Chapter 12 Aqueous Solutions. Malone and Dolter - Basic Concepts of Chemistry 9e2 Setting the Stage...

Chapter 12Chapter 12

Aqueous SolutionsAqueous Solutions

Malone and Dolter - Basic Concepts of Chemistry 9e

2

Setting the Stage – Water as a Solvent

Most of the surface of the earth is covered with water with a high concentration of dissolved compounds and suspended matter.

Our blood is also mostly water, with a wide variety of dissolved and suspended substances.

Water is a remarkable solvent, and is central to the understanding of chemistry.


3

Setting a Goal - Part ASolutions and the Quantities Involved

You will learn how water acts to dissolve specific types of compounds and how concentrations of solutes are expressed and used.


4

Objective for Section 12-1

Describe the role of water in the formation of aqueous solutions of ionic compounds, strong acids, and polar compounds.


5

Professor’s Joke (with Apologies)

One day, a biologist, a chemist and a physicist were

exploring the ocean in a boat for the very first time.

The physicist said, “I’m going to determine the physical

laws of the ocean”. Off he went into the water, but he

never returned. Later, the biologist said, “ I’m going to

search the water for living organisms”. He too entered

the ocean and was never seen again.

The chemist waited a while and then wrote down in

his note book, “Physicists and biologists are soluble in

water”.


6

12-1 The Nature of Aqueous Solutions

Miscible – two liquids are miscible if they mix to form a homogeneous solution.

Immiscible – two liquids are immiscible if they do not mix to form a solution (they separate into two phases like oil and water).

Our focus will be on solids dissolving to form a solution (although there are other possibilities).


7

Solids Dissolving in Water

Consider that when a salt such as NaCl dissolves in water

It loses Na+ - Cl- electrostatic interaction It loses H2O - H2O interactions It gains ion-dipole interaction (forces

between the ions and the dipoles of the solvent molecules)

For dissolution to occur, the latter interaction must be stronger that the sum of the first two.


8

Interaction of Water and Ionic Compounds


9

Dissolution of Polar Compounds

Some polar compounds (e.g. strong acids) dissolve by ionization, producing dissolved ions.E.g. HCl(g) + aq H+(aq) + Cl-(aq)

or HCl(g) + H2O(l) H3O+(aq) + Cl-(aq)


10

Dissolution of Polar Compounds

Many polar compounds dissolve by forming strong dipole-dipole interactions, but without ionization.

E.g. methanol in water


11

Objectives for Section 12-2

Define terms associated with solubility.

Predict the effects of temperature and pressure on solubility.


12

12-2 The Effects of Temperature and Pressure on Solubility

The solubility of a solute is defined as the maximum amount that dissolves in a given amount of solvent at a specific temperature (usually g/100 mL at 25 ºC).

Saturated solutions are those that have the maximum amount of solute dissolved.

Unsaturated solutions are those with less than the maximum amount.


13

Supersaturated Solutions

Such solutions can sometimes be created where more solute is dissolved than the standard solubility.

Such solutions are metastable, and solid can crystallize out if the solution is disturbed in certain ways.


14

Concentration

The amount of solute present in a given amount of solvent or solution is the concentration.

Is expressed in a number of ways (one way is shown below).

g of solute

100 g solvent


15

Temperature and Solubility

Most solids and liquids are more soluble in water at higher temperatures.

This is the basis of recrystallization – a method for purifying solids.

Gases are less soluble in water at higher temperatures (consider oxygen in water).


16

Effect of Pressure on Solubility

The solubility of solids and liquids is not affected to any significant extent by the external pressure.

The solubility of gases is strongly affected by pressure.


17

Henry’s Law

Henry’s Law states that the solubility of a gas is proportional to the partial pressure of that gas above the solution.

Solubility = kPgas

The constant is known as Henry’s Law Constant.


18


Perform calculations of concentration involving percent by mass.


19

12-3 Concentration: Percent by Mass

In general, concentration can be expressed in two ways:


20

Percent by Mass

Percent by mass expression of concentration is a very useful one.


21

% and Other Units by Mass

% by mass is equivalent to parts per hundred.

Extremely low concentrations are better expressed as parts per million (ppm) or parts per billion (ppb).


22

Parts per Million (ppm) and Parts per Billion (ppb)

Sometimes it is more convenient to express concentration in ppm or ppb.


23

Calculating ppm and ppb

A sample of water contains 8.8 x 10-5% Pb2+ by mass. Calculatethis concentration in ppm and ppb.

8.8 x 10-5% 8.8 x 10-7 g solute /g solution

8.8 x 10-7 g solute

g solutionx 106 ppm = 0.88 ppm

8.8 x 10-7 g solute

g solutionx 109 ppm = 880 ppb


24


Perform calculations of concentration involving molarity.

Perform calculations involving the dilution of concentrated solutions.


25

12-4 Concentration: Molarity

Most common concentration unit used by chemists.

Number of moles of solute (n) per volume in liters (V) of solution.


26

Dilution of Concentrated Solutions

Materials are often shipped as concentrated solutions and diluted where it is used (think of orange juice).

We often dilute concentrated solutions in the course of being chemists.

Suppose we need to make 1.00 L (Vd)of 0.555 M (Md) HCl from the concentrated acid (11.3 M) (Mc).

What volume must we measure?


27

Diluting Concentrated Solutions

Number of moles of solute in dilute solution = nd = Md × Vd

Number of moles of solute in concentrated solution = nc = Mc × Vc

The number of moles of solute is constant (nd = nc), so Mc × Vc = Md × Vd


28


Solution to our dilution of concentratedHCl problem:


29



30


Perform calculations involving titrations and other solution stoichiometry.


31

12-5 Stoichiometry Involving Solutions

Cd(NO3)2 (aq) + Na2S(aq) CdS(s) + 2 NaNO3 (aq)

What mass of CdS is formed when 158 mL of a 0.122 M Na2S solution reacts with excess Cd(NO3)2.

?

Solution:

0.158 L Na2S x0.122 mol

1 L Na2Sx

1 mol CdS

1 mol Na2S

144.5 g CdS

1 mol CdS= 2.79 g

Na2S x


32

General Procedure: Summary Chart


33

Titrations

Titrations are usually volume-to-volume stoichiometry problems where a concentration is the unknown variable.

The end of the reaction is generally detected by the addition of an indicator (a substance that changes color when the reaction is just complete).


34

Calculating an Unknown Concentration from Titration Data

What is the concentration of an HCl solution if 25.0 mL reacts with 38.5 mL of 0.145 M K3PO4 solution?


35

Alternative solution to problem

3HCl K3PO4

MA = ?VA = 25.0 mLnA = 3

MB = 0.145 MVB = 38.5 mLnB = 1

MAVA

nA

MBVB

nB=

Hence MA =(0.145 M)(38.5 mL)(3)

(25.0 mL)(1)

= 0.668 M


36

Setting a Goal – Part BThe Effects of Solutes on the Properties of Water

You will learn how the physical properties of aqueous solutions differ from those of pure water.


37


Explain the differences between nonelectrolytes, strong electrolytes, and weak electrolytes.


38

12-6 Electrical Properties of Solutions

Conductivity Conductor – a substance that allows the flow

of electrons Insulator – a substance that resists the flow of

electrons Electrolytes – substances whose aqueous

solutions conduct electricity. It is the presence of ions in solution that

allows the flow of electrons.


39

Physical Properties of Solutions

Nonelectrolytes – compounds whose aqueous solutions do not conduct electricity (e.g. glucose, sucrose).

Strong electrolytes – species that yield solutions that are good conductors (e.g. Na+Cl-, H2SO4).

Weak electrolytes – species that yield solutions that conduct electricity poorly.

(e.g. HF, CH3COOH)


40


Explain how the presence of a solute in water affects its physical properties.

Calculate the boiling and melting points of aqueous solutions of nonelectrolytes and electrolytes.


41

12-7 Colligative Properties of Solutions

Those properties that depend on the number of particles in solution are called colligative properties.

They include elevation of boiling point and depression of freezing point.

They arise due to lowering of vapor pressure.


42

Vapor Pressure Lowering

Vapor pressure – presence of a nonvolatile solute lowers the equilibrium vapor pressure from that of the pure solvent.

A nonvolatile solute is one that has practically no vapor pressure at the relevant temperatures.

The nonvolatile solute prevents solvent molecules from getting to the surface of the liquid.


43

Boiling Point Elevation

Boiling point – lowered vapor pressure of a solution results in a higher boiling point than the pure solventTb = Kbm Tb is number of degrees the boiling point

is raised in Celsius or KelvinKb = boiling point elevation constantm = molality = mol solute/kg solvent (temperature independent concentration unit)


44

Freezing Point Depression

The lowered vapor pressure as a result of the solute lowers the freezing pointTf = Kfm Tf is the number of degrees the freezing point is loweredKf = freezing point depression constantm = molality


45

Calculating Molality

What is the molality of a solution of 468 g of glucose in 350 g water (molar mass of glucose = 342.5 g/mol)?

Solution


46

Calculating the Boiling Point of a Solution

What is the boiling point of a solution of 468 g of glucose in 350 g of water?

Solution


47

Osmotic Pressure

Tendency for a solvent to move through a thin porous membrane from a dilute solution to a more concentrated solution.

The membrane is said to be semipermeable in that small solvent molecules can pass through, but large hydrated species like ions cannot.


48

Osmotic Pressure

Solvent molecules pass through the membrane in both directions, but the rate at which they diffuse to the more dilute side of the membrane is lower.

The solutes hold back the solvent due to the solute-solvent interactions.

This uneven passage of water molecules causes the water level to rise on the concentrated side and drop on the dilute side.


49

Osmosis

The difference in height of solution caused by osmosis creates a pressure on the solute side, which eventually counterbalances the osmosis until equilibrium is established.

The extra pressure required to establish the equilibrium is the osmotic pressure.


50

Colligative Properties: Summary

Electrolytes yield a stronger effect than nonelectrolytes, since colligative properties depend on the number of particles.

For example, NaCl yields two particles when it dissolves, while glucose (a nonelectrolyte) yields only one particle.

Hence, NaCl is twice as effective as, say, glucose, at the same molality.

Chapter 12 Aqueous Solutions. Malone and Dolter - Basic Concepts of Chemistry 9e2 Setting the Stage...

Documents

Transcript of Chapter 12 Aqueous Solutions. Malone and Dolter - Basic Concepts of Chemistry 9e2 Setting the Stage...