1 / 42

Review of Basic Concepts, Molarity, Solutions, Dilutions and Beer’s Law

Review of Basic Concepts, Molarity, Solutions, Dilutions and Beer’s Law. Chapter 4 4.5. Aqueous Solutions. In Chemistry, many reactions take place in water. This is also true for Biological processes. Reactions that take place in water are said to occur in an aqueous solution .

rupali
Download Presentation

Review of Basic Concepts, Molarity, Solutions, Dilutions and Beer’s Law

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Review of Basic Concepts, Molarity, Solutions, Dilutions and Beer’s Law Chapter 4 4.5

  2. Aqueous Solutions • In Chemistry, many reactions take place in water. This is also true for Biological processes. • Reactions that take place in water are said to occur in an aqueous solution. • Three types of reactions take place in aqueous solutions: Precipitation, Acid-Base and Redox.

  3. Properties of Aqueous Solutions • Solution-a homogeneous mixture of two or more substances. • Solute-a substance in a solution that is present in the smallest amount. • Solvent-a substance in a solution that is present in the largest amount. • In an aqueous solution, the solute is a liquid or solid and the solvent is always water.

  4. Properties of Aqueous Solutions • All solutes that dissolve in water fit into one of two categories: electrolyte or non-electrolyte. • Electrolyte-a substance that when dissolved in water conducts electricity • Non-electrolyte-a substance that when dissolved in water does not conduct electricity. • To have an electrolyte, ions must be present in water.

  5. Electrolytic Properties of Aqueous Solutions • NaCl in water. • What happens? • NaCl(s)→ Na+(aq) + Cl–(aq) • Completely dissociates

  6. Strong vs. Weak Electrolytes • How do you know when an electrolyte is strong or weak? • Take a look at how HCl dissociates in water. • HCl(s)→ H+(aq) + Cl–(aq)

  7. Electrolytic Properties of Aqueous Solutions

  8. Electrolytic Properties of Aqueous Solutions

  9. Hydrated Ions

  10. Electrolytic Properties of Aqueous Solutions • What about weak electrolytes? • What makes them weak? • Ionization of acetic acid • CH3COOH(aq)↔ CH3COO–(aq) + H+(aq)

  11. Electrolytic Solutions

  12. Precipitation Reactions • Precipitation Reaction-a reaction that results in the formation of an insoluble product. • These reactions usually involve ionic compounds. • Formation of PbI2: • Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)

  13. Preciptate

  14. Precipitate

  15. Precipitation Reactions • How do you know whether or not a precipitate will form when a compound is added to a solution? • By knowing the solubility of the solute! • Solubility-The maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature. • Three levels of solubility: Soluble, slightly soluble or insoluble.

  16. Precipitation Reactions

  17. Determining Solubility • Determine the level of solubility for the following: (1) Ag2SO4 (2) CaCO3 (3) Na3PO4

  18. Acid-Base Reactions • Acids-generally have a sour taste, change litmus from blue to red, can react with certain metals to produce gas, conduct electricity. • Bases-generally have a bitter taste, change litmus from red to blue, feel slippery, conduct electricity. • BrØnstead Acid-proton donor • BrØnstead Base-proton acceptor

  19. Acid-Base Reactions • Acid or Base? • HCl(aq) + H2O(l)→ H3O+(aq) + Cl–(aq) • NH3(aq) + H2O(l)→ NH4+(aq) + OH–(aq)

  20. Acid-Base Reactions • Look at the following compounds and decide whether they are a BrØnstead Acid or a BrØnstead Base. • HBr • NO2– • HCO3–

  21. Acid-Base Reactions

  22. Oxidation-Reduction Reactions • Can also be called Redox reactions. • Considered electron-transfer reactions. • Occur in steps called half-reactions. • Half-Reactions- Two parts to a redox reaction that explicitly show the electrons involved and where they are transferred.

  23. Oxidation Reduction Reactions • Oxidation Reaction-refers to the half-reaction that involves the loss of electrons. • Reduction Reaction-refers to the half-reaction that involves the gain of electrons. OILRIG • Oxidizing agent-the compound or ion in a redox reaction that donates electrons. • Reducing agent-the compound or ion in a redox reaction that accepts electrons.

  24. Oxidation-Reduction Reactions

  25. Concentration of Solutions • Concentration of a Solution-amount of solute present in a given quantity of solvent or solution. • We will use Molarity, M for measurement. Molarity can also be called Molar Concentration. • Molarity–the number of moles of solute per liter of solution. • Molarity- moles of solutes/ liters of solution • Or n/v • Moles- grams of sample/ molecular weight of sample • Or g/ mw

  26. Concentration of Solutions • How many moles are there in 24.0g of C? moles = g/mw moles = 24.0g C/ 12.0g C moles = 2.0 There are 2.0 moles of C in 24.0g of C.

  27. Concentration of Solutions • How many grams are in 2.0 moles of Boron? moles= g/MW 2.0 moles = grams/ 10.81g Boron 2.0 moles x 10.81g Boron = grams Grams = 21.62 • There are 21.62 g of Boron in 2.0 moles of Boron.

  28. Concentration of Solutions • What is the Molarity of a 1L solution containing 9.0g HCl? 9.00g HCl x 1 mol HCl/ 18.00g HCl = 0.5 mol HCl M = n/v M = 0.5 mol HCl/ 1L M = 0.5 The concentration of the solution is 0.5M.

  29. Preparation of Solutions • Now that you know how to calculate M, n and v, what does that mean? • You can make your own solutions! • What are the steps in making a proper solution?

  30. Preparation of Solutions

  31. Concentration of Solutions • How many grams of Potassium Dichromate, K2Cr2O7, are required to prepare a 250mL solution with a concentration of 2.16M? 250mL x 1L/ 1000mL = .250L M= n/v n= M x v n= 2.16M x .250L n= 0.54 mol moles = g/MW Grams = moles x MW Grams = 0.54 mol K2Cr2O7 x 294.2 g K2Cr2O7 Grams = 159 159 grams of K2Cr2O7 are needed to prepare the requested solution.

  32. Concentration of Solutions • In a biochemical assay, a chemist needs to add 0.381g of glucose to a reaction mixture. Calculate the volume in millimeters of a 2.53M glucose solution that she should use for this addition. moles = g/MW moles = 0.381g C6H12O6/ 180.2g C6H12O6 moles = 2.114 x 10 –2 mol C6H12O6 M = n/v v = n/M v = 2.114 x 10 –2 mol C6H12O6 / 2.53M C6H12O6 v = 8.36mL She should use 8.36mL of the 2.53M glucose solution.

  33. Preparation of Solutions • Explain the process of creating 1L of 3.0M KCl. M = n/v n = M x v n = 3.0M x 1L n = 4.0 mol of KCl needed moles= g/MW Grams = moles x MW Grams = 4.0 mol KCl x 36.0g KCl Grams = 144g KCl Weigh out 144g of KCl. Put in a 1L flask. Add enough dH20 to dissolve KCl. Fill flask to 1L meniscus.

  34. Dilution of Solutions • Dilution-the procedure for preparing a less concentrated solution from a more concentrated one. • Dilutions can be made in increments of 10, 20, 50 or any other value. • Serial Dilution-theprocess of diluting a solution by removing part of it, placing this in a new flask and adding water to a known volume in the new flask.

  35. Dilution of Solutions • When you want to dilute a solution, what happens to the number of moles present in the solution? • Do they increase? • Decrease? • Stay the same?

  36. Dilution of Solutions

  37. Dilution of Solutions

  38. Dilution of solutions • Since moles are constant before and after dilution, we can use the following formula for calculations. • MiVi = MfVf

  39. Dilution of Solutions • Describe how you would prepare 800mL of a 2.0M H2SO4 solution, starting with a 6.0M stock solution of . 800mL x 1L/ 1000mL = 0.800L MiVi = MfVf 6.0M x Vi = 2.0M x 0.800L 6.0M x Vi = 1.6M x L Vi = 1.6M x L/ 6.0M Vi = 0.26L 0.26L of the 6.0M H2SO4 solution should be diluted to give a final volume of 800mL.

  40. Concentration of Solutions • There are several ways to determine the concentration of a solution. • In this week’s lab, we will be using spectroscopy to help us identify the molar concentration of an unknown substance. • Spectroscopy is helpful because it gives us: • Amount of light transmitted through a solution • Amount of light absorbed by a solution • Beer’s Law- a relationship between proportionality constant, path length of radiation going through solution and concentration of the solution. • A = abc • A = -log(T)

  41. Beer’s Law • If we know the value of T, we can solve the previous equation and figure out what A is. • Absorbance vs. wavelength • Absorbance vs. concentration • Use excel and be sure to add in your equation for the line. • Calculation of A or C: • A = mc

  42. Beer’s Law

More Related