chapter 15 solutions. 1.to understand the process of dissolving 2.to learn why certain substances...

Chapter 15

Solutions

1. To understand the process of dissolving2. To learn why certain substances dissolve in water3. To learn qualitative terms describing the concentration of

a solution 4. To understand the factors that affect the rate at which a

solid dissolves   

Objectives Section 1 – Homogenous and Heterogeneous Solutions

Forming solutions

• Solution = homogeneous mixture

• Can be solid, liquid, or gas

• Solute =

• Solvent = – Aqueous solution

Solubility

• Solubility of ionic substances– Strong attractive forces that hold ionic crystal

together are overcome by the strong attraction between the ionic crystal and the water molecule

– When ionic compounds dissolve they break into individual ions

• NaCl• CaCl2

• Solubility of polar substances– Polar molecules can form hydrogen bonds with

water and dissolve– Like dissolves like

• Substances insoluble in water– Nonpolar molecules will not dissolve in water– Water-water hydrogen bonds keep the water from

mixing with the nonpolar molecules

• How substances dissolve– A “hole” must be made in the water structure for

each solute particle. – The lost water-water interactions must be

replaced by water-solute interactions. – “like dissolves like”

Solution Composition

• Saturated – solution contains as much solute as will dissolve at that temperature– If more solute is added it will not dissolve

• Unsaturated – solution that has not reached the limit of solute that will dissolve in it. – If more solute is added it will dissolve

• Supersaturated - occurs when a solution is saturated at an elevated temperature and then allowed to cool but all of the solid remains dissolved • Contains more dissolved solute than a saturated solution• Very unstable

Solution Composition: An Introduction

• A supersaturated solution is clear before a seed crystal is added.

Solution Composition: An Introduction

• Crystals begin to form in the solution immediately after the addition of a seed crystal.

Solution Composition: An Introduction

• Excess solute crystallizes rapidly.

• Solution concentration = the amount of solute in a given amount of solution

• Qualitative measurements of solution concentration– Concentrated – lots of solute– Dilute – not a lot of solute

Solution Composition: An Introduction

Which solution is more concentrated?

Solution Composition: An Introduction

Which solution is more concentrated?

Factors affecting the rate of dissolving

• Surface area – dissolving occurs at surface of substance being dissolved

• Stirring – removes newly dissolved particles from the solute surface and continuously exposes the surface to fresh solvent

• Temperature - molecules moving more rapidly, more interaction between solvent and solute

1. To understand mass percent and how to calculate it 2. To understand and use molarity 3. To learn to calculate the concentration of a solution

made by diluting a stock solution

Objectives Section 2 – Concentration of Solutions

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

• Cow’s milk typically contains 4.5% by mass of lactose, calculate the mass of lactose in 175 g of milk.

Solution Composition: Molarity

• What is the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in 1500 mL of solution?

• Give the concentration of the ions in a 0.50 M solution of Co(NO3)2

• How man moles of Ag+ ions are present in 25 mL of a 0.75 M AgNO3 solution?

Solution Composition: Molarity • Standard solution - a solution whose concentration is

accurately known • To make a standard solution

– Weigh out a sample of solute.

– Transfer to a volumetric flask.

– Add enough solvent to mark on flask.

Molarity

• To make a 0.5-molar (0.5M) solution, first add 0.5 mol of solute to a 1-L volumetric flask half filled with distilled water.

Molarity

• Swirl the flask carefully to dissolve the solute.

Molarity

• Fill the flask with water exactly to the 1-L mark.

• A chemist needs 1.00 L of 0.200 M K2Cr2O7 solution. How much solid K2Cr2O7 (molar mass = 294.2 g/mol) must be weighed out to make this solution?

Dilution

• Water can be added to an aqueous solution to dilute the solution to a lower concentration.

• Only water is added in the dilution – the amount of solute is the same in both the original and final solution.

• Diluting a solution – Transfer a measured amount of original solution to

a flask containing some water. – Add water to the flask to the mark (with swirling)

and mix by inverting the flask.

Making Dilutions

• The total number of moles of solute remains unchanged upon dilution, so you can write this equation.

• M1 and V1 are the molarity and volume of the initial solution, and M2 and V2 are the molarity and volume of the diluted solution.

• If we want to prepare 500. mL of 1.00 M acetic acid from a 17.5 M stock solution, what volume of the stock solution is required?

1. To learn to solve stoichiometric problems involving solution reactions

2. To do calculations involving acid-base reactions3. To learn about normality and equivalent weight 4. To use normality in stoichiometric calculations 5. To understand the effect of a solute on solution properties

Objectives Section 3 – Properties of Solutions

Stoichiometry of Solution Reactions

• Calculate th 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.

• When Ba(NO3)2 and K2CrO4 react in aqueous solution the yellow solid BaCrO4 is formed. Calculate the mass of BaCrO4 that forms when 3.50 x 10-3 mol of solid Ba(NO3)2 is dissolved in 265 mL of 0.0100 M K2CrO4 solution.

Neutralization Reactions • An acid-base reaction is called a neutralization reaction. • Steps to solve these problems are the same as before.

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

Normality

• Unit of concentration– One equivalent of acid – amount of acid that furnishes 1

mol of H+ ions – One equivalent of base – amount of base that furnishes 1

mol of OH ions – Equivalent weight – mass in grams of 1 equivalent of acid

or base

Normality

Normality

• Phosphoric acid, H3PO4 can furnish three H+ ions per molecule. Calculate the equivalent weight of H3PO4

Normality

• To find number of equivalents

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

Colligative properties

• a solution property that depends on the number of solute particles present

• The presence of solute “particles” causes the liquid range to become wider. – Boiling point increases – Freezing point decreases

• Vapor pressure lowering – – Vapor pressure:

– Adding a nonvolatile solute to a solution lowers the solvent’s vapor pressure

• Boiling point elevation –– Boiling point

– Because vapor pressure is lowered, boiling point increases

• Freezing point depression – solute particles interfere with attractive forces holding solvent particles together