Unit 2: Equilibrium Icons are used to prioritize notes in this section. Make some notes: There are SOME important items on this page that should be copied.

Unit 2: Equilibrium Icons are used to prioritize notes in this section.
Make some notes: There are SOME important items on this page that should be copied into your notebook or highlighted in your printed notes. Copy as we go: There are sample problems on this page. I EXPECT YOU to copy the solutions into a notebook— Even if you have downloaded or printed the notes!!! Look at This: There are diagrams or charts on this page. Look at them and make sure you understand them. (but you don’t need to copy them) Extra Information: This page contains background information that you should read, but you don’t need to copy it. Review: There is review material on this page. It is up to you to decide if you want to make notes or highlight it, depending on how well you remember it. R

Unit 4 (formerly Module 5)
Equilibrium

Equilibrium 11.0 Overview:
Equilibrium is the concept of a system remaining “in balance”. A system in equilibrium does not change at the macroscopic level (the level that we can detect with our senses). True equilibrium should not be confused with homeostasis or “steady-state”, a process by which living organisms attempt to maintain a consistent internal conditions by absorbing or excreting materials from and to their environment.

Equilibrium vs. Homeostasis
Equilibrium usually exists in a closed system, where materials cannot easily enter or leave. Examples: A reversible chemical reaction occurs inside a closed container. The reaction appears to have stopped, but at a molecular level changes are still going on. A liquid in a sealed bottle does not appear to evaporate. Homeostasis usually exists in an open system, where materials can enter and leave. Examples: A dog lives in a kennel. It eats and excretes roughly equal amounts, and therefore maintains a steady weight and internal conditions. A cell in a human body maintains a balance of nutrients.

The Temporary Nature of Equilibrium
Although we study equilibrium as if it is an unchanging state, in reality dynamic processes as well as static forces may act upon the equilibrium. Eventually something will upset the equilibrium, temporarily or permanently throwing it “out of balance” Often a new equilibrium will be re-established after the original equilibrium is upset.

Examples A precariously balanced rock formation can endure in “equilibrium” for centuries. Suddenly the balanced rock falls, temporarily disturbing the equilibrium. A reversible chemical reaction inside a beaker has reached a state of equilibrium and appears to have stopped. A researcher adds more of one of the reactants, upsetting the equilibrium. The reaction temporarily resumes until a new equilibrium is established.

The Old Man of the Mountain
For two hundred years a precarious rock formation in New Hampshire was said to resemble the face of an old man. It had become a symbol of New Hampshire, appearing on postcards, road signs and coins On May 3, 2003 The rock face collapsed. In terms of equilibrium, we could say that this was a static equilibrium that endured for centuries, until it was disturbed by a spring storm. Afterwards a new equilibrium was established, that unfortunately no longer resembled a face.

Qualitative Aspects of Chemical Equilibrium
Chapter 11 Qualitative Aspects of Chemical Equilibrium

Qualitative Aspects of Equilibrium
11.1 Qualitative Aspects of Equilibrium What is an equilibrium? What properties does it have? How can we distinguish dynamic, and static equilibria and tell them apart from simple steady states?

Note: Gravity is not considered to be a dynamic force.
Static Equilibrium Static equilibrium exists when a system remains unchanging without any active, dynamic processes involved One rock sitting on another is at static equilibrium, even if it is balanced precariously. No dynamic processes are acting on it, so it remains unchanged. Static equilibrium is a bit boring. We seldom deal with it in chemistry. Note: Gravity is not considered to be a dynamic force. It is a static force.

Dynamic Equilibrium Dynamic equilibrium is the result of two opposing, active processes occurring at the same rate. No visible changes take place, but there are constant changes in the particles at a microscopic level. There are several types of dynamic equilibrium of interest to chemists, which are described on the following slides including: Phase Equilibrium Solubility Equilibrium Chemical Equilibrium

A dynamic equilibrium is a bit like a hockey game.
Barring penalties, there is always the same number of players on the ice, but some players are constantly leaving the bench as others return to it. 19 11

Phase Equilibrium (1st type of dynamic equilibrium)
Phase equilibrium is a dynamic equilibrium that occurs when a single substance is found in several phases or states within a system as the result of a physical change. Example: In a closed bottle a water may exist as both a liquid and a gas at the same time (eg. Water vapour above liquid water). As water molecules evaporate from the liquid phase, other water molecules condense from the gaseous phase.

Solubility Equilibrium (2nd type of dynamic equilibrium)
Solubility Equilibrium occurs when a solute is dissolved in a solvent, and an excess of the solute is in contact with the saturated solution. Example: If you add too much sugar to a cup of tea, the tea becomes saturated with sugar and no more appears to dissolve. In fact, some molecules of sugar are dissolving as other molecules recrystallize back into solid sugar.

Chemical Equilibrium (3rd type of dynamic equilibrium)
Chemical equilibrium occurs when two opposing chemical reactions occur at the same rate, leaving the composition of the system unchanged. Example: Dinitrogen tetroxide (N2O4) and nitrogen dioxide (NO2) can exist in the same container. Each can change into the other, and at equilibrium they do so at the same rate. N2O4  2 NO2 This is the most important type of equilibrium in chemistry!!

Irreversible and Reversible Reactions
11.2 Irreversible and Reversible Reactions Some chemical reactions are easily reversed, like the electrolysis of water. Others, such as the burning of wood are impossible to reverse under laboratory conditions.

Irreversible Chemical Reactions
An irreversible reaction is a reaction that can only occur in one direction, from reactants to products. This definition is assumed to refer to reactions occurring under normal laboratory conditions. In a lab, it is easy to burn a piece of wood. It is impossible, under laboratory conditions, to turn ash, smoke, carbon dioxide and water back into wood. The growth of a tree does allow wood to be produced from materials that might include wood ashes, but growing a tree takes decades, and requires countless changes involving many complex chemical mechanisms and multiple organic catalysts (enzyme systems). Burning is therefore NOT considered reversible!

Remember! Irreversible = One Way
Reactants  Products Reactants can become products, but products cannot turn back into reactants.

Reversible Chemical Reactions
Some reactions are easily reversed using common laboratory procedures. For example, it is possible to decompose water into hydrogen and oxygen in a electrolytic cell, and equally possible to synthesize water from hydrogen and oxygen in a fuel cell: 2H2O + electrical energy  2 H2 + O2 2H2 +O2  2H2O + electrical energy

Remember! Reversible = Both Ways
Reactants Products Reactants can become products, and products can also turn back into reactants. Also show as: reactants  products

Reversibility and Equilibrium
Only reversible reactions can produce a true dynamic chemical EQUILIBRIUM

A Chemical System is at Equilibrium if it meets these Criteria:
The System is Closed, for example by being sealed inside a container so material cannot enter or leave. The Change is reversible. The reaction or change can proceed in both direct and reverse directions. There is no Macroscopic Activity. Nothing seems to be happening – the properties of the system are constant. These unchanging properties can include: colour, amount of undissolved solute, concentration, pressure etc. There is Molecular Activity. Reactions continue at the microscopic or molecular level

Definitions of some easily confused terms
Macroscopic: Occurring at the level we can detect with our senses, as opposed to microscopic. Observable changes. Microscopic: Occurring at a level below what we can see. Too small to observe without instruments. Dynamic Equilibrium: A balance that involves two opposing active processes that are occurring at the same rate. This contrasts with Static Equilibrium and Steady State (Homeostasis). Static Equilibrium: A balance that does not involve active processes. Steady State (including Homeostasis): An apparent balance that occurs in an open system. An unchanging set of properties is maintained, but materials enter and leave the system.

Read all the questions, make sure you understand them, be prepared to answer them verbally next class.

Le Châtelier’s Principle
11.4 Le Châtelier’s Principle Henri Louis Le Châtelier ( ) was a French chemist who is most famous for his studies of chemical equilibrium. In addition he studied metal alloys and, with his father, was involved in the development of methods of purifying aluminum

Equilibrium is not eternal
An equilibrium can exist for a long time, only to change (be upset) when certain conditions change. After it is upset, there is a period of adjustment, then a new equilibrium is established. New Equilibrium Original Equilibrium Adjusting... Equilibrium Upset

Some factors that might upset an equilibrium.
Which of these factors do you think might affect the amount of reactant and product at equilibrium? Maybe Temperature? Maybe Pressure? Maybe Concentration of reactants and products? Maybe Catalyst? Most of them do, but one does not. We’ll see which one doesn’t a bit later!

Henri LeChâtelier studied many of these factors to see how they could effect a system at equilibrium. He found some factors could favour the direct reaction, increasing the amount of product. Others could favour the reverse reaction, increasing the amount of reactant. Regardless, eventually an equilibrium was re-established, but with new amounts of product and reactant.

When an Equilibrium is Upset...
Henri LeChatelier stated the following generalization: “If the conditions of a system in equilibrium change, the system will react to partially oppose this change until it attains a new state of equilibrium.” In other words: The system will establish a new equilibrium. .

Will any reversible reaction will eventually reach equilibrium?
Answer: yes, as long as it in a closed system. Will the amount of product always equal the amount of reactant at equilibrium? Short answer: NO! they are not always equal. At equilibrium the amount of “reactant” and “product” may vary depending on several factors.

The Effect of a Catalyst
Adding a catalyst to a system already at equilibrium will have NO EFFECT. Adding a catalyst to a system that has not yet reached equilibrium will cause it to reach equilibrium faster. Why? A catalyst increases both forward and backward rates equally, so the final result will be the same, but the process of reaching equilibrium will be faster.

Effect of Temperature Increasing the temperature will favour the endothermic reaction. Decreasing the temperature will favour the exothermic reaction.

Effect of Pressure Increasing the pressure may favour the reaction that produces fewer gas particles Decreasing the pressure may favour the reaction that produces more gas particles. Note: only reactants or products that are in the gaseous state are counted towards the effects of pressure.

Effects of Concentration
The effects of concentration of the reactants and products are most important of all. Increasing or decreasing the concentration of Reactants WILL have an effect. Remember: Pure solids(s) and liquids(l) do NOT have a variable concentration. Before we can see the effects of changing a concentration we should remember what LeChatelier said...

Remember LeChatelier’s Principle:
“If the conditions of a system in equilibrium change, the system will react to partially oppose this change until it attains a new state of equilibrium.” In other words, any “stress” or change that you make to the system will cause it to react in a way that tries to (partially) undo the change that you made.

Changes in Concentration
If you increase the concentration of a reactant (gas or aqueous), the equilibrium will shift to use up some of the reactant you added. If you increase a product (gas or aqueous), the equilibrium will shift to reduce the product and make more reactant H2(g) + I2(g) 2HI(g) If you increase the concentration of Hydrogen… …The system will react to reduce the amount of Hydrogen… …by shifting the reaction towards the product

H2 + I2  2HI  Sudden increase in[H2]. The equilibrium changes:
[HI] goes way up, [I2] goes down. The equilibrium changes: [HI] goes way up, [I2] goes down. This causes [H2] to adjust towards its original level

Note: Adding more of a undissolvable solid or pure liquid normally has NO EFFECT on a system at equilibrium. It can only affect the equilibrium if the concentration changes, and usually only gases and aqueous solutions have variable concentration. However: adding water to an aqueous solution can change its concentration by dilution. Adding solid to a solution that is not saturated MIGHT increase its concentration (if it dissolves!).

Changes in Temperature
Increasing the temperature causes the equilibrium to shift in the direction that absorbs heat (the endothermic direction). Decreasing the temperature shifts the equilibrium in the exothermic direction. 2SO2 + O2 2SO3 + Heat If you increase the temperature… …The system will react to reduce the temperature… …by shifting the reaction towards endothermic side

 Sudden increase in temperature.
2SO2 + O2  2SO3 + heat  Sudden increase in temperature.  The endothermic reaction kicks in, getting rid of some SO3, and creating more SO2 and more O2 and cooling things off  This causes a lowering of the temperature, moving it towards its original level

Changes in Pressure (only affects systems where one or more materials are gases)
Increasing the pressure causes the equilibrium to shift in the direction that has the fewest gas molecules. Decreasing the pressure shifts the equilibrium in the direction that produces more gas molecules. 8 molecules 4 molecules If you increase the pressure… …the system will create fewer molecules… …to reduce the pressure.

N2 + 3 H2  2 NH3 Sudden increase in pressure.
Pressure partially adjusts towards the original level.. When the pressure increases, the equilibrium adjusts, making more NH3 (since 2 molecules NH3 are fewer particles than 3 molecules of H2 plus 1 molecule of N2)

I2(s) I2(g) Special Note Re. Gases
Only gases can be affected by pressure. If only one side of an equation has gases, then... Increasing pressure will favour the side with no gases. Decreasing the pressure will favour the side with gases. I2(s) I2(g) Decreased pressure Increased pressure

, Questions 1 to 4

The Quantitative Aspects of Equilibrium
Chapter 12 The Quantitative Aspects of Equilibrium

12.1 Chapter 12 In this section we will explore the mathematical aspects of equilibrium, including: The Equilibrium Constant (Kc) The Equilibrium Law

The Equilibrium Constant and Equilibrium Law Expressions
The equilibrium constant (Kc) is: A number derived from an equilibrium law expression. a ratio between the concentration of products and the concentration of reactants of a reversible reaction at equilibrium, but… With each aqueous or gaseous product and reactant raised to the power of its corresponding coefficient

Deriving the Equilibrium Law
Warning: Math Content Ahead The next two slides show how the equilibrium law was derived from the rate law. If you want to understand the relationship between rates and equilibrium you should follow this. If you just want to use the equilibrium law to find Kc, you can skip forward three slides.

A Generalized Reversible Reaction with reactants and products
Forward reaction = dir Reverse reaction = rev aA + bB ↔ cC + dD reactants products Forward rate: rdir = kdir [A]a [B]b Reverse rate: rrev = krev [C]c [D]d At equilibrium, rdir = rrev so, through the magic of algebra… Products on top Reactants on bottom

Simplifying the Rate Constant
The two separate rate constants (kdir and krev) are often replaced by a single equilibrium constant, Kc: Becomes:

The Equilibrium Law aA + bB ↔ cC + dD
The lowercase letters represent coefficients, the uppercase letters are chemical formulas, the square brackets mean concentration. Kc is the equilibrium constant For a chemical equation of the type: aA + bB ↔ cC + dD The equilibrium law expression is: “Products” always go on the top! “Reactants” always go on the bottom!

Variants of the Equilibrium Constant
keq These symbols are all used for Equilibrium Constants in different text books or for different types of reaction. Kc Ka I usually use Kc, since your text book and the study guide both use it. The old textbook used keq. In problems involving acids and bases, Ka and Kb are often used. Ksp is used for problems involving the solubility product. Kb Ksp

What does the Equilibrium Constant mean?
It is a ratio between the amount of reactant and product that exist at equilibrium. If Kc is greater than 1, then the direct reaction is favoured, and there is more product than reactant at equilibrium* If Kc is less than 1, then the reverse reaction is favoured, and there is more reactant than product at equilibrium* If Kc is 0, the reaction is impossible. if Kc is infinite (∞) the reaction is spontaneous and irreversible so as much reactant as possible will change to product Reactions we call irreversible have high Kc value (>1010) Reactions that don’t normally occur have Kc values near 0 (<10-10) *this is a slight over-simplification, since the formula can be complicated, but it is generally true.

Effect of Temperature on an Equilibrium Constant
An equilibrium constant relates to concentrations, and only remains constant if the other conditions, such as temperature, remain fixed. If the temperature were to change, so would the value of Kc How the value of Kc might change depends on the type of reaction (exothermic or endothermic) and how the temperature changed (increased or decreased)

Table Showing Effects of Temperature on Equilibrium Constants
Type of reaction Temperature change Favoured Reaction Change in Kc Exothermic* Increase Reverse () Decrease Direct () Endothermic *Exothermic means either ΔH < 0 or that Reactants  Products + Energy Notice that any change in temperature that favours the direct reaction will cause the value of Kc to increase.

Fixed Concentrations Some materials have “fixed” concentrations, ie. Their concentrations cannot change in an equilibrium. Examples: An undissolved solid, (it can’t have a concentration unless it dissolves.) A pure liquid. (a pure substance always has its maximum concentration) These cases, which include all substances with the (s) and (l) phase markers, are not included in equilibrium calculations.

Example What is the equilibrium expression for the following equation:
CN1-(aq) + H2O(l)  HCN(aq) + OH1- (aq) Answer: Not this: Why? Because H2O (liquid water) is a pure substance and therefore has a fixed concentration. Substances with fixed concentration are not included in an equilibrium expression.

Copy the following equations, and write the equilibrium law expression for each
H2(g) + I2(g)  2 HI(g) 2 BrCl(g) Cl2(g) + Br2(g) CO2(g) + H2O(l) H2CO3(aq)

Calculating Equilibrium Concentrations
12.1.4 Calculating Equilibrium Concentrations To calculate the concentrations of reactants and products at equilibrium we sometimes use a table that records the initial concentration, the change in concentration and the final equilibrium concentration of each reactant or product. We call such a table an I.C.E. table The ICE method

The I.C.E. method I.C.E. stands for:
The I.C.E. method is a technique that can help solve some equilibrium problems I.C.E. stands for: Initial concentration [A]I Change in concentration Δ[A] Equilibrium concentration [A]E

I C E Reactant + Reactant  Product + Product Reactant 1 Reactant 2
Identify the molar ratios (based on the coefficients of the equation) Eliminate any unused ratios, such as those based on liquids and undissolved solids. Write the chemical equation of the reaction. Show all reactants and Products Reactant + Reactant  Product + Product Molar ratio Molar ratio Molar ratio Molar ratio Reactant 1 Reactant 2 (if needed) Product 1 Product 2 I (Initial) C (Change) E (Equilibrium) Concentration Before Concentration Before Concentration Before Concentration Before Concentration Difference (Δ[C]) Negative Ratio (reactant) Negative Ratio (reactant) Positive Ratio (product) Concentration After Sum Sum Sum Fill in the missing squares in “C” row. They will be ratios to the one you know The numbers will be negative for reactants, positive for products. Use the Equilibrium Concentrations to answer any further questions: such as finding Kc , Ka Kb or Ksp) Make a table with 3 rows labelled I, C, and E, and enough columns for all the reactants and products you are interested in Add to find the equilibrium values (adding a negative number is like subtracting) Enter the information you already know into the table. If no values are given for the initial product concentrations, it is usually safe to assume they are zero. Look for a column that you can complete!

ICE method: Step by Step
Write the chemical equation of the reaction. Show all reactants and Products Identify the molar ratios (based on the coefficients of the equation) Eliminate any unused ratios, such as those based on liquids and undissolved solids . Make a table with 3 rows labelled I, C, and E, and enough columns for all the reactants and products you are interested in Enter the information you already know into the table. If concentrations are not given as mol/L you may have to convert them. If not told otherwise, you may assume that the initial product concentrations are zero. Look for a column that you can complete! Fill in the missing squares in “C” row. They will be ratios to the one you know The numbers will be negative for reactants, positive for products. Add to find the equilibrium values (adding a negative number is like subtracting) Use the Equilibrium Concentrations to answer any further questions (such as finding Kc , Ka or Kb)

Sample Problem (see p. 316 and 317 in Text Book)
Copy this work! At a given temperature, 10 moles of nitrogen oxide (NO) and 8 moles of Oxygen (O2) are placed in a 2 litre container. After a given period of time, the following equilibrium is obtained: 2 NO(g) + O2(g)  2NO2(g) Once equilibrium is attained, 8 moles of NO2 have been produced. Calculate the equilibrium constant. Preliminary work: It is necessary to convert to moles per litre! Initial concentrations [NO]I = 10 mol in 2 L = 5 mol/L [O2]I = 8 mol in 2L = 4 mol/L assume: [NO2] I = 0 mol in 2L = 0 mol/L Final Concentration: [NO2]E = 8 mol in 2L = 4 mol/L

I C E 5 mol/L 4 mol/L 0 mol/L +4 mol/L -4 mol/L -2 mol/L 4 mol/L
Fill in the missing squares in “C” row. They will be ratios to the one you know The numbers will be negative for reactants, positive for products. Look for a column that you can complete! Add to find the equilibrium values (adding a negative number is like subtracting) Enter the information you already know into the table Make a table with 3 rows labelled I, C, and E, and enough columns for all the reactants and products you are interested in Write the chemical equation of the reaction. Show all reactants and Products Identify the molar ratios (based on the coefficients of the equation) 2 NO(g) O2(g)  2NO2(g) 2 1 2 NO O2(g) NO2 I (Initial C) C (Change) E (Equilibrium) 5 mol/L [NO]i 4 mol/L i[O2] 0 mol/L [NO2]i +4 mol/L Δ [NO2] -4 mol/L (-ratio) -2 mol/L (-ratiot) 4 mol/L [NO2]e 1 mol/L 2 mol/L

Finishing the Problem Now you need to find the Kc value. To do this, use the values from the “E” line of your table (the equilibrium concentrations) and substitute them into the Kc formula: The value of the equilibrium constant is 8 (although you do not need to give the units of an equilibrium constant, since our concentrations were in moles/Litre, and our time is assumed to be in seconds it would be safe to call it mol/(L∙s))

A Tougher Sample Problem (see p. 316 and 317 in Text Book)
At 1100K the equilibrium constant of the following reaction is 25 H2(g) + I2(g)  2HI(g) 2 moles of hydrogen and 3 moles of iodine are placed in a 1 L container. What is the concentration of each substance when the reaction attains equilibrium.. Initial concentrations: [H2] = 2 mol/L, [I2] = 3 mol/L This time we have no numbers for equilibrium concentrations, we will have to use algebraic variable for some of the values. We can use x to represent the change in Hydrogen Concentration Let: Δ[H2] = -x (why negative? Because hydrogen is a reactant!)

I C E 2 mol/L 3 mol/L 0 mol/L -x mol/L -x mol/L +2x mol/L 2-x mol/L
Fill in the missing squares in “C” row. They will be ratios to the one you know The numbers will be negative for reactants, positive for products. Add to find the equilibrium values (adding a negative number is like subtracting) Make a table with 3 rows labelled I, C, and E, and enough columns for all the reactants and products you are interested in Look for a column that you can complete! Write the chemical equation of the reaction. Show all reactants and Products Enter the information you already know into the table Identify the molar ratios (based on the coefficients of the equation) H2(g) I2(g)  HI(g) 1 1 2 H2 I2(g) HI I (Initial C) C (Change) E (Equilibrium) 2 mol/L 3 mol/L 0 mol/L -x mol/L -x mol/L +2x mol/L 2-x mol/L 3-x mol/L 2x mol/L

Continuing the Problem
Set up the Kc formula, substituting in the expression from line “E” So.. This equation can be rearranged in the following steps:  Carried over to next slide

Using the Quadratic formula
 Equation to be solved Quadratic formula, where: a=21, b=125, c=150 substitute Two solutions to formula But only one of the solutions will give a real answer!

Choosing the Best Answers
Two solutions to formula Try finding the correct concentrations using the solutions Solve using 4.29 Solve using 1.67 Concentration cannot be negative! At equilibrium: The concentration of hydrogen will be 0.33 mol/L, The concentration of iodine will be 1.33 mol/L and The concentration of hydrogen iodide will be 3.34 mol/L

Simplifying I.C.E. tables (the 5% rule)
Sometimes, when doing an ICE table you may have to subtract a very small value from a relatively large value, for example 2.0 mol/L – 1.0x10-4 mol/L. In this case, don’t bother doing the subtraction, since by the time you change it to show significant digits, the result will be the same: 2.0 mol/L – mol/L = mol/L ≈ 2.0 mol/L. In fact, as a rule of thumb, if the number you subtract is less than 5% of the original number, you can skip the subtraction.

Page 318 Questions #1 to 8, (Equilibrium Law and Kc)
Questions #19 to 21, (I.C.E. method) Optional Assignments: Link to another worksheet

12.2 Acids and Bases The concept of acids and bases has been with us for a long time, but there has always been difficulty with the exact definitions. In this section we will look at four theories about acids and bases (Don’t worry, you only need to remember 2 of them) We will find out the difference between “strength”, “concentration” and “acidity” of an acid. We will also explore the concepts of pH and pOH a bit deeper than you did in grade 9 or 10. We will explore the mathematical relationships between concentration and pH

Properties of Acids & Bases
Acids are solutions that: Turn litmus red, but leave phenolphthalein clear. React with active metals to give off H2 gas React with carbonate salts to give off CO2 gas Taste sour (if safe to taste) Have low pH numbers (below 7) Bases are solutions that: Turn litmus blue, and turn phenolphthalein purple. Taste bitter (if safe to taste) Seldom react with metals or carbonates Have High pH numbers (above 7) Emulsify fats and oils

Theories of Acids and Bases Hypothesis#1: Lavoisier’s Mistake (optional item)
Lavoisier (1776) dealt mostly with strong oxyacids, like HNO3 and H2SO4. He claimed that: An acid is a substance that contains oxygen. This hypothesis is now known to be completely wrong, but it did give oxygen its name: “oxy” (meaning acid)+ “gen” (former or creator)

Theories of Acids and Bases (Hypothesis#2: Arrhenius’ Theory)
Arrhenius (c. 1884) claimed that: An acid is a substance that dissociates in water to produce H+ ions A base is a substance that dissociates in water to produce OH- ions Arrhenius’ theory is still used to this day as a “simplified” way of explaining acids and bases. It explains most of the properties of acids... Why their formulas usually begin with H, why they give off hydrogen when reacting with metals, etc. But this theory does not account for acidic and basic salts– substances that act like acids and bases, but don’t have an H or OH in their formula, or for anhydrous acids, or for reactions that occur outside of water.

Theories of Acids and Bases (Hypothesis#3: Brønsted-Lowry)
Brønsted and Lowry (1923) proposed: An acid is a substance from which a proton (H+) can be removed. An acid is a proton donor. A base is a substance that can cause a proton to be removed from an acid. A base is a proton acceptor. Although a bit complicated for explaining simple acid/base reactions, this is the main theory in use today.

Brønsted-Lowry and Conjugates
One of the results of the Brønsted-Lowry theory is that in a reaction, each acid has a corresponding base and each base has a corresponding acid (called their conjugates) H+ H+ HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq) Acid (strong) Base (weak) Conjugate Acid of H2O Conjugate Base of HCl Becomes Becomes

Simple way of finding conjugates
The formula of a conjugate base is the formula of the acid with one H removed, and more negative charge. The conjugate base of HBr is Br- The conjugate base of HCN is CN- The conjugate base of water (H2O) is OH- The first conjugate base of H3PO4 is H2PO4- H3PO4 can have other conjugates, such as HPO42-, PO43- The formula of a conjugate acid is the usually the negative ion, with an H added in front of it and one step more positive: The conjugate acid of F- is HF The conjugate acid of water (H2O) is H3O+

Theories of Acids and Bases (Hypothesis#4: Lewis Acids)
(optional item) Gilbert Lewis (c.1940) proposed: An acid is a substance that can accept an electron pair to form a covalent bond. A base is a substance that can donate a pair of electrons to form a covalent bond. This theory is not explained in the new textbook, and will probably not be required for examinations. It is given here as optional enrichment material. The Lewis theory is the only one that can explain the properties of acidic & basic salts.

Comparing the Hypotheses
Theory An Acid is... A Base is... Details Lavoisier A substance with oxygen Incorrect hypothesis Arrhenius A substance that dissociates in water to produce H+ ions A substance that dissociates in water to produce OH- ions. All acids have H All bases have OH Bronsted-Lowry A substance from which an H+ ion can be removed (a proton donor) A substance which can remove the H+ from an acid (a proton acceptor) Bases can have any negative ion. Lewis A substance that can accept a pair of electrons to form a covalent bond A substance that can donate a pair of electrons to form a covalent bond Acids may have H+ or some other + ions, a base may have OH- or some other – ions Incorrect Hypothesis Optional Hypothesis

Assignment Find the Conjugate base of each of these acids: 1) HCl
2) HNO3 3) HBr Find the conjugate acid of each of these bases: 1) CN- 2) HCO3- 3) H2O

Dissociation Both theories of acids tell us that acids are compounds that “dissociate” in water to give off H+ ions (protons), which immediately attach to water molecules to become H3O+ ions Eg. HCl added to water breaks up into: H+ and Cl- ions, the H+ ions join H2O as shown by: HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq) (monoproteic acid) Different acids can dissociate differently: H2SO4(aq) + 2H2O(l)  2H3O+(aq) + SO4-(aq) (diproteic acid) H3PO4 (aq) + 3H2O  3H3O+(aq) + PO4-(aq) (triproteic acid)

Dissociation of Water H2O  H+ + OH-
In distilled water, at any given moment about 2 molecules out of every billion ( %) exist as ions: H+ and OH- This means there is an equilibrium reaction occurring. H2O  H+ + OH- Since so few of the molecules are ionized, the K value of this equilibrium must be tiny! (see calculation 3 slides from now) This also means that water can act as either a very weak acid or a very weak base! (it can give off H+ or OH-)

Equivalence of H+ and H3O+
Some textbooks use H+ and other books use H3O+ to represent the hydrogen ion content of acids. Although H3O+ better represents the actual state of the ions in a water solution, H+ is simpler to write. As long as the solvent is water, the two are the same [H+]=[H3O+] In most cases, when water is present

Similarity of Ka Kb Kw and Kc
Ka is called the acidity constant Kb is called the basicity constant Kw is the ionization constant of water but all three are calculated the same way as Kc and are used for similar purposes. We usually use Ka or Kb in acid-base reactions Tables of Ka values help us compare the “natural strength” of various acids.

The Ionization Constant of Water (Kw)
Water is H2O, but at any given temperature a tiny fraction (≈ 2/ 109) of water molecules dissociate into ions: H2O(l)  H+(aq) + OH-(aq) Kw represents a ratio between the ionized and unionized molecules:

Because of the equivalence of H+ and H3O+ in water solutions, you could also write:
At room temperature [H3O+] and [OH-] in pure water are both about 1.0×10-7 mol/L So... The value of Kw varies a bit with temperature (see p 328), but has an average value of 1.0 × at 25°C.

Effect of Temperature on Kw
At lower temperatures, Kw becomes smaller. Less water molecules dissociate when it is cold At higher temperatures, Kw becomes larger. More water molecules dissociate when it is hot. Kw affects the pH of water. We say that the pH of pure water is 7, but in reality it varies by temperature: At the boiling point (100°C) , the pH of pure water will be close to 6! At the freezing point (0°C) it will be about 7.5!

The Equilibrium Constant of an Acid
The acidity constant of an acid (Ka) represents the fraction of the acid which will dissociate and release H+ (ie. H3O+) ions. It is one of two factors that affect what the pH of an acid solution will be. (the other factor is the concentration of the acid) Acids with low Ka values are considered naturally “weak” acids. Acids with high Ka values are considered naturally “strong” acids.

In this case, we can calculate the Ka value this way:
If we consider an acid to have the formula HA then its dissociation can be represented by HA(aq) + H2O(l)  H3O+(aq) + A-(aq) In this case, we can calculate the Ka value this way: NOTE: Because of the equivalence of [H+] and [H3O+] in some texts, this could be shown as 

The Ionization Percentage of an Acid
If you ever want to work out the percentage of an acid that is ionized (like if you had too much time on your hands, or were bored or something) you can use this formula:

Copy the problem and solve:
Johnny dissolves 3.4 mols of dried weak acid powder in 500 mL of water. HA(aq) + H2O(l)  H3O+(aq) + A-(aq) When he measures the acidity he discovers that the acid has 4.5x10-2 mol/L of H3O+ ions. What is the Ka of this acid? What is the percentage ionization of this acid?

12. pH, [H3O+] and Ka of Acids There are three different factors that affect the apparent strength of an acid. In this section we will discover some of the ways of measuring acids and bases, including: Strength vs. concentration vs. acidity What are pH and pOH? pH (by the chart) pH (by calculation) Applying the I.C.E. Method to Acids and Bases For finding [H+], [OH-]concentrations, Ka, Kb.

Measures of an Acid The natural strength Ka The concentration mol/L
There are three different ways of measuring acids: The natural strength Ka The concentration mol/L The degree of Acidity pH

Natural Strength of an Acid
A “strong” acid is one that dissociates completely (100%) when dissolved in water. Strong acids have Ka values close to infinity (huge numbers > 1010). Strong acids dissociate irreversibly. Typical Reaction: HA + H2O  H3O+ + A- A “weak” acid is one that does NOT dissociate completely when dissolved in water. Weak acids have small Ka values, usually less than 1. Weak acids dissociate reversibly, creating a possible equilibrium. Typical reaction: HA + H2O  H3O+ + A-

“Strong” Acids and “Weak” Acids Acids shown in red, Conjugate Base in blue
Strong Acid Name Reaction Ka (>>1) Hydrochloric Acid HCl(aq) +H2O H3O Cl- ∞ Hydrobromic Acid HBr(aq) +H2O  H3O Br- Hydroiotic Acid HI(aq) +H2O H3O I- Nitric Acid HNO3(aq) +H2O  H3O+ + NO3- Sulphuric Acid H2SO4+2H2O  2H3O+ + SO4- Weak Acid Name Reaction Ka (<1) Iodic HIO3+H2O  H3O++IO3- 1.7x10-1 Oxalic Acid H2C2O4+H2O H3O++HC2O4 5.8x10-2 Formic Acid HCOOH +H2O H3O++HCOO- 1.8x10-4 Acetic Acid CH3COOH +H2O H3O++CH3COO- 1.7x10-5 Citric Acid H3C6H5O7+H2O H3O++H2C6H5O7 3.2x10-7 Hydrogen Peroxide H2O2 + H2O  H3O+ + HO2- 2.4x10-12 Water H2O + H2O  H3O+ OH- 1.0x10-14

Concentration of an Acid
The concentration of an acid is the number of moles of an acidic substance that have been dissolved in a volume of water. This is determined by the concentration formula: 6 mol/L is considered a very concentrated acid 0.1 mol/L is considered a fairly dilute acid Where: n = number of moles of acidic solute V= volume of solution in Litres

Strength, Concentration and Acidity
The strength and concentration of an acid together determine its degree of acidity STRENGTH The natural strength of an acidic compound, determined by its Ka value Concentration The number of moles per litre of the acidic compound in solution. Degree of Acidity The effective strength of an acid, based on its [H3O+] concentration and Usually recorded as its pH

Degree of Acidity The degree of acidity is the EFFECTIVE strength of the acid, that is, how effective the acid is at reacting with other materials. Acidity can be measured using: the H3O+ concentration, or more commonly a measure called the pH

Who invented pH and pOH? (optional background information)
The degree of acidity of a solution is directly related to its [H3O+] concentration. Unfortunately this is often a small number expressed in scientific notation, such as 1.3x10-5 mol/L or 2.3x10-13 mol/L. Not easy numbers to remember, compare or write. In 1909 Søren Sørensen suggested an easier way to record the degree of acidity of solutions. It was a logarithmic scale called pH. Each level of the pH scale represents a 10 fold difference in H3O+ (or H+) concentration. ie. A acid of pH 2 has 10 times more [H+] than one that’s pH3.

Or: (since [H3O+] is equivalent to [H+] in water solutions)
pH and pOH pH is a measure of the degree of acidity, given by the following formula: pH = - log [H3O+] Or: (since [H3O+] is equivalent to [H+] in water solutions) pH = - log [H+] Although less used, there is also a measure of the degree of alkalinity, called pOH: pOH = - log [OH-]

pH and Concentration of H+ ions For solutions of an exact pH you may use the chart below:
For simplicity I used [H+] instead of [H3O+] Acids Bases pH [H+]* [OH-]* pOH 10 0 10-14 14 8 10-8 10-6 6 1 10-1 10-13 13 9 10-9 10-5 5 2 10-2 10-12 12 10 10-10 10-4 4 3 10-3 10-11 11 10-0 7 10-7 *Concentration in mol/L Temperature = 25°C

What about “in-between” pH values?
See next slide for suggestions The formula for pH is: pH = -log [H+] On your calculator you must find out how to calculate the negative logarithm of a number! The formula for [H+] concentration is: [H+]=10 –pH or… [H+]=log-1(-pH) Sometimes log-1 is called antilog: [H+] =antilog (-pH) Sometimes log-1 is called inverse log: [H+] =invlog (-pH) Sometimes log-1 is called 10x: [H+] =10(-pH) On your calculator find out raise 10 to a negative number or how to calculate the inverse logarithm (antilog) of a negative number.

Question: Find the pH if the H3O+ concentration is 1.40x10-8 mol/L
Solution: pH = – log(1.40 x 10 – 8) (TI 83 instructions) Type: log (1.40 x 10 ^ (-) 8) Enter (–) log ^ (–) Enter The answer should be: pH=7.85… (I’ve rounded to 3 Sig.Fig.) Question: Find the H3O+ concentration if the pH is 4.30 Solution: [H+] =10 – 4.30 (TI-83 instructions) 10 ^ (-) 4.3 ^ (–) Enter The answer should be: pH=5.01… x mol/L(I’ve rounded to 3 Sig.Fig.)

Using the Windows Calculator
Switch to scientific view To find pH, use: 1.4 Exp = To find [H3O+] use: = or: = 7.85… 5.01…x10-5 Finding the pH if H3O+ concentration is 1.40x10-8 mol/L Finding [H3O+] concentration if pH is 4.30

Traditional Scientific Calculators
Question: Find the pH if the H3O+ concentration is 1.40x10-8 mol/L Solution: pH = – log(1.40 x 10 – 8) Try: (1.4 Exp 8 +/- ) log +/- = Or: (1.4 Exp 8 +/- ) log +/- = Exp +/- log +/- EE +/- log +/- Question: Find the H+ concentration if the pH is 4.30 Solution: [H+] = Try: 10 yx /- = Or: /- Inv log = Or /- 2ndF 10x = Or: 10 ^ (-) 4.3 ×10-5

A) [H+]=1.0x10-8 mol/L B) [OH-]=1.0x10-3 mol/L
1. Find the [H+] and [OH-] of the following pH solutions without using a calculator: A) pH=4 B) pH=12 C) pH=9 D) pH=7 E) pH=8 2. Find the pH of the following solutions without a calculator: A) [H+]=1.0x10-8 mol/L B) [OH-]=1.0x10-3 mol/L C) [H+]=1.0x10-5 mol/L D) [OH-]=1.0x10-9 mol/L 3. Find the [H+] of these solutions using the 10x or antilog function of your calculator: [H+]=10-pH A) pH= B) pH= C) pH= D) pH= E) pH=7.1 4. Find the pH of the following solutions using the log function of your calculator: pH= - log [H+]. A) [H+]=4.5x10-3 mol/L B) [H+]=3.4x10-8 mol/L C) [H+]=3.0x10-7mol/L D) [H+]=2.5x10-2 mol/L

Problem 5. 2 sig. digits 5. Calculate the Ka value of a monoproteic, weak acid if a 0.10 mol/L (initially) solution has a pH of 5.5 when it reaches equilibrium. Assume: HA H+ + A- Hint: You must find the concentration of [H+] ions before you do the problem. 2 sig. digits

Solutions to Problems 1- 4
1. Answers A) 1x10-4, 1x10-10 mol/L B) 1x10-12, 1x10-2 mol/L C) 1x10-9, 1x10-5 mol/L D) 1x10-7, 1x10-7 mol/L E) 1x10-8, 1x10-6 mol/L 2. Answers: A) pH=8 B) pOH-=3 pH= 11 C) pH=5 D) pOH-=9 pH=5 3. Answers in mol/L A) 2.0x10-4 mol/L B) 1.6x10-10 mol/L C) 6.3x10-7 mol/L D) 1.0x10-4 mol/L E) 7.9x10-8 mol/L 4. Answers: pH= - log [H+]. A) pH= B) pH=7.5 C) pH=6.5 D) pH=1.6

Solution to Problem 5 Calculate the Ka value of a monoproteic weak acid if a 0.1 mol/L (initially) solution has a pH of 5.5 when it reaches equilibrium. Assume: HA H+ + A- Hint: You must find the concentration of [H+] ions before you do the problem. Find the [H+]: [H+] = log-1 (-5.5) or [H+] = [H+] = 3.162x10-6 mol/L For an extremely accurate solution, use the I.C.E. method to find the equilibrium concentrations… (continued on next slide) For a quicker solution, using the 5% rule, see later slides

HA  H+ + A- 1 : 1 : 1 HA H+ A- I (initial) 0.10 C (change)
Round to 2 Sig. Digits : : HA H+ A- I (initial) 0.10 C (change) E (equil.) 3.163x10-6 -3.163x10-6 +3.163x10-6 +3.163x10-6 3.163x10-6 9.9997x10-2 Ka = [H+][A-] = (3.163x10-6 x 3.163x10-6) mol/L = x10-10 [HA] x10-2 mol/L ≈ 1.0x10-10

The Short-cut: An alternative solution using 5% rule
Ka = [H+][A-] = (3.163x10-6 x 3.163x10-6) mol/L = x10-10 [HA] mol/L = 1.0x10-10 Since x 10-6 is less than 5% of 0.1, we can use the 5% rule and simply write 0.1, instead of subtracting (0.1 – ) and using an ICE table HA H+ A- I (initial) 0.10 C (change) E (equil.) 3.163x10-6 After rounding to the correct number of significant digits, our answer is the same as doing it the hard way! assume 0.10 0.10

See the solution on page 333 (I told you, it’s only half-evil)
A Half-Evil Example Calculate the pH of an aqueous solution of formic acid HCOOH at 0.20 mol/L if its acidity constant is 1.8x10-4.The equation of this reaction is as follows HCOOH(aq) + H2O(l)  H3O+(aq) + HCOO-(aq) See the solution on page 333 (I told you, it’s only half-evil)

Assignments Page 339, Questions 4 to 23

Unit 4: Chapter 12.2.5 Solubility Product Constant
Calculating the solubility constant Examples

Solubility A saturated solution contains:
Dissolved ions  Solid solute crystals A saturated solution contains: Dissolved solute (usually ions) in the solution Some non-dissolved solute (crystal)at the bottom

In a saturated solution...
There is an equilibrium situation: Solute(s)  ion+(aq) + ion–(aq) Some of the solute dissociates (dissolves into ions) and at the same time, some of the ions crystallize back into solid. Dissolved Ions Solid Solute

Solubility The solubility of a substance is the maximum amount of the substance that dissolves in a given volume of the solvent Usually given in g/L or sometimes in g/100mL For calculation purposes you should use molar solubility (mol/L) To convert g/100 mL to g/L, multiply by 10 To convert g/L to mol/L, change grams to moles using the mole formula: mass (g) Moles molar mass(g/mol)

Solids generally have a higher solubility at higher temperatures
The solubility of a substance partly depends on the temperature of the water you are dissolving it in. Standard tables give solubility at 25°C Solids generally have a higher solubility at higher temperatures Gases generally have a lower solubility at higher temperatures. There are a few exceptions to these generalizations!

What Is Ksp The solubility product constant is a number used to compare the solubilities of different solutes. The higher the Ksp, the more of the solute can be dissolved before the solution becomes saturated. Low Ksp values mean very little of the substance will dissolve. Ksp is calculated in a similar way to other equilibrium constants, but it’s a bit easier!

“Soluble” and “Insoluble” Substances
A “soluble” substance is a substance with a high Ksp value. At room temperature, a large amount of the substance can be dissolved in water. A truly insoluble substance doesn’t exist, but any substance with a very low Ksp is said to be “insoluble”, since so little of it will dissolve in water that it can barely be measured. See table 8.11 on page 423 to find out what common substance will be “soluble” and “insoluble”

General Formula for Ksp
For a solute that dissociates like this: XmYn(s)  mX+(aq) +nY-(aq) The Ksp can be calculated by this way: Ksp = [X+]m[Y-]n

Calculating Ksp Example 1: Ksp = [Ba2+][SO42-] Example 2:
BaSO4(s)  Ba2+(aq) + SO42-(aq) At equilibrium there will be a certain concentration of Ba2+ ions and SO42- ions Ksp = [Ba2+][SO42-] Example 2: CaCl2(s) Ca2+(aq) + 2Cl-(aq) At equilibrium there will be a certain concentration of Ca2+ ions and Cl- ions Ksp = [Ca2+][Cl-]2 Eliminate solid. : : Eliminate solid. : :

Dissociates equally (1:1)
: 1 Ksp = [Ba2+][SO32-] If the solubility of BaSO3 is g/L, what is the Ksp of this compound? Since it dissociates equally, the concentration of both ions will be equal to the moles of BaSO3 that dissolved ( g/L). We have to convert that to mol/L MBaSO3 = (16.0) = g/mol nBaSO3 = g/L / g/mol =1.15x10-5 mol So... [Ba2+] = [SO32-] = 1.15x10-5 mol/L Ksp = [Ba2+][SO32-] =(1.15x10-5)(1.15x10-5) = 1.32x10-10 mol2/L2

Example The solubility of silver carbonate (Ag2CO3) is 3.6x10-3 g/100mL at 25C. Calculate the value of the solubility product constant of silver carbonate. Data: Solubility = 3.6×10-3 g/100mL Molar Solubility =? [+ ions] = ? [– ions] = ? Ksp= ? we want the solubility in g/L, so multiply by 10... 3.6×10-3 g/100mL = 3.6×10-2 g/L Now we need it in mol/L, so... n= 3.6×10-2 g /L 275.8 g/mol n= 1.3×10-4 mol/L Molar solubility = 1.3x10- 4 mol/L Use solubility: m = g (per litre) M = 2(107.8)+12+3(16) Values from periodic table and formula of Ag2CO3. Carry over to the next slide.

Ksp=(2.6×10- 4 mol/L)2(1.3×10-4 mol/L) Ksp =8.8×10-12 mol3/L3
Info carried over from previous slide The molar concentration of CO3 2-(aq) in solution will equal the molar solubility! Data: Solubility = 3.6x10-3 g/100mL Molar Solubility= 1.3x10-4 mol/L [CO32-] [Ag+] Ksp=? Equation: Ag2CO3(s)  2 Ag+ + CO32- Ksp = [Ag+]2 [CO32-] [CO32-]=1.3×10- 4 mol/L [Ag+] = 2.6×10- 4 mol/L Ksp=(2.6×10- 4 mol/L)2(1.3×10-4 mol/L) Ksp =8.8×10-12 mol3/L3 : : = 1.3x10- 4 mol/L = 2.6x10- 4 mol/L [Ag+] is double [CO32-]. Multiply by 2 

At the annual chemistry Christmas party, a careless chemist spills a whole bottle of calcium fluoride, CaF2, into a 2 litre punch bowl filled with fruit punch. Most of the calcium fluoride dissolves, but a little powder settles to the bottom of the punch bowl. The Ksp of calcium fluoride is 3.4*10-11 CaF2 dissociates: CaF2(s)  Ca2+(aq) + 2F-(aq) The lethal dose of fluoride ions is 1g. Will all of the eight chemists at the party die if each drinks a cup of the punch? (1 cup ≈ 250mL or ¼ L.) Would one chemist die if he drank all the punch?

Ksp = [Ca2+] [F-]2 The dissociation formula is: CaF2(s)  Ca2+(aq) + 2F-(aq) So there will be twice as much F- produced as Ca2+ Let’s choose a variable to represent the concentration of Ca2+ ions produced at equilibrium. Double that variable to represent the F- ion concentration. Let x = [Ca2+] and 2x = [F-] Note: The units of x will be mol/L

Ksp = [Ca2+] [F-]2 Simplify x(2x)2 Reverse the equation and divide by 4 to isolate the x3 Simplify: Cube root of both sides... ...Gives us “x”, but we aren’t finished...

But x represents [Ca2+], and since the [F-] is twice as high: So to change this to grams per litre we must multiply by the molar mass of F-, which is 19.0 g/mol (note: this is not F2) A cupful (250mL) is one quarter of this, so...

Each cup contains only 0.00194 g of fluoride ions, so nobody would die from drinking one cup.
If somebody drank the whole punchbowl (2 litres = 8 cups) he would still only get g of fluoride ions. Still well below the deadly limit. The chemists would survive and have sparkling white teeth! Note: Although this dosage might not be lethal it could have long-term side effects. Never drink contaminated punch! The maximum recommended concentration of fluoride in water is 1 mg/L, about one eighth what is in the punch.

Exercises on Solubility
Page 340 # 36 to 39

Unit 2: Equilibrium Icons are used to prioritize notes in this section. Make some notes: There are SOME important items on this page that should be copied.

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Unit 2: Equilibrium Icons are used to prioritize notes in this section. Make some notes: There are SOME important items on this page that should be copied.

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