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Chapter# 4

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  1. Chapter# 4 Solution Chemistry

  2. Solutions • Solutions are homogeneous mixtures of two or more substances. • The solvent is the substance in greatest quantity. • Solutes are the other ingredients in the mixture. • Solutions can exist in all states of matter.

  3. Solution Examples • Margarine • Tap Water • Steel • 18 Carat Gold • Air • Sterling Silver

  4. Solution Examples Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper. What is the solvent in 18 ct gold ?

  5. Solution Examples Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper. What is the solvent in 18 ct gold ? Gold

  6. Solution Examples Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper. What is the solvent in 18 ct gold ? Gold What are the solutes?

  7. Solution Examples Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper. What is the solvent in 18 ct gold ? Gold What are the solutes? Silver and Copper

  8. Solution Properties • Some general propertiesof solutions include: • Solutions may be formed between solids, liquidsor gases. • They are homogenous in composition • They do not settle under gravity • They do not scatter light (Called the Tyndall Effect) Solute particles are too small to scatter light and therefore light will go right through a solution like is shown on the next slide.

  9. Tyndall Effect Laser light reflected by a colloid. In a solution you would not see any red light.

  10. Solutions

  11.  104.5o   Aqueous Solutions Water is the dissolving medium

  12. Some Properties of Water • Water is “bent” or V-shaped. • Water is a molecular compound. • Water is a polar molecule. • Hydration occurs when ionic compounds dissolve in water.

  13. SOLUTION CONCENTRATION The ratio of the amount of solute to amount of solution, or solvent is defined by the concentration. solute solute Concentration = = solvent solution There are various combinations of units that are used in these rations. Ratio X 102 X 103 X 106 X 109 g solute ppb (w/w) ppm (w/w) ppt (w/w) % (w/w) = g solution g solute ppt (w/v) ppm (w/v) ppb (w/v) = % (w/v) mL solution mL solute = % (v/v) ppt (v/v) ppb (v/v) ppm (v/v) mL solution

  14. SAMPLE SOLUTION PROBLEMS • 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. • Find the mass of water and salt required to make 333 g of a • 44.6 % (w/w) solution. 25.2 g NaCl

  15. SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 25.2 g NaCl 33.6g H2O + 25.2 g NaCl

  16. SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl 33.6g H2O + 25.2 g NaCl

  17. SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O + 25.2 g NaCl

  18. SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O + 25.2 g NaCl 44.6 g NaCl 100 g solution

  19. SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O + 25.2 g NaCl 44.6 g NaCl 333 g solution 100 g solution

  20. SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O + 25.2 g NaCl 44.6 g NaCl 333 g solution 100 g solution

  21. SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O + 25.2 g NaCl 44.6 g NaCl 333 g solution = 149 g NaCl 100 g solution

  22. SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O + 25.2 g NaCl 44.6 g NaCl 333 g solution = 149 g NaCl 100 g solution Mass of water?

  23. SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O + 25.2 g NaCl 44.6 g NaCl 333 g solution = 149 g NaCl 100 g solution Mass of water? 333 g solution – 149 g NaCl = 184 g H2O

  24. SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O + 25.2 g NaCl 44.6 g NaCl 333 g solution = 149 g NaCl 100 g solution Mass of water? 333 g solution – 149 g NaCl = 184 g H2O

  25. SAMPLE SOLUTION PROBLEMS • How many grams of NaCl are required to dissolve in 88.2 g of water to make a 29.2% (w/w) solution. • A sugar solution is 35.2%(w/v) find the mass of sugar contained in a 432 mL sample of this sugar solution.

  26. SOLUTION CONCENTRATION The solution concentrationcan also be defined using moles. The most common example is molarity (M). Themolarityof a solution is defined as: “The number of moles of solute in 1 L of solution” and is given the formula: Moles solute Molarity (M) = Liters solution

  27. MOLARITY SAMPLE PROBLEMS • A student dissolves 25.8 g of NaCl in a 250 mL volumetric flask. Calculate the molarity of this solution. (picture of volumetric flask is on the next slide) • Find the mass of HCl required to form 2.00 L of a 0.500 M solution of HCl. • A student evaporates the water form a 333 mL sample of a 0.136 M solution of NaCl. What mass of salt remains?

  28. SOLUTION PREPARATION In the lab we would use a piece of glassware called a volumetric flask to prepare this solution.

  29. SOLUTION PREPARATION

  30. VOLUMETRIC FLASK

  31. SOLUTION DILUTION Often we will want to make a dilute solution from a more concentrated one. To determine how to do this we use the formula : C1V1 = C2V2 Where: C1 = concentration of more concentrated solution V1 = volume required of more concentrated solution C2 = concentration of more dilute solution V2 = volume of more dilute solution We can use any unitsin this equation but they must be the same on both sides.

  32. DILUTION PROBLEM How would one prepare 50.0 mL of a 3.00 M solution of NaOH using a 7.10 M stock solution? C1V1 = C2V2 (7.10 M)V1 = (3.00 M) (50.0 mL) This means that you add 21.1 mL of the concentrated stock solution to a 50.0 mL volumetric flask and add water until the bottom of the meniscus touches the line on the volumetric flask.

  33. DILUTION PROBLEM How would one prepare 50.0 mL of a 3.00 M solution of NaOH using a 7.10 M stock solution? C1V1 = C2V2 (7.10 M)V1 = (3.00 M) (50.0 mL) (3.00 M) (50.0 mL) (7.10 M)V1 = (7.10 M) (7.10 M) V1 = 21.1 mL This means that you add 21.1 mL of the concentrated stock solution to a 50.0 mL volumetric flask and add water until the bottom of the meniscus touches the line on the volumetric flask.

  34. Reaction Driving Forces • Formation of a solid • Formation of water • Transfer of electrons • Formation of a gas • Formation of a weak electrolyte Five Driving Forces Favor Chemical Change

  35. Types of Aqueous Solutions • The solute is the solution component in the smallest amount while the solvent is the larger component of a solution. • Solutes whose solutions conduct electricity are called electrolytes • Solutes whose solutions do not conduct electricity are called nonelectrolytes • Electrolytes are solutes that form ions when they dissolve. Ionic solutes or acids usually form solutions that conduct electricity. Solutions are homogeneous mixtures of a solute and a solvent.

  36. Solution Conductivity Strong electrolyte Weak electrolyte Nonelectrolyte

  37. Solution Formation Water is one of the best solvents known. It is able to dissolve ionic solutes, such as sodium chloride, to produce solutions that conduct electricity. Molecules, containing a positive and negative regions, are called polar. Water is an example of a polar molecule and can dissolve ionic solutes by the positive region of water attracting to the negative ion of an ionic solute thus separating the crystal lattice in to a solution of solvated ions.

  38. Acid-Base Reactions Acids undergo characteristic double replacement reactions with oxides, hydroxides, carbonates and bicarbonates. 2HCl (aq) + CuO (s) CuCl2 (aq) + H2O (l) 2HCl (aq) + Ca(OH)2 (aq) CaCl2 (aq) + 2H2O (l) 2HCl (aq) + CaCO3 (aq) CaCl2 (aq) + H2O (l) + CO2 (g) 2HC l (aq) + Sr(HCO3)2 (aq) SrCl2 (aq) + 2H2O (l) + 2CO2 (g)

  39. Acid-Base Reactions Bases undergo a double replacement reaction with acids called neutralization: NaOH (aq) + HCl (aq)  H2O (l) + NaC l (aq) In words this well known reaction is often described as: “acid plus base = salt plus water” We previously discussed this reaction when describing types of reactions.

  40. Acid-Base Reactions We have discussed the double replacement reactions and ionic equations before. Since the acids and bases undergo double replacement reactions called neutralizationreactions, then they can have ionic equations too. Molecular equation: HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l) Total ionic equation: H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq) Na+ (aq) + Cl- (aq) + H2O (l) Net ionic equation: H+ (aq) + OH- (aq)  H2O (l)

  41. Acid-Base Reactions Another property of acids is their reaction with certain metals to produce hydrogen gas, H2 (g). Zn (s) + 2HC l (aq) H2 (g) + ZnCl2 (aq) This is an example of a single replacement reaction and is a redox reaction. Total ionic equation: Zn (s) + 2H+ (aq) + 2Cl- (aq)  H2 (g) + Zn2+ (aq) + 2Cl- (aq) Net ionic equation: Zn (s) + 2H+ (aq)  H2 (g) + Zn2+ (aq)

  42. Solution Stoichiometry Consider the following balanced equation: CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq) • Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride.

  43. Solution Stoichiometry Consider the following balanced equation: CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq) • Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO3 L solution

  44. Solution Stoichiometry Consider the following balanced equation: CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq) • Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO3 L solution L solution 103 mL

  45. Solution Stoichiometry Consider the following balanced equation: CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq) • Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO3 L solution moles AgCl 143.45 g AgCl 33.2 mL moles AgCl moles AgNO3 L solution 103 mL

  46. Solution Stoichiometry Consider the following balanced equation: CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq) • Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO3 L solution 143.45 g AgCl moles AgCl 33.2 mL moles AgCl moles AgNO3 L solution 103 mL

  47. Solution Stoichiometry Consider the following balanced equation: CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq) • Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO3 L solution 143.45 g AgCl moles AgCl 33.2 mL moles AgCl moles AgNO3 L solution 103 mL = 0.476 g AgCl

  48. Solution Stoichiometry Consider the following balanced equation: CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq) • Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride. • Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and 200.0 mL of a 0.200 M solution of calcium chloride solution. • Find the volume of the excess reactant.

  49. Acid-Base Reactions • Bronsted-Lowry acids are proton (H+) donors. • Bronsted-Lowry bases are proton acceptors. • Free hydrogen ions don’t exist in water because they strongly associate with a water molecule to create a hydronium ion (H3O+).

  50. Acid-Base Reactions • A neutralization reaction takes place when an acid reacts with a base and produces a solution of a salt and water. • A salt is made up of the cation characteristic of the base and the anion characteristic of the acid. • Example: HCl + NaOH ---> NaCl + H2O