Chapter 15: Solutions

15.1 SolubilityObjectives: To understand the process of dissolving.To learn why certain substances dissolve in water.

Chapter 15.1: Solutions

Solution: Homogeneous mixture in which the components are uniformly intermingled.

Examples: air, shampoo, orange soda, coffee, gasoline, cough syrup, etccc…

Solvent: substance present in the largest amount.

Solute: the other substance.

Aqueous solution: water as the solvent.

Figure 15.1: Dissolving of solid sodium chloride.

Figure 15.2: Polar water molecules interacting with positive and negative

ions of a salt.

Chapter 15.1: Solutions

Solubility of Polar Substances:

Ethanol Water

Can form hydrogen bonds.Makes it compatible with water.

Figure 15.3: The polar water molecule interacts strongly with the polar O—H bond in ethanol.

Figure 15.4: Structure of common table sugar.

Contains many polar O-H bonds each of which can hydrogen-bond to a water molecule.

Figure 15.5: A molecule typical of those found in petroleum.

Similar electronegativityShare electrons ==Nonpolar

Like dissolves Like

• Nonpolar solutes dissolve in nonpolar solvents.

15.2: Solution Composition: An Introduction

• Objective: To learn qualitative terms associated with concentration of a solution.

15.2: Solution Composition: An Introduction

• Saturated solution: contains as much solute as will dissolve at that temperature.

• Unsaturated: solution that has not reached the limit of solute.

• Supersaturated solution: contains more dissolved solid than a saturated solution will hold at that temperature. A supersaturated solution is very unstable.

• Concentrated solution: relatively large amount of solute• Dilute Solution: small amount of solute.

15.3: Factors Affecting the Rate of Dissolution

• Objectives:

1) To understand the factors that affect the rate at which a solute dissolves.

15.3: Factors Affecting the Rate of Dissolution

• Three factors affect the speed of the dissolving process.– Surface area– Stirring– Temperature

15.4: Solution Composition: Mass Percent

• Objective:

To understand the concentration term mass percent and learn how to calculate it.

15.4: Solution Composition: Mass Percent

Mass percent = mass of solute x 100%

mass of solution

Mass percent = grams of solute x 100%

grams of solute + grams of solvent

15.4: Solution Composition: Mass Percent

A solution is prepared by mixing 2.5 g of calcium chloride w/ 50.0g of water.

Calculate the mass percent of calcium chloride in this solution.

4.76%

15.4: Solution Composition: Mass Percent

A solution is prepared by mixing 1.00 g of ethanol with 100.0 g of water. Calculate the mass percent of ethanol in this solution.

0.990%

15.4: Solution Composition: Mass Percent

A solution of milk contains 4.5% by mass of lactose. Calculate the mass of lactose present in 175 g of milk.

7.9 g

15.5: Solution Composition: Molarity

Objectives:

1) To understand molarity.

2) To learn to use molarity to calculate the number of moles of solute present.

15.5: Solution Composition: Molarity

To measure the concentration by volume.concentration is defined as the amount of

solute in a given volume of solution.

Molarity (M) describes the amount of solute in moles and the volume of the solution in liters.

Molarity is the number of moles of solute per volume of solution in liters.

15.5: Solution Composition: Molarity

Molarity is the number of moles of solute per volume of solution in liters.

M=molarity= moles of solute = mol

liters of solution L

15.5: Solution Composition: Molarity

Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.5L of solution.

0.192 M

15.5: Solution Composition: Molarity

Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl into enough water to make 26.8 ml

1.59 M

15.5: Solution Composition: Molarity

Give the concentrations of all the ions in each of the following solutions.

a. 0.50 M Co(NO3)2

b. 1 M FeCl3

0.50 M Co, 1.0M NO3

1 M Fe, 3.0 M Cl

15.5: Solution Composition: Molarity

To analyze the alcohol content of a certain wine, a chemist needs 1.00 L of an aqueous 0.200 M K2Cr2O7. How much solid K2Cr2O7 (molar mass =294.2 g) must be weighed out to make this solution?

0.200 mol, 58.8 g

Figure 15.7: Steps involved in the preparation of a standard aqueous solution.

Standard solution: is a solution whose concentration is accurately known.

15.5: Solution Composition: Molarity

How many moles of Ag+ ions are present in 25 ml of a 0.75 M AgNO3 solution?

0.50 M Co, 1.0M NO3

1 M Fe, 3.0 M Cl

15.6: Dilution

Objective: To learn to calculate the concentration of a solution made by diluting a stock solution.

15.6: Dilution

Objective: To learn to calculate the concentration of a solution made by diluting a stock solution.

15.6: Dilution

Stock solution is purchased. Often need to add water or another solvent to get to the desired concentration.

Moles of solute after dilution=moles of solute before dilution

# of moles of solute stays the same, but more water is added increasing the volume, so the molarity decreases.

M= moles of solute

volume (L)

15.6: Dilution

Volume of dilute molarity of moles of solute

Solution (liters) x dilute soln. = present

15.6: Dilution

What volume of 16M sulfuric acid must be used to prepare 1.5 L of a 0.10M H2SO4 solution?

9.4 ml

15.6: Dilution

What volume of 19M sodium hydroxide must be used to prepare 1.0L of a 0.15M NaOH solution?

7.9 ml

15.6: Dilution

What volume of water is needed to prepare 500.0 ml of a 0.250 M Ca(NO3)2 solution from a 5.00 M Ca(NO3)2 solution?

475 ml

15.7: Stoichiometry of Solution Reactions

Objectives: To understand the strategy for solving stoichiometric problems for solution reactions

15.7: Stoichiometry of Solution Reactions

Step 1: Write the balanced equation for the reaction. For reactions involving ion, it is best to write the net ionic equation.

Step 2: Calculate the moles of reactants.Step 3: Determine which reactant is limiting.Step 4: Calculate the moles of other

reactants or products, as required.Step 5: Convert to grams or other units, if

required.

15.7: Stoichiometry of Solution Reactions

Calculate the mass of solid NaCl that must be added to 1.50 L of a 0.100 M AgNO3 solution to precipitate all of the Ag+ ions in the form of AgCl. Calculate the mass of AgCl formed.

21.5 g AgCl

15.7: Stoichiometry of Solution Reactions

Calculate the mass of solid sodium sulfate that must be added to 250.0 ml of a 0.2 M solution of barium nitrate to precipitate all of the barium ions in the form of barium sulfate. Also calculate the mass of barium sulfate forms.

7.10 g Na2SO4; 11.7 g BaSO4

15.7: Stoichiometry of Solution Reactions

Calculate the mass of sodium iodide that must be added to 425.0 ml of a 0.100 M Pb(NO3)2 solution to precipitate all of the Pb2+ ions as PbI2. Also calculate the mass of PbI2 forms.

12.7 g NaI; 19.6g PbI2

15.7: Stoichiometry of Solution Reactions

Calculate the mass of the white solid CaCO3 that forms when 25.0 ml of a 0.100 M Ca(NO3)2 solution is mixed with 20.0 ml of a 0.150M Na2CO3 solution.

0.250 g CaCO3

15.7: Stoichiometry of Solution Reactions

Calculate the mass of the blood-red solid Ag2CrO4 that forms when 50.0 ml of a 0.250M AgNO3 solution is mixed with 30.0 ml of a 0.250M K2CrO4 solution.

2.07 g Ag2CrO4

15.8: Neutralization Reactions

Objective: To learn how to do calculations involved in acid-base reactions.

Neutralization reaction: acid-base reaction

When just enough strong acid is added to react exactly with the strong base, theacid is neutralized. One product is water.

15.8: Neutralization Reactions

What volume of a 0.100 M HCl solution is needed to neutralize 25.0 ml of a 0.350 M NaOH solution?

8.75 x 10-2 L

15.8: Neutralization Reactions

What volume of a 0.150 M HNO3 solution is needed to neutralize 45.0 ml of a 0.550 M KOH solution?

165 ml

15.8: Neutralization Reactions

What volume of a 1.00 x 10-2 M HCl solution is needed to neutralize 35.0 ml of a 5.00 x 10-3 M Ba(OH)2 solution?

35.0 ml

15.9: Normality

Objectives: To learn about normality and equivalent weight.

To learn to use these concepts in stoichiometric calculations.

15.9: Normality

Normality:

Equivalent of an acid: amt. of acid that can furnish 1 mol of H+ ions.

Equivalent of a base: amt of base that can furnish 1 mol of OH- ions.

Equivalent weight: mass in grams of 1 equivalent of that acid or base.

15.9: Normality

Normality: is defined as the number of equivalents of solute per liter of solution.

Normality=N= # of equivalents= equivalents1 liter of soln. L

Molar Mass Equivalent Wt (g)HCl 36.5 g 36.5 gH2SO4 98.0 49.0gH3PO4 98.0 32.7 g

15.9: Normality

A solution of sulfuric acid contains 86 g of H2SO4 per liter of solultion. Calculate the normality of this solution.

1.8 N H2SO4

15.9: Normality

Arsenic acid, H3AsO4, can furnish three H+ ions per molecule. Calculate the equivalent weight of H3AsO4.

47.31 g

15.9: Normality

Calculate the equivalent weight of HBr.

80.91 g

15.9: Normality

A solution of phosphoric acid contains 50.0g of H3PO4 per liter of solution. Calculate the normality of this solution.

1.53 N

15.9: Normality

A solution of hydrochloric acid contains 25.0 g of HCl per liter of solution. Calculate the normality of this solution.

0.686N

15.10: The Properties of Solutions: Boiling Point and Freezing Point

Objective: To understand the effect of a solute on solution properties.

1.0M NaCl solution freezes at about -1oC.

Boils at 104oC.

Figure 15.9: A bubble in the interior of liquid water surrounded by solute

particles and water molecules.

Figure 15.10: Pure water.

Colligative property: depends on the numberof solute particles.