Lambert’s law a – Molar absorptivity Units of molar absorptivity:

46. “A” is unitless but “a” has units M -1 cm -1

47. Definition of molar absorptivity a - is a measure of the amount of light absorbed per unit concentration and pathlength A compound with a high molar absorptivity is very effective at absorbing light (of the appropriate wavelength)

48. Finding unknown concentrations of solution using Beer’s law Diameter of the test tube = path length Light source Sample test tube Detector Diameter = d cm

49. Finding unknown concentrations of solution using Beer’s law c1c1 A1A1 c2c2 A2A2 c3c3 A3A3 c4c4 A4A4 c5c5 A5A5 Solutions of known concentration and their absorbances Absorbance Concentration C1,a1 C2,a2 C3,a3 C4,a4 C5,a5 Best-fit line

50. Absorbance Concentration c 1,A 1 c 2,A 2 c 3,A 3 c 4,A 4 c 5,A 5 Best-fit line x x m n AmAm AnAn cmcm cncn x y Equation of the best-fit line: Y=mX + z m = slope = intercept C= mA + Z concentration Absorbance

51. C= mA + Z C unknown = mA unkown + Z If using Microsoft excel: Please refer to section 3 of MATHCHEM 1.0 Obtain the equation of best-fit line

52. Sample Problem Find the molar absorptivity of the a 0.5 M solution whose absorbance is 0.77,When measured in a tube of path length 2 cm? Given: Concentration of the solution = 0.5 M =c Absorbance = 0.77= A Path length = 2 cm = b To be found: Molar absorptivity of the solution= ?= a

53. Sample Problem Methodology: Beer’s law Calculation:

54. Summary 1. Beer’s Law: and 2. Wave length is a property of light. It is the distance between two consecutive crests or troughs in a wave. The typical unit of wavelength is nm or Ǻ. 3.Path length is NOT a property of light, It is the distance that light travels in the sample. Typical unit of path length is cm. 4. If we know the absorbance of solutions of known concentration, we can use Beer’s law to find the concentration of a solution, for which we know only the value of absorbance.