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Chapter# 4 Solution Chemistry.

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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 properties of solutions include:
Solutions may be formed between solids, liquids or 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 Aqueous Solutions Water is the dissolving medium 104.5o  

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 = = solution solvent There are various combinations of units that are used in these rations. Ratio X 102 X 103 X 106 X 109 g solute % (w/w) ppt (w/w) ppm (w/w) ppb (w/w) = g solution g solute = % (w/v) ppt (w/v) ppm (w/v) ppb (w/v) mL solution mL solute = % (v/v) ppt (v/v) ppm (v/v) ppb (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 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 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 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 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 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 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 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 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 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 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 concentration can also be defined using moles. The most common example is molarity (M). The molarity of 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 M solution of HCl. A student evaporates the water form a 333 mL sample of a 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 units in 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
Five Driving Forces Favor Chemical Change Formation of a solid Formation of water Transfer of electrons Formation of a gas Formation of a weak electrolyte

35 Types of Aqueous Solutions
Solutions are homogeneous mixtures of a solute and a solvent. 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.

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) H2O (l) 2HCl (aq) CaCO3 (aq)  CaCl2 (aq) H2O (l) CO2 (g) 2HC l (aq) + Sr(HCO3)2 (aq)  SrCl2 (aq) + 2H2O (l) CO2 (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 neutralization reactions, 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) HC 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) H+ (aq) Cl- (aq)  H2 (g) Zn2+ (aq) Cl- (aq) Net ionic equation: Zn (s) H+ (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 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 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 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 M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO3 L solution moles AgCl g AgCl 33.2 mL moles AgCl L solution moles AgNO3 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 M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO3 L solution moles AgCl g AgCl 33.2 mL moles AgCl L solution moles AgNO3 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 M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO3 L solution moles AgCl g AgCl 33.2 mL moles AgCl L solution moles AgNO3 103 mL = 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 M solution of silver nitrate and an excess of calcium chloride. Find the mass of silver chloride formed from 33.2 mL of a M solution of silver nitrate and mL of a 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

51 Strong Acids and Bases A strong acid or strong base is completely ionized in aqueous solution. HCl, HBr, HI, HNO3, HClO4 and H2SO4 are all strong acids. All other acids are assumed to be weak acids. A weak acid or weak base only partially ionize in aqueous solution. Amphiprotic substances can behave as either a proton acceptor or a proton donor.

52 Types of Equations Molecular Equations have reactants and products written as undissociated molecules. HCl + NaOH ---> NaCl + H2O Overall Ionic Equations show all the species, both ionic and molecular present in the reaction. H+ + Cl- + Na+ + OH Na+ + Cl- + H2O

53 Continued Strong acids and strong bases are written as the corresponding ions in an overall ionic equation. Net Ion Equations describe the actual reaction taking place. H+ + OH > H2O The Na+ and Cl- ions are spectator ion is this reaction, because they are unchanged by reaction taking place.

54 Precipitation Reactions
Reactions in which a solid product forms from the reactants in solution. Solubility guidelines for ionic compounds allows the prediction of the formation of solid products.

55 Solubility Guidelines
All compounds containing the following ions are soluble: Cations:Group I ions (alkali metals) and NH4+ Anions:NO3- and CH3COO- (acetate) Compounds containing the following anions are soluble except as noted: Group 17 ions (halides), except with Ag+, Cu+, Hg22+, Pb2+, SO42- All other compounds are insoluble except the following group 2 (alkaline earth) hydroxides: Ba(OH)2, Ca(OH)2, and Sr(OH)2

56 Precipitation Precipitation is the formation of a solid when two solutions are combined.

57 Precipitate Formation
Does a precipitate form when sodium chloride is mixed with silver nitrate? If so what is the precipitate? NaCl Na+ + Cl- AgNO Ag+ + NO3- (Salt water solution) (Silver nitrate solution)

58 Net Ionic Equations Soluble ionic compounds are called strong electrolytes and completely ionize in aqueous solution. Write the balanced net ionic equation when sodium sulfate reacts with barium acetate.

59 Types of Solutions Unsaturated solution does not contain the maximum concentration A saturated solution contains the maximum concentration of solute that can dissolve in it. A supersaturated solution contains more than the quantity of a solute that is predicted to be soluble in a given volume of solution at a given temperature.

60 Solubility Curves of Various Solutes

61 Supersaturated Solution
Sodium acetate precipitates from a supersaturated solution.

62 Supersaturated Solutions
By forming a solution at a high temperature then slowly cooling it we can form supersaturated solutions that contain more solute than in a saturated solution. These kinds of solutions are very unstable and tend to separate out the excess solute with the slightest disturbance.

63 Titration Titration is an experimental procedure to determine the concentration of an unknown acid or base. The figure on the left shows the glassware for a titration experiment. A buret clamp holds the buret to a ring stand and below the buret is a flask containing the solution to be titrated, which includes an indicator. The purpose of the indicator is to indicate the point of neutralization by a color change.

64 Key Titration Terms A titration is a volumetric method used to determine the concentration of an unknown solution by reacting it with a standard solution. A standard solution is a solution of known concentration. The equivalence point in a titration is reached when enough standard solution has been added to completely react with the unknown solution. The end point in a titration is reached when the indicator changes color.

65 Titration The picture on the left shows the tip of a buret, with air bubble, which is not good, and also shows the stop-cock. Note the position of the stop-cock is in the “off” position. This picture shows the color of the phenolphthalein indicator at the end-point. In this experiment a mL aliquot of M NaOH titrant is added to 5.00 mL of an unknown HCL solution. The acid solution in the beaker starts out clear and becomes pink when all of the HCL has been consumed. NaOH + HCl  NaCl + HOH

66 How can we calculate the concentration of acid in the beaker?
Titration How can we calculate the concentration of acid in the beaker?

67 How can we calculate the concentration of acid in the beaker?
Titration How can we calculate the concentration of acid in the beaker? Normal procedure, yes, a conversion. Steps 1-4, again!

68 Titration How can we calculate the concentration of acid in the beaker? Normal procedure, yes, a conversion. Steps 1-4, again! 0.100 mole NaOH L NaOH solution

69 Titration How can we calculate the concentration of acid in the beaker? Normal procedure, yes, a conversion. Steps 1-4, again! 0.100 mole NaOH 10-3 L solution 23.00 mL soln mole HCl mL solution mole NaOH L NaOH solution L 0.460 M HCl

70 Indicators Indicators are weak organic (carbon containing) acids of various colors depending on the formula of the acid. Below is a generic acid. HA  H+ + A- colorless pink Describe the color change when a strong acid is added?

71 Indicators Indicators are weak organic (carbon containing) acids of various colors depending on the formula of the acid. Below is a generic acid. HA  H+ + A- Less pink colorless pink Describe the color change when a strong acid is added?

72 Indicators Indicators are weak organic (carbon containing) acids of various colors depending on the formula of the acid. Below is a generic acid. HA  H+ + A- colorless pink Describe the color change when a strong acid is added? Describe the color change when a strong base is added? Less pink

73 Indicators Indicators are weak organic (carbon containing) acids of various colors depending on the formula of the acid. Below is a generic acid. HA  H+ + A- colorless pink Describe the color change when a strong acid is added? Describe the color change when a strong base is added? Less pink Darker pink

74 Indicators Indicators are weak organic (carbon containing) acids of various colors depending on the formula of the acid. Below is a generic acid. HA  H+ + A- colorless pink Describe the color change when a strong acid is added? Describe the color change when a strong base is added? Describe the color change when the pH is lowered? Less pink Darker pink

75 Indicators Indicators are weak organic (carbon containing) acids of various colors depending on the formula of the acid. Below is a generic acid. HA  H+ + A- colorless pink Describe the color change when a strong acid is added? Describe the color change when a strong base is added? Describe the color change when the pH is lowered? Less pink Darker pink Less pink

76 Indicators Indicators are weak organic (carbon containing) acids of various colors depending on the formula of the acid. Below is a generic acid. HA  H+ + A- colorless pink Describe the color change when a strong acid is added? Describe the color change when a strong base is added? Describe the color change when the pH is lowered? Describe the color change when the pH is raised? Less pink Darker pink Less pink

77 Indicators Indicators are weak organic (carbon containing) acids of various colors depending on the formula of the acid. Below is a generic acid. HA  H+ + A- colorless pink Describe the color change when a strong acid is added? Describe the color change when a strong base is added? Describe the color change when the pH is lowered? Describe the color change when the pH is raised? Less pink Darker pink Less pink Darker pink

78 Color versus pH of Many Different indicators

79 How can we make an indicator?

80 How can we make an indicator?
Step Two Step Three Step One Red Cabbage Cook the Cabbage Filter the Juice

81 What color is the juice after filtering?

82 Colors of cabbage juice at various pH values
What color is the juice after filtering? The color of pH 6, 7, or 8 Colors of cabbage juice at various pH values

83 CH#4 REVIEW

84 Titraction of HCl and NaHO
To the left is a plot that shows the pH of an HCl solution as a function of the added volume of M NaOH. Which of the following plots would correspond to the same titration but using M NaOH? © 2008 W. W. Norton & Company Inc. All rights reserved. A) B) C) Titraction of HCl and NaHO

85 Titraction of HCl and NaHO
Consider the following arguments for each answer and vote again: The shape of both titration curves is the same, but the pH for titration with the stronger base should be higher at every point on the curve. The titration curve stays the same up to the equivalence point, but the pH will be higher when there is excess strong base. The curve will be shifted to the left because only half the volume of M NaOH will be required to reach the equivalence point. Answer: C Titraction of HCl and NaHO

86 Dilution of Acidic Acid
An acetic acid (HAc) solution of pH 3.0 is diluted by a factor of 10 with water. What is the new pH of the solution? © 2008 W. W. Norton & Company Inc. All rights reserved. A) < 4.0 B) 4.0 C) > 4.0 Dilution of Acidic Acid

87 Dilution of Acidic Acid
Consider the following arguments for each answer and vote again: Dilution of an acetic acid solution will drive the HAc/Ac- equilibrium toward further ionization of HAc. Therefore, the final pH will be between 3 and 4. Dilution of an acidic solution with pH 3.0 by a factor of 10 will result in a decrease in the H3O+ concentration from M to M, giving a pH of 4.0. Dilution of a weak acid solution with pH 3.0 will give a less acidic solution than the dilution of a strong acid solution with the same pH. Answer: A Dilution of Acidic Acid

88 pH of HCl,H2SO4,and NaOH Solutions
Which of the following three solutions would have the highest pH? © 2008 W. W. Norton & Company Inc. All rights reserved. A) M NaOH B) M H2SO4 C) M HCl pH of HCl,H2SO4,and NaOH Solutions

89 pH of HCl,H2SO4, and NaOH Solutions
Consider the following arguments for each answer and vote again: Of the three solutions, only the NaOH solution is basic, so its pH must be the highest. Although H2SO4 is a strong acid, it dissociates in water to form the base SO42-, making its pH higher than that of the other two solutions. Since HCl dissociates completely in water, the concentration of H3O+ is 10‑12 M for this solution. Therefore, the pH of the solution is 12. Answer: A pH of HCl,H2SO4, and NaOH Solutions

90 THE END


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