handout powerpoint chem 301 pharchm1

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REVIEW SESSIONSANALYTICAL CHEMISTRY/ INORGANIC PHARMACEUTICAL CHEMISTRY (CHEM 301/ PHARCHM1)

Prepared by:Chemistry Review Group/ Pharmacy Review GroupScholia Tutorial ClubUST Faculty of Pharmacy

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PREREQUISITE

Chemistry - study of the property of matter and the changes it undergoes.

Topics in Chemistry which involved studying the property of matter are:▫ Classification▫ Energetics and Thermodynamics▫ Electrochemistry▫ Chemical and physical bonds▫ Molecular geometry▫ Solutions and solubility

Changes in matter are otherwise termed as chemical reactions.

© Scholia Tutorial Club, University of Santo Tomas Faculty of Pharmacy

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PREREQUISITE

•Qualitative Chemistry – branch of chemistry that deals with the identification of what matter is by examination of its properties and changes undergone.

•Quantitative Chemistry – branch of chemistry that deals with the identification of how much matter there is by examination of its properties and changes undergone


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PREREQUISITE

•Qualitative and Quantitative Chemistries are combined into what is known as Analytical Chemistry

•Pharmaceutical Chemistry – branch of pharmacy and chemistry that deals with the use of the properties and changes of significant compounds to bring out a certain use in the pharmacy practice.


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PREREQUISITE

•Concentration Expressions•Recall that the concentration (amount of

solute present) in a solution can be expressed in the following ways:

Molarity = g solute/ mm solute (L solvent)Molality = g solute/ mm solute (kg

solvent)


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PREREQUISITE

Normality = Equivalents solute/ L solvent

Also, dilution is important in case the concentration is meant to be altered.

M1V1 = M2V2


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REVIEW PROPER #1: COLLIGATIVE PROPERTIES OF SOLUTIONS•Colligative – adj. – dependent on

quantity instead of nature•Colligative properties are specific for

solutions, which depend on the number of solute particles instead of their nature.


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REVIEW PROPER #1: COLLIGATIVE PROPERTIES OF SOLUTIONS

1. Vapor Pressure – the pressure exerted by the vapor being given off by the liquid at a given temperatureRule: The lowering of vapor pressure of a solvent is equal to the product of mole fraction solute and vapor pressure of pure solvent.

ΔP = P⁰ x MF solute


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REVIEW PROPER #1: COLLIGATIVE PROPERTIES OF SOLUTIONS• Ex. A solution contains 1.75 moles of sucrose

in a liter of water. What is the vapor pressure lowering of the solvent? What is the vapor pressure of the solution? P⁰ = 23.72⁰ C

ΔP = P⁰ x MF soluteWhere : ΔP = vapor pressure lowering; and

P⁰ = vapor pressure of solventΔP = P (Vp solution) – P ⁰ (Vp solvent)

MF solute = moles solute/ moles solution* moles solution = moles solute + solvent© Scholia Tutorial Club, University of Santo Tomas Faculty of Pharmacy

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Moles solvent = 100g/ 18 = 55.56 moles water

Moles solution = 57.31MF solute = 1.75/57.31 = 0.031ΔP = 23.72(0.031) = 0.727 mmHg

Vapor pressure soln = P⁰ - lowering= 23.72 – 0.727 = 23.0 mmHg


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REVIEW PROPER #1: COLLIGATIVE PROPERTIES OF SOLUTIONS• 2. Boiling Point Elevation – the pressure

exerted by the vapor being given off by the liquid at a given temperature

• Rule: The boiling point elevation of the solvent is directly proportional to the molecular concentration of the solution.

• ΔTb = kb(m)Where kb = molal bpt constant

m = molalityΔTb = bpt elevation

• ΔTb = Tb (Bpt. Solution) – Tb ⁰ (Bpt. Solvent)


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Ex. What is the boiling point elevation of 60.0g of water that contains 2.00g of glycerol, C3H5(OH)3? Molality = 0.362, Bpt constant of water = 0.51⁰CΔTb = kb(m)ΔTb = (0.51⁰C)(0.362) = 0.185⁰CThus, if bpt solvent is 100⁰C, the bpt of

the entire solution becomes 100.185⁰C.


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REVIEW PROPER #1: COLLIGATIVE PROPERTIES OF SOLUTIONS3. Freezing Point Depression – point where

liquid turns into solid.Rule: The freezing point depression is directly

proportional to the molal concentration of the solution.

ΔTf = kf(m)Where Tf = Fpt. Depression

m = molalitykf = molal fpt constant

ΔTf = Tf ⁰ (Fpt. solvent) - Tf (Fpt. Solution) © Scholia Tutorial Club, University of Santo Tomas Faculty of Pharmacy

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REVIEW PROPER #1: COLLIGATIVE PROPERTIES OF SOLUTIONS•Ex. Calculate the freezing point

depression of 0.46 molal diphenyl-naphthalene solution in 200g of benzene. Fpt = 5.5°C Constant = 5.12°C

ΔT = (5.12°C)(0.46m) = 2.3°CT = Fpt - ΔT = 5.5°C - 2.3°C= 3.2°C


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4. Increase in Osmotic Pressure (π) – pressure that is required to stop osmosis

π = MRTWhere M = molarity

R = gas constantT = temperature (K)


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REVIEW PROPER #1: COLLIGATIVE PROPERTIES OF SOLUTIONS•Ex: What is the osmotic pressure of a

selection at 29.0⁰C that contains 15.0g of glucose in 80.0mL solution?

π = MRT= [150g / (180g/n x 0.08L)] x 0.0821L-

atm/mol-K x 302Kπ = 25.8 atm


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END OF REVIEW PROPER.


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REVIEW PROPER #2: CHEMICAL KINETICSA. RATE: DEFINITION AND

DETERMINATION•Rate – any quantity divided by time. (Q/t)•Rate of Reaction mathematically is

defined in a way that the quantity given is the amount of compound (product or reactant) over time.

•Reactants are used up, and become lesser overtime. Therefore their rate should have a negative sign.© Scholia Tutorial Club, University of Santo Tomas Faculty of Pharmacy

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REVIEW PROPER #2: CHEMICAL KINETICS•Products are formed, and become greater

overtime. Therefore their rate should have a positive sign. This sensible argument is justified by the basic law of conservation of mass.

•Q = Amount of reactant (usually expressed in concentrations as molar [M])


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REVIEW PROPER #2: CHEMICAL KINETICS

(If we use the reactionA + B ---> AB, and use AB as Q):

Rate of reaction of product = Q/t

= AB(in M) / t(in S)* SINCE AB IS A PRODUCT, ITS RATE SHOULD

HAVE A POSITIVE SIGN.Therefore, if we get the rate of any product (either

A or B),Rate of reaction of reactant = Q/t

= (-) A or B(in M) / t(in S)


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REVIEW PROPER #2: CHEMICAL KINETICSExercise:NO2 is produced at a rate of 0.0072 M/s.

2N2O5 ---> 4NO2 + O2

Find rate of oxygen production.Find rate of N2O5 usage.


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REVIEW PROPER #2: CHEMICAL KINETICSB. CONCEPTS AND CORRESPONDING

FACTORS AFFECTING RATE I. Rate Law: Factor of ConcentrationThe Rate Law is a core of Chemical kinetics

simply because it is the single most essential concept that explains the speed of the feasible reaction.

Concentration – the quantity of matter in a given mixture. © Scholia Tutorial Club, University of Santo Tomas Faculty of Pharmacy

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REVIEW PROPER #2: CHEMICAL KINETICS•Rate Law – an equation that relates the

rate of reaction to concentration of reactants.

•Concentration – since the rate law depends on the concentration of reactants, a change on reactant changes the rate law and the rate law constant.


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REVIEW PROPER #2: CHEMICAL KINETICS• The rate law can be described by its order.• The reaction has a order, based on the units

of reactants involved in the rate law. Individual reactants have their own order (first order, second order) and the overall rate law has also an order (the combination of all individual orders)

Ex. Rate = k[A]2[B]Individual orders: 2nd order for A; 1st order for

BOverall order: 3rd order© Scholia Tutorial Club, University of Santo Tomas Faculty of Pharmacy

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REVIEW PROPER #2: CHEMICAL KINETICSExercise: Given the following:NO + H2 -> products

Give the rate law, orders of reactants, overall order, constant k.

Solve for the rate if NO = 0.031M, andH2 = 1.2 M


Trial NO [M] H2[M] Rate [m/s]

1 5 x 10-3 2 x 10-3 1.2 x 10-5

2 10 x 10-3 2 x 10-3 5 x 10-5

3 10 x 10-3 4 x 10-3 10 x 10-5

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REVIEW PROPER #2: CHEMICAL KINETICSII/III. Collision Theory: Factors of

Temperature and OrientationThe Collision Theory is as important as Rate Law because it tells about the possibility of the feasible reaction.Temperature – amount of thermal energy possessed by a body of matter.Orientation – position of two or more interacting bodies of matter.


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REVIEW PROPER #2: CHEMICAL KINETICSCollision Theory – states that two bodies of

matter must make contact to each other in order to have a chance of making a chemical reaction happen.

- Not all collisions are effective; thus, ineffective collisions do not induce a chemical reactions (Although a physical one may take place).

- The factors that decide whether the collision is effective are the two factors under this theory:


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REVIEW PROPER #2: CHEMICAL KINETICS1. Temperature – enough (thermal) energy

is needed upon collision to favor the formation of new products. This required energy to make the chemical reaction possible is called the energy of activation. (EA)


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REVIEW PROPER #2: CHEMICAL KINETICS2. Orientation – colliding particles must

be properly oriented in order to amplify the attraction between each other, making more possible the achievement of EA.


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REVIEW PROPER #2: CHEMICAL KINETICSIV. Catalysis: Factor of CatalystsA catalyst is a substance that helps in increasing

the rate of reaction. The following are characteristic of most catalysts:

1. A catalyst speeds rate of reactions thru lowering of the activation energy.

2. It is not used up in the reaction.3. It does not affect chemical equilibrium (refer to

Proper #3)4. It usually acts by physical forces, forming a

temporary complex with the target reactant to stabilize transition state.


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REVIEW PROPER #2: CHEMICAL KINETICS•Catalysts can be generally categorized

into two: Heterogenous and Homogenous. Heterogenous catalysts exist in a phase different from that of the reactant (metal catalyst in gas production), while homogenous catalysts exist in a phase same with that of the reactant (liquid catalyst in aqueous/liquid reactant).


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REVIEW PROPER #2: CHEMICAL KINETICS•Special catalysts called enzymes are the

most efficient ever known, existing within biological systems such as that of the human body. The reactants on which these catalysts act upon are called substrates.


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REVIEW PROPER #2: CHEMICAL KINETICS•C. REACTION MECHANISMS:

LOGICAL LEVEL OF REACTION STUDIES

•Reaction mechanisms explain specifically what happens to reactants in a multi-step reaction until they become products.


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REVIEW PROPER #2: CHEMICAL KINETICS•Chemical kinetics covers reaction

mechanisms because the study of reaction rates is not the same in multi-step reactions compared to that of a simpler single-step reaction.


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REVIEW PROPER #2: CHEMICAL KINETICSAnatomy of a simple multi-step (2-step)

reaction:A + B ---> C + DD + E ----> F + BA + E ---> C + F

A, E = reactants D = intermediate

F, C = products B = catalyst


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REVIEW PROPER #2: CHEMICAL KINETICSIntermediate – a product in one step of the reaction,

but not a product in the overall reaction. A product not seen in the final product side.

ex. ATP and NADPH in photosynthesis Catalyst – in reaction mechanisms, is seen as a

reactant in one step that does not become used, and thus is seen at the final product side. A reactant not seen in the final reactant side. ex. Concentrated sulfuric acid, Palldium

NOTE: Catalysts and intermediates cannot participate in the rate law as they did not participate per se in the reaction.


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REVIEW PROPER #2: CHEMICAL KINETICSII. Reaction mechanisms and the Rate

LawThe integration of chemical kinetics and

reaction mechanisms relies on three general rules:

The individual steps must add up to the overall balanced equation.

The individual steps are bound to all rules set up by chemical kinetic principles; and

The reaction mechanism is derived from the rate law, not the other way around.


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REVIEW PROPER #2: CHEMICAL KINETICSIndividual steps have individual ratesNot all steps of a multi-step reaction have

significant duration. A two-step reaction might have a 0.001 second step, followed by a 5 second step. Definitely, the second step in the example is far more useful in determining the kinetics of the reaction.


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REVIEW PROPER #2: CHEMICAL KINETICSRate-determining step (RDS) – the slower/slowest

individual step of a reaction mechanism. It can be concretized through the rate law of the multi-step reaction. Thus, the concentration of reactants in the RDS are the only ones considered in the entire reaction’s rate law.

ex.NO + O2 <----> NO3 (fast)

NO3 + NO -----> 2NO2 (slow)

Because the second reaction is the slower reaction and

thus the RDS, the rate law will be:Rate = k [NO]


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REVIEW PROPER #2: CHEMICAL KINETICSExercises: 1. Give the rate law for the following

reaction mechanism:N2O -----> N2 + O (slow)

N2O + O ----->N2 + O2

2. Give the rate law for the following reaction mechanism:

A + B <--> ABAB --->A + C (slow)


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REVIEW PROPER #3: INTRODUCTION TO EQUILIBRIUMA. DEFINITION•Equilibrium - state wherein the amount

of products and reactants stay the same in an equal reaction process.

•Unless products can return to the reactant compounds, equilibrium cannot happen because it only happens on reversible reactions. (Double arrow <-----> is used to denote a reversible reaction). © Scholia Tutorial Club, University of Santo Tomas Faculty of Pharmacy

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REVIEW PROPER #3: INTRODUCTION TO EQUILIBRIUM• Equilibrium on reversible reactions are achieved

only when NET production of either products or reactants is zero. Therefore, there will be a flat graph of concentration vs. time on an energy diagram for reactions in equilibrium (Kinetics, Rate). If the reaction is returned to progress, there will be changes to maintain equilibrium and the graph will deviate temporarily from a flat graph.

• Mathematical/quantitative study of equilibrium depends highly on equilibrium constants.


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REVIEW PROPER #3: INTRODUCTION TO EQUILIBRIUM


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REVIEW PROPER #3: INTRODUCTION TO EQUILIBRIUMB. LAW OF MASS ACTION AND THE

EQUILIBRIUM-CONSTANT

Law of Mass Action – states that at a given temperature, a chemical reaction will have a stable ratio of products and reactants by determining the concentrations (l) or partial pressures (g) of the reactant-product mixture. The numerical representation of the ratio between products formed and reactants retained is shown by the equilibrium constant Keq.


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Keq can give us ideas regarding the direction of the reaction in equilibrium.

If the Keq is high, the reaction will favor formation of products.

If the Keq is low, the reaction will not favor formation of reactants.


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REVIEW PROPER #3: INTRODUCTION TO EQUILIBRIUMNotes:1. The coefficient of the equation will serve as the

exponent for the moles of the given species.Ex. If 2A + B <-> 2C, then Keq = [C]2 /[A]2[B]

2. The equilibrium constant is stoichiometric in nature; it does not aim to discuss reactions charted against time or against reaction mechanisms. It aims to measure, not to dissect reaction.

3. Pure substances (pure solids and liquids) have a concentration of 1, and are not included in the constant expression. Reactions with such species are called as heterogeneous.Ex. If 2A(g) + B(s) <-> 2C(g), then Keq = [C]2 /[A]2


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REVIEW PROPER #3: INTRODUCTION TO EQUILIBRIUMC. MANIPULATION OF KeqThe Keq can be mathematically manipulated depending on

the changes in the given reaction. For example, we use the given original equation:

2SO2 + O2 <--> 2SO3 Keq = 0.15

1. Backward reaction = reciprocal of Keq

2SO3 <--> 2SO2 + O2 Keq = 1/0.15 = 6.67

2. Division of equation into two = Square root of KeqSO2 + 1/2 O2 <--> SO3 Keq = √0.15 = 0.39

3. Multiplication of two = Square of Keq4SO2 + 2O2 <--> 4SO3 Keq = (0.15)2 = 0.023


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REVIEW PROPER #3: INTRODUCTION TO EQUILIBRIUMExercises:1. If the equilibrium constant for the formation of

ethenol from acetaldehyde is 0.000006, what will the constant be when acetaldehyde becomes the product?

2. If twice the amount of 2mL formic acid turning into formate has the constant of 0.000177, what is the constant of 2mL formic acid turning into formate?


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REVIEW PROPER #3: INTRODUCTION TO EQUILIBRIUMD. LE-CHATELIER’S PRINCIPLE

- states that a disturbance in equilibrium (producing a significant net amount of product or reactant away from equilibrium) will produce a progress in reaction that will return the reaction back to equilibrium.


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REVIEW PROPER #3: INTRODUCTION TO EQUILIBRIUMConcentration – changes in one side of the

reaction will cause the inverse change in the other side.

ex. 3A + 2C <-> 2BIncrease in A will shift the reaction towards

the right.Increase in B will shift the reaction towards

the left.Decrease in C will shift the reaction towards

the left.© Scholia Tutorial Club, University of Santo Tomas Faculty of Pharmacy

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REVIEW PROPER #3: INTRODUCTION TO EQUILIBRIUMTemperature – the nature of the reaction should

be first known. If it is exothermic, the heat will be on the product side. If it is endothermic, the heat will be on the reactant side. Thus, any increase in temperature will shift the reaction toward the opposite side of the reaction, and any decrease in temperature will shift the reaction toward its side of the reaction.


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Ex. 3AC <-> 3A + C + heat

Increase in temperature will shift the reaction towards the left, and vice-versa.

3AC + heat <-> 3A + CIncrease in temperature will shift the

reaction towards the right, and vice-versa.


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REVIEW PROPER #3: INTRODUCTION TO EQUILIBRIUMVolume and Pressure – factor applicable

only to gases. The more numerous between reactants in products will have more volume. Thus, increasing pressure will decrease the volume on the greater side and causes a shift to that side; and vice-versa.


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Ex. 2A + 5C <-> 6FReactants = 7 moles; Products = 6 moles

If pressure is applied, the reaction will shift towards the left. If pressure is relieved, the reaction will shift towards the right.


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REVIEW PROPER #3: INTRODUCTION TO EQUILIBRIUMCatalysts and Inert Gases – catalysts do

not affect equilibrium. Recall that catalysts lower EA in a progressing reaction, but a reaction in equilibrium is already done with progression. (refer to introduction to equilibrium). Inert gases and anything that is not used up or produced in the reaction will not affect the equilibrium as well.


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REVIEW PROPER #3: INTRODUCTION TO EQUILIBRIUMExercises:1. In the human lungs, carbon dioxide is quickly turned

into carbonic acid. If a person inhales twice the amount of carbon dioxide, what will happen to the carbonic acid content of the body?

2. The exothermic conversion of magnesium carbonate to magnesium oxide takes place extremely slowly in room temperature. If the heat is increased in the equilibrated mixture, what will happen?

3. Nitrogen and hydrogen gases react to form ammonia. If two moles of ammonia are produced from every three moles of hydrogen and one mole of nitrogen in a closed system, what will happen if pressure is applied to the mixture? Vice-versa? © Scholia Tutorial Club, University of Santo Tomas Faculty of Pharmacy

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REVIEW PROPER #3: INTRODUCTION TO EQUILIBRIUME. REACTIONS NEARING

EQUILIBRIUMReactions that will proceed to equilibrium but are not et there can be represented b finding the product of the concentration/partial pressures of the products formed so far. That product is called the Reaction quotient (Qc)


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REVIEW PROPER #3: INTRODUCTION TO EQUILIBRIUMSeveral inferences can be made, using the basis of

Le Chatelier’s principle on concentration:

• If Qc > Keq, the reaction has too many products, and will shift the equilibrium to the left (towards

reactants)• If Qc < Keq, the reaction has too many reactants,

and will shift the equilibrium to the right (towards products)

• Qc = Keq when the reaction is in equilibrium


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REVIEW PROPER #3: INTRODUCTION TO EQUILIBRIUM•Exercise: Given that the Keq for a

reaction is 21, give the inference when the products A and B are both 5.3M while the only reactant is 1.85M.


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REVIEW PROPER #4: SPECIFIC EQUILIBRIUMCan be divided into two sub-propers:1. Water, Acid, and Base Equilibria

(Kw, Ka, Kb, pH, pOH, their antilogs, %I)with common-ion effect and buffers

2. Solubility Equilibria(Ksp) with common-ion effect and the new ion product


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WATER, ACIDS, AND BASES

A. WATER, ACIDS, AND BASES: STRENGTH I. Review and Introductiona. Definitions

Arrhenius – acids produce hydrogen ions in solution, while bases produce hydroxide ions in solution.

Bronsted-Lowry – acids donate protons, and bases receive protons.

Lewis – acids receive electrons, and bases donate electrons.


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b. Strength of acids and bases: pH/pOHAcids and bases have a certain strength,

chemically on their ability to donate and accept electrons, respectively.

By visual observation, we can determine

the strength of an acid by its ability to leave out a proton/hydrogen ion and of a base by its ability to attract a proton/hydrogen. © Scholia Tutorial Club, University of Santo Tomas Faculty of Pharmacy

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Mathematically, we can calculate acid or base strength by a numerical value of hydrogen or hydroxide ions released per amount of substance by pH and pOH In a solution respectively.

pH = -log [H+]pOH = -log [OH-]


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Conversely, if we are given the pH or pOH of the certain acid or base, we can derive the [H+] or [OH-] by using the antilog function.

[H+] = antilog(-pH)[OH-] = antilog(-pOH)

Diversity of the acids and bases allows them to be separated based on their strength. There are the strong acids and bases, and there are

the weak.© Scholia Tutorial Club, University of Santo Tomas Faculty of Pharmacy

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Very important note: Strong acids and bases completely dissociate in water. All molecules of the acid/base are disintegrated into its constituent ions. Weak acids and bases dissociate only partially in water, and thus some of the acid/base still remains: therefore, equilibrium exists in weak acids/bases. Consequently, equilibrium constants for these weak acids/bases exist.


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Exercises:1. If the pH of an acid is 4.7, what is its

[H+]?2. If the [OH-] of a base is 3.2 x 10-2, what is

its pOH?3. What is the hydrogen ion concentration of

an impure water sample with a pH of 10.48?

4. If the negative logarithm of the hydroxide ion concentration, of an acid is 11.23, find [OH-]


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c. Conjugate acids and basesConjugate – adj. – joined together; coupled.A conjugate acid is the acid minus a

proton. (ex. H2CO3 is an acid; HCO3- is its conjugate acid). Inversely related, a conjugate base is a base plus a proton. (NH3 is a base; NH4+ is its conjugate base). Conjugates result from acid/base equilibrium reactions.


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Exercises:1. What is the conjugate base of ammonia?2. In the reaction KHSO4 + K2CO3 <-->

K2SO4 + KHCO3, which is the acid reactant? Its conjugate base. The base reactant? Its conjugate acid?


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II.kW: Autoionization of Water and EquilibriumPure water is almost stable and non-reactive, especially strongly bonded together by hydrogen bonds. However, their occasional disturbances in charge due to induced dipoles allow them to produce ions within themselves (Thus, auto- or self-ionization).


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The reaction is as follows:

H2O + H2O <-> H3O+ + OH-

The Keq of water by 25 deg. C is 1.0 x 10-14. This is

also called the ion-product constant for water (kW), Thus, we can get the concentration of ions produced.

H+ = 1.00 x 10-7; pH = 7OH- = 1.00 x 10-7; pOH = 7

Thus, pH and pOH of water is both 7. This is called the neutral pH.


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WATER, ACIDS, AND BASESIII. Ka and Kb: Acid/Base Dissociation Constants a. Acid Dissociation and Base Dissociation

constantsThe acid dissociation constant (Ka), like any

equilibrium constant, is solved by dividing products by reactants. Specifically, the product acid dissociation is the conjugate acid, while the reactant is obviously the original acid.

Ka = [A-][H+][HA]


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The base dissociation constant (Kb) can similarly be calculated as follows:

Kb = [HB][OH-][B-]


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Because weak acids and bases release only a small amount of hydrogen or hydroxide ions respectively, the term percent ionization (%I) is used to compare how much ions are released from the original acid/base.

%I = [H+] or [OH-]

[HA] or [B-]© Scholia Tutorial Club, University of Santo Tomas Faculty of Pharmacy

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Exercises (assume all acids and bases are monoprotic):

1. The Ka of phosphoric acid at is 7.2 x 10-3. Find its pKa.

2. The Kb of methylamine is 4.4 x 10-4. Find its pKb.

3. If a 0.19M base has a pH of 10.88, find its Kb. Find %I.

4. If the hydrogen ion concentration of a 0.77M weak acid is 1.32 x 10-4, what is its pKa?


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b. Dissociation constants and Ion concentrations: Ka, Kb, Kw, [H+], [OH-] and their negative logarithmsBecause the conjugate acid or base in the ionization of an acid or base is accompanied with either a hydrogen ion or a hydroxide ion, by getting the Ka or Kb, we can get the molarity of either H+ or OH- produced, and ultimately the pH and pOH.


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WATER, ACIDS, AND BASESExercises (assume all acids and bases are monoprotic):1. If the Kb of Diethylamine is 8.6 x 10-4, what is its pKb?2. Ascorbic acid has a pKa of 4.10. What is its Ka? 3. Phenol, an acidic compound with a Ka of 1.3 x 10-10, should

have a Kb of what?4. If formic acid has a pKa of 3.77, find its Kb and %I [H+],

considering it is 1M.5. A 0.75M monoprotic acid has a [H+] of 2.3 x 10-6. Find its Ka.6. Given that urea has a Kb of 1.5 x 10-14, find its pKa.7. Ethylamine has a Kb of 5.6 x 10-4. Find its pOH.8. Benzoic acid has a Ka of 6.5 x 10-5. Find its hydroxide ion

concentration and pOH, considering it is 1M.9. A 0.35M base has a pOH of 4.24. Find its OH- concentration,

Kb, pKb, pKa, H+ concentration, pH and %I [OH-].© Scholia Tutorial Club, University of Santo Tomas Faculty of Pharmacy

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IV. Common-ion effect on Acid-Base EquilibriaIn pure weak acid solution (consider an acid first), the Ka will be equal to the product of the hydrogen and conjugate base. If more conjugate base is added, the equilibrium will have to shift to the direction of the acid, because of Le Chatelier’s principle.


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•Ex. A 0.5M acetic acid mixed with 0.1M sodium acetate should have what difference in pH compared to that of the acid alone?

1.8 x 10-5 = x2/0.5 – x*remove x if the constant is less than 10-4

= 3 x 10-3M H+, pH is 2.5 for acetic acid alone1.8 x 10-5 = x(0.1)/0.5= 9 x 10-5, pH if 4.04

There is 1.54 units in pH difference.


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Exercises:1. A 0.25M acetic acid mixed with 0.1M of

its conjugate base has how much difference in terms of %I to 0.25M acetic acid alone? Ka = 1.8 x 10-5

2. If a 0.12M acid solution contains 0.032M of its conjugate base, what is the difference in terms of pOH to 0.12M acid solution alone? pKa = 9.22


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V. Buffers and the Henderson-Hasselbach FormulaBuffers are systems that limit drastic changes in pH by absorbing produced free hydrogen or hydroxide ions within themselves. This is achieved by mixing an acid with its conjugate base. These two buffer components will then have a specific pH different to the pH of each individual component. Difference in pH of the buffer to its components is key to finding the concentration of buffer when acid concentration is given and vice-versa.© Scholia Tutorial Club, University of Santo Tomas Faculty of Pharmacy

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The Henderson-Hasselbach equation can be used to determine buffer component ratios, utilizing the differences in pH of buffer and its component acid/base.

pH = pKa + log[A-]

[HA]


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Exercises:1. What is the pH of a 0.2M NH3/NH4CL buffer? Kb

NH3 = 1.8 x 10-5

2. Find the ratio of an acetic acid/sodium acetate if the pH is 4.63. (pKa acetic acid = 4.74). If 0.25M acetic acid was used, how many moles of sodium acetate was used?

3. If the the pH of a secondary phosphate buffer is 12.34, what is the ratio of acid to conjugate base? (pKa secondary phosphate = 12.38)

4. A lactic acid buffer contains equal amounts of acid and conjugate base. If its Ka is 1.4 x 10-4, what is its pH?


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END OF REVIEW SUB-PROPER.


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IONS AND PRECIPITATES

i. Definition and conceptsA solution is a homogenous mixture that consists of a solute and solvent. The process by which the solute successfully dissolves in solvent is solvation.

Solubility is highly dependent on polarity and energy. Polarity is the chemical basis of solubility that uses the dipoles of the solute and solvent to predict whether solvation will take place or not. Energy is important since even in physical interactions, enough energy must be used to overcome the intermolecular forces between like particles.


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II. ksp: Solubility and Equilibrium

Much like equilibrium constants, the same formula goes to calculating the constant for solutions. Specifically, the product is/are the ions, while the reactant is the solute.

Ksp = Product of ions


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Recall that in heterogenous equilibrium, solid products have a concentration equivalent of 1. Thus, if we add the solute, the denominator would simply be 1.

NOTE: If an ion has a coefficient, such will be used as both coefficient and exponent for the particular ion.Ex. AB <-> A + 2B, then Ksp = [A][2B]2


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Exercises:1. AB dissolves into equal portions of A and

B ions. Find the Ksp if each ion has a concentration of 5.86 x 10-7 M.

2. If cupric hydroxide dissociates into one cupric ion and two hydroxide ions, and its ksp is 2.2 x 10-20, then what is the concentration of cuprous and hydroxide ions?


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III. Molar Solubility and Solubility

Molar solubility is defined as the number of moles of salt that a liter of solution can handle before it becomes saturated. (n/L or M)

Solubility is considered as the amount of grams of ions produced that can saturate a liter of solution. © Scholia Tutorial Club, University of Santo Tomas Faculty of Pharmacy

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Ex. Find molar solubility, ksp and solubility of AgBr if this salt saturates a liter of solution with 1.65 x 10-4 grams. MM = 187.8 g/n

Answers:Molar solubility = 8.8 x 10-7 M AgBrKsp = 7.7 x 10-13

Solubility = 1.65 x 10-4 g/L Ag+ and Br-© Scholia Tutorial Club, University of Santo Tomas Faculty of Pharmacy

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Exercises:1. If barium sulfate dissociates into barium

and sulfate ions, find molar solubility and solubility if the given mass to saturate the solution is 2.5 x 10-3 g per liter. MM = 233 g/n

2. Use the exercise problem #2 for Ksp to find molar solubility and solubility. MM = 97.5 g/n


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IV. Common-ion effectIn a salt solution that is saturated, the ksp will be equal to the product of the ions produced. If one of the two ions (usually the anion) is added, the equilibrium will have to shift to the direction of the salt, because of Le Chatelier’s principle. The other ion would have to change quantity to retain the ksp.


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Exercise: Given that the solution will form a precipitate, the Ksp values for AgI, AgBr and AgCl are 1.6 x 10-10, 7.7 x 10-12, and 8.3 x 10-17, respectively. If silver ions are slowly added to the solution containining 0.05n of all these anions, which will precipitate first? Which has the greatest molar solubility?


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V. Precipitate and solubility predictionIn a salt solution that is saturated, the ksp will be equal to the product of the ions produced. If two other salts with the cation and anion components are mixed together, their resultant ions may be measured and a new constant may be derived. This is usually denoted as Q. Because the target salt would have its own ksp, Q and ksp can be compared:


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•If Q > Ksp, precipitation, supersaturation occurs and will proceed the reduction of products.

•If Q = Ksp, no precipitation, equilibrium is achieved and reaction progress stops.

•If Q < Ksp, no precipitation, unsaturation occurs and will proceed the addition of products.


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Ex. If 0.01M NaOH is added to a liter of 0.1M CuCl2, will Cu(OH)2 precipitate? Ksp = 1.6 x 10-19

Resultant molarities:MNaOH = (10)(0.01)/1010 = 9.9 x 10-5 M

MCuCl2 = (1000)(0.01)/1010 = 9.9 x 10-3 M

Q = [Cu][OH]2 = 9.7 x 10-11; Q>Ksp, thus precipitate will form


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Exercises:1. If PbCl2 has a ksp of 1.7 x 10-5, will a

precipitate form if 3g of PbBr2 is added to a liter of 0.77g of NaCl? MM PbBr2 = 367.2; MM NaCl = 58

2. What if 0.2M of 250mL PbBr2 was used in problem #1?


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END OF REVIEW SUB-PROPER.END OF REVIEW PROPER.

END OF REVIEW.


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