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Unit 2 equilibrium icons are used to prioritize notes in this section

Unit 2: EquilibriumIcons 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

Unit 4(formerly Module 5)

Equilibrium


Equilibrium

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.

11.0

Equilibrium


Equilibrium vs homeostasis

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

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

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

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.


Chapter 11

Chapter 11

Qualitative Aspects of Chemical Equilibrium


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?

11.1

Qualitative Aspects of Equilibrium


Static equilibrium

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

  • 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


Unit 2 equilibrium icons are used to prioritize notes in this section

  • 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 1 st type of dynamic equilibrium

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 2 nd type of dynamic equilibrium

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 3 rd type of dynamic equilibrium

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

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.

11.2

Irreversible and Reversible Reactions


Irreversible chemical reactions

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

Remember!Irreversible = One Way

Reactants  Products

Reactants can become products, but products cannot turn back into reactants.


Reversible chemical reactions

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

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

Reversibility and Equilibrium

Only reversible reactions can produce a true dynamic chemical EQUILIBRIUM


A chemical system is at equilibrium if it meets these criteria

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

Definitionsof 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.


Unit 2 equilibrium icons are used to prioritize notes in this section

  • Page 287

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


Le ch telier s principle

Henri Louis Le Châtelier (1850-1937) 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

11.4

Le Châtelier’s Principle


Equilibrium is not eternal

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

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!


  • Unit 2 equilibrium icons are used to prioritize notes in this section

    • Henri LeChâtelierstudied 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

    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.

      .


    Unit 2 equilibrium icons are used to prioritize notes in this section

    • 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

    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

    Effect of Temperature

    • Increasing the temperature will favour the endothermicreaction.

    • Decreasing the temperature will favour the exothermic reaction.


    Effect of pressure

    Effect of Pressure

    • Increasingthe 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

    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

    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

    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


    H 2 i 2 2hi

    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


    Unit 2 equilibrium icons are used to prioritize notes in this section

    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

    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


    2so 2 o 2 2so 3 heat

    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

    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.


    N 2 3 h 2 2 nh 3

    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)


    Special note re gases

    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


    Unit 2 equilibrium icons are used to prioritize notes in this section

    • Page 304, Questions 1 to 4


    Chapter 12

    Chapter 12

    The Quantitative Aspects of Equilibrium


    Chapter 121

    In this section we will explore the mathematical aspects of equilibrium, including:

    The Equilibrium Constant (Kc)

    The Equilibrium Law

    12.1

    Chapter 12


    The equilibrium constant and equilibrium law expressions

    The Equilibrium Constantand 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

    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

    aA + bB↔cC + dD

    reactantsproducts

    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…

    A Generalized Reversible Reaction with reactants and products

    Forward reaction = dir

    Reverse reaction = rev

    Products on top

    Reactants on bottom


    Simplifying the rate constant

    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

    The Equilibrium Law

    For a chemical equation of the type:

    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

    The equilibrium law expression is:

    “Products” always go on the top!

    “Reactants” always go on the bottom!


    Variants of the equilibrium constant

    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

    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 Kcvalue (>1010)

      • Reactions that don’t normally occur have Kcvalues 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

    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

    Table Showing Effects of Temperature on Equilibrium Constants

    *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

    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

    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.


    Unit 2 equilibrium icons are used to prioritize notes in this section

    • 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

    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

    12.1.4

    Calculating Equilibrium Concentrations

    The ICE method


    Unit 2 equilibrium icons are used to prioritize notes in this section

    The I.C.E. method

    • 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 concentrationΔ[A]

      • Equilibrium concentration[A]E


    Unit 2 equilibrium icons are used to prioritize notes in this section

    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

    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.

    Make a table with 3 rows labelled I, C, and E, and enough columns for all the reactants and products you are interested in

    Use the Equilibrium Concentrations to answer any further questions:

    such as finding Kc , Ka Kb or Ksp)

    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.

    Add to find the equilibrium values (adding a negative number is like subtracting)

    Look for a column that you can complete!


    Ice method step by step

    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 undissolvedsolids

    .

    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

    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


    Unit 2 equilibrium icons are used to prioritize notes in this section

    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.

    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!

    Enter the information you already know into the table

    Add to find the equilibrium values (adding a negative number is like subtracting)

    Identify the molar ratios (based on the coefficients of the equation)

    Write the chemical equation of the reaction. Show all reactants and Products

    2 NO(g) + O2(g) 2NO2(g)

    2

    1

    2

    5 mol/L

    [NO]i

    4mol/L

    i[O2]

    0mol/L

    [NO2]i

    +4 mol/L

    Δ[NO2]

    -4 mol/L

    (-ratio)

    -2 mol/L

    (-ratiot)

    4mol/L

    [NO2]e

    1mol/L

    2mol/L


    Finishing the problem

    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

    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 havenonumbers for equilibrium concentrations, we will have to use algebraic variable for some of the values.

    We can use xto represent the change in Hydrogen Concentration

    Let: Δ[H2] = -x(why negative? Because hydrogen is a reactant!)


    Unit 2 equilibrium icons are used to prioritize notes in this section

    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

    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!

    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) 2HI(g)

    1

    1

    2

    2 mol/L

    3 mol/L

    0mol/L

    -x mol/L

    -x mol/L

    +2x mol/L

    2-x mol/L

    3-xmol/L

    2xmol/L


    Continuing the problem

    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

    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

    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 0.33mol/L,

    The concentration of iodine will be 1.33mol/L and

    The concentration of hydrogen iodide will be 3.34mol/L


    Simplifying i c e tables the 5 rule

    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 – 0.0001 mol/L = 1.9999 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.


    Unit 2 equilibrium icons are used to prioritize notes in this section

    • Page 318

      • Questions #1 to 8, (Equilibrium Law and Kc)

      • Questions #19 to 21, (I.C.E. method)

      • Optional Assignments:

      • Link to another worksheet


    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

    12.2

    Acids and Bases


    Properties of acids bases

    Properties of Acids & Bases

    • Acids are solutions that:

      • Turn litmusred, 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 litmusblue, 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

    Theories of Acids and BasesHypothesis#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

    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 br nsted lowry

    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.


    Br nsted lowry and conjugates

    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

    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

    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

    Comparing the Hypotheses

    Incorrect Hypothesis

    Optional Hypothesis


    Assignment

    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

    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

    Dissociation of Water

    • In distilled water, at any given moment about 2 molecules out of every billion (0.00000002%) 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 h 3 o

    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 k a k b k w and k c

    Similarity of Ka KbKw and Kc

    • Ka is called the acidity constant

    • Kb is called the basicity constant

    • Kwis 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 k w

    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:


    Unit 2 equilibrium icons are used to prioritize notes in this section

    • 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 × 10-14 at 25°C.


    Effect of temperature on k w

    Effect of Temperature on Kw

    • At lower temperatures, Kw becomes smaller. Less water molecules dissociate when it is cold

    • At higher temperatures, Kwbecomes 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 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.


    Unit 2 equilibrium icons are used to prioritize notes in this section

    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 


    Unit 2 equilibrium icons are used to prioritize notes in this section

    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:


    Unit 2 equilibrium icons are used to prioritize notes in this section

    • 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?


    Ph h 3 o and k a 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.

    12.

    pH, [H3O+] and Ka of Acids


    Measures of an acid

    Measures of an Acid

    • There are three different ways of measuring acids:

      • The natural strengthKa

      • The concentrationmol/L

      • The degree of AciditypH


    Natural strength of an acid

    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” Acids and “Weak” AcidsAcids shown in red, Conjugate Base in blue


    Concentration of an acid

    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

    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 effectivestrength of an acid,

    based on its [H3O+] concentration and

    Usually recorded as itspH


    Degree of acidity

    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

    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 SørenSørensen 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.


    Ph and poh

    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

    pH and Concentration of H+ ionsFor solutions of an exact pH you may use the chart below:

    For simplicity I used [H+] instead of [H3O+]

    Acids

    Bases

    *Concentration in mol/L

    Temperature = 25°C


    What about in between ph values

    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.


    Unit 2 equilibrium icons are used to prioritize notes in this section

    Question: Find the pH if the H3O+ concentration is 1.40x10-8mol/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 10-5mol/L(I’ve rounded to 3 Sig.Fig.)


    Using the windows calculator

    Using the Windows Calculator

    7.85…

    • Switch to scientific view

      To find pH, use: 1.4 Exp 8 =

      To find [H3O+] use: 4.30 =

      or: 10 4.30 =

    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

    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

    +/-

    7.85...

    EE

    +/-

    log

    +/-

    Question: Find the H+ concentration if the pH is 4.30

    Solution: [H+] = 10 - 4.30

    Try:10 yx 4.30 +/- =

    Or:4.30 +/- Inv log =

    Or4.30 +/- 2ndF 10x =

    Or: 10 ^ (-) 4.3

    5.01... ×10-5


    Unit 2 equilibrium icons are used to prioritize notes in this section

    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/LB) [OH-]=1.0x10-3 mol/L

    • C) [H+]=1.0x10-5mol/L D) [OH-]=1.0x10-9mol/L

      3.Find the [H+] of these solutions using the 10x or antilog function of your calculator: [H+]=10-pH

    • A) pH=3.7 B) pH=9.8 C) pH=6.2 D) pH=4.0 E) pH=7.1

      4.Find the pH of the following solutions using the log function of your calculator: pH= - log [H+].

    • A) [H+]=4.5x10-3mol/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

    Problem 5.

    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


    Unit 2 equilibrium icons are used to prioritize notes in this section

    Solutions to Problems 1- 4

    1.Answers

    • A) 1x10-4, 1x10-10 mol/LB) 1x10-12, 1x10-2 mol/L

    • C) 1x10-9, 1x10-5mol/L D) 1x10-7, 1x10-7mol/L

    • E) 1x10-8, 1x10-6mol/L

      2.Answers:

    • A) pH=8B) 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) 7.9x10-8mol/L

      4.Answers: pH= - log [H+].

    • A) pH= 2.3 B) pH=7.5

    • C) pH=6.5 D) pH=1.6


    Solution to problem 5

    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+] = 10 -5.5

      • [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


    Unit 2 equilibrium icons are used to prioritize notes in this section

    HA  H+ + A-

    Round to 2 Sig. Digits

    1 : 1 : 1

    -3.163x10-6

    +3.163x10-6

    +3.163x10-6

    3.163x10-6

    9.9997x10-2

    Ka = [H+][A-] = (3.163x10-6x3.163x10-6) mol/L=1.0004 x10-10

    [HA] 9.9997x10-2 mol/L

    ≈ 1.0x10-10


    The short cut an alternative solution using 5 rule

    The Short-cut:An alternative solution using 5% rule

    Ka = [H+][A-] = (3.163x10-6x3.163x10-6) mol/L=1.0004 x10-10

    [HA]mol/L

    = 1.0x10-10

    Since 3.163 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 – 0.000003163) and using an ICE table

    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


    A half evil example

    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

    Assignments

    Page 339, Questions 4 to 23


    Unit 4 chapter 12 2 5 solubility product constant

    Unit 4: Chapter 12.2.5Solubility Product Constant

    Solubility

    Calculating the solubility constant

    Examples


    Solubility

    Solubility

    • A saturated solution contains:

      • Dissolved solute (usually ions) in the solution

      • Some non-dissolved solute (crystal)at the bottom

    Dissolved ions 

    Solid solute crystals


    In a saturated solution

    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


    Solubility1

    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)


    Unit 2 equilibrium icons are used to prioritize notes in this section

    • 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 k sp

    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

    “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 k sp

    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 k sp

    Calculating Ksp

    Eliminate solid.

    1 : 1 : 1

    • Example 1:

    • 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.

    1 : 1 : 2


    Unit 2 equilibrium icons are used to prioritize notes in this section

    Dissociates equally (1:1)

    1 :

    1

    • Ksp = [Ba2+][SO32-]

      • If the solubility of BaSO3 is 0.0025g/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 (0.0025 g/L).

      • We have to convert that to mol/L

      • MBaSO3 = 137.3 + 32.1 + 3(16.0) = 217.4 g/mol

      • nBaSO3 = 0.0025 g/L/ 217.4 g/mol=1.15x10-5mol

      • So... [Ba2+] = [SO32-] = 1.15x10-5mol/L

      • Ksp = [Ba2+][SO32-]

        =(1.15x10-5)(1.15x10-5) = 1.32x10-10 mol2/L2


    Example1

    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 = 3.6×10-3g/100mL

    Molar Solubility =?

    [+ ions] = ?

    [– ions] = ?

    Ksp= ?

    we want the solubility in g/L, so multiply by 10...

    3.6×10-3g/100mL = 3.6×10-2g/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.


    Unit 2 equilibrium icons are used to prioritize notes in this section

    Info carried over from previous slide

    The molar concentration of CO32-(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- 4mol/L

    [Ag+]= 2.6×10- 4mol/L

    Ksp=(2.6×10- 4 mol/L)2(1.3×10-4 mol/L)

    Ksp =8.8×10-12 mol3/L3

    1 : 2 : 1

    =1.3x10- 4 mol/L

    =2.6x10- 4 mol/L

    [Ag+] is double [CO32-]. Multiply by 2 


    Unit 2 equilibrium icons are used to prioritize notes in this section

    • 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?


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    • 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


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    • 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...


    Unit 2 equilibrium icons are used to prioritize notes in this section

    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...


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    • 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 0.0155 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

    Exercises on Solubility

    • Page 340 # 36 to 39


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