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24-Feb-16 1Chemistry The Molecular Nature of Matter and Change Kinetics: Rates and Mechanisms of Chemical Reactions 16.1 Factors That Influence Reaction Rate 16.2 Expressing the Reactio

Lecture Notes GENERAL CHEMISTRY 2

(The contents of the lectures may be changed without notice)

HÀ NỘI, 01 - 2016

24-Feb-16 1

Chemistry

The Molecular Nature of

Matter and Change

Kinetics: Rates and Mechanisms of Chemical Reactions

16.1 Factors That Influence Reaction Rate

16.2 Expressing the Reaction Rate

16.3 The Rate Law and Its Components

16.4 Integrated Rate Laws: Concentration Changes over Time

16.5 The Effect of Temperature on Reaction Rate

16.6 Explaining the Effects of Concentration and Temperature

16.7 Reaction Mechanisms: Steps in the Overall Reaction

16.8 Catalysis: Speeding Up a Chemical Reaction

16.1 Factors That Influence Reaction Rate

1 Reactant Concentration: C ↑→ reaction rate ↑

2 Physical State: the greater the surface area per

unit volume, the more contact it makes with other

reactants, and the faster the reaction.

3 Temperature: T ↑→ reaction rate ↑

• The term k is the rate constant, which is specific for a

given reaction at a given temperature.

• The exponents m and n are individual reaction orders

and are determined by experiment.

• m, n are also called to be reaction orders with respect to

→ the reaction order with respect to NO is 1.

→ the reaction order with respect to O 3 is 1.

→ the reaction is second order overall.

-2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

vrxn= k[NO]2[H2]1

→ the reaction is second order with respect to NO.

→ the reaction is first order with respect to H 2

→ the reaction is third order overall.

Exercise 1

Determine the individual order with respect to each reactant

and the overall reaction order from the given rate law

To determine the values of m and n,

• we run a series of experiments in which one reactant concentration changes while the other

is kept constant,

• and we measure the effect on the initial rate in each case.

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Experiment

Initial Rate (mol/L·s)

Example 3: Determine the individual order with respect to each

reactant and the overall reaction order from the given table

below:

aA + bB → products

From the data, we make 3 mathematical equations, solve the

Initial Rate (mol/L·s) [O 2 ] [NO]

(a) What is the reaction order with respect to A (red), B (blue)?

The overall order?

(b) Write the rate law for the reaction

(c) Predict the initial rate of experiment 4

is 1.

For reactant B (blue): experiments 1 and 3 show that when the number of particles of B doubles (while A remains constant), the rate quadruples The order with respect to B

16.4 Integrated Rate Laws

First-order rate equation:

Second-order rate equation:

Zero-order rate equation:

The value of k is easily determined from

experimental rate data

The units of k depend on the overall reaction order.

Units of k

Example 5:

First-order reaction

[A]t& time ln[A]t& time 1/[A]t & time

Reaction Half-life

concentration of a reactant to drop to half its initial value.

Second-order reaction

Example 6:

Find the rate constant of the reaction, if substance A (green) decomposes to two other substances, B (blue) and C (yellow), in a first-order gaseous reaction The following scenes represent the reaction mixture at the start of the reaction (at 0.0 s) and after 30.0 s

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Exercise 3

Draw the molecular scenes of the reaction mixture at t = 60.0 s and

90.0 s, if substance A (green) decomposes to two other substances, B

(blue) and C (yellow), in a first-order gaseous reaction The following

scenes represent the reaction mixture at the start of the reaction (at 0.0

s) and after 30.0 s

Zero Order First Order Second Order

Rate law rate = k rate = k[A] rate = k[A]2

Units for k mol/L·s 1/s L/mol·s

Half-life [A]o/2k ln2/k 1/(k[A]o) Integrated rate law [A]t = -kt + [A]0 ln[A]t = -kt + ln[A]0 1/[A]t = kt +1/[A]0

Plot for straight line [A]t vs t ln[A]t vs t 1/[A] tvs t Slope, y intercept -k, [A]0 -k, ln[A]0 k, 1/[A]0

Overview about Reaction Rate

16.5 Temperature and the Rate Constant

• Temperature has a dramatic effect on reaction rate For many reactions, an increase of 10°C will double or triple the rate Experimental data shows that k increases exponentially as T increases

• The Arrhenius equation:

Activation Energy

• In order to be effective, collisions between particles must

exceed a certain energy threshold

• When particles collide effectively, they reach an activated state

The energy difference between the reactants and the activated

• The lower the activation energy, the faster the reaction

E a

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∆Hrxn> 0 ∆Hrxn< 0

∆Hrxn= Ea,for– Ea,revExothermic reaction, ∆Hrxn< 0

∆Hrxn= Ea,for– Ea,rev

Endothermic reaction, ∆Hrxn> 0

Calculating Activation Energy

At two different temperatures

Temperature

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must collide in order to react

larger number of collisions, hence increasing reaction rate

numbers of reactant particles, not their sum

Collision Theory

Temperature and Collision Energy

kinetic energy of the particles This leads to more frequent

collisions and reaction rate increases

Reaction rate therefore increases

Transition State Theory

formation of a transition state or activated complex.

partial bonds It is a transitional species partway between

reactants and products

of maximum potential energy

activation energy

Transition State & Intermediate State

reaction which is formed between the reaction when reactants change into products

new molecule(involves breaking of bonds of reactants and formation of new ones)

intermediate has a discrete lifetime (be it a few nanoseconds or many days), whereas a transition state lasts for just one bond vibration cycle

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•Transition states are local energy maximums and have partial

bonds This might be one of the reasons why they can’t be

isolated as intermediates

16.7 Reaction Mechanisms

Reaction Mechanisms

reaction steps that make up the overall equation

elementary steps because each one describes a single

molecular event.

the number of particles involved in the reaction

reaction stoichiometry – reaction order equals molecularity for

an elementary step only.

Rate Laws for General Elementary Steps

The molecularity of each step equals the total number

(a) Write the overall balanced equation.

(b) Determine the molecularity of each step.

(c) Write the rate law for each step.

SOLUTION:

a) Make the sum of the elementary steps to find the overall equation.

2NO 2Cl(g) → 2NO2(g) + Cl2(g)

b) The molecularity of each step equals the total number of reactant particles.

Step(1) is unimolecular Step(2) is bimolecular.

c) We write the rate law for each step using the molecularities as reaction

orders.

rate step1= k1 [NO 2 Cl]

ratestep2= k2[NO2Cl][Cl]

The Rate-Determining Step of a Reaction

rate-limiting step

law for the overall reaction

been proposed to occur by a two-step mechanism:

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16.8 Catalysis: Speeding up a Reaction

without itself being consumed in the reaction

pathway that has a lower total activation energy than the

uncatalyzed reaction

reactions, and does not affect either ∆H or the overall yield for a

the reactant and product, chemists recognize two categories of catalyst:

+ homogeneous: the catalysis is same physical state as the reactants and products

+ heterogeneous: the catalysis is NOT same physical state

Catalysis

The Molecular Nature of

Matter and Change

Equilibrium: The Extent of Chemical Reactions

17.1 The Equilibrium State and the Equilibrium Constant (K)

17.2 The Reaction Quotient (Q) and the Equilibrium Constant (K)

17.3 Expressing Equilibria with Pressure Terms: Kcand Kp

17.4 Reaction Direction: Comparing Q and K

17.5 How to Solve Equilibrium Problems

17.6 Le Chatelier’s Principle

17.1 The Equilibrium State

& Equilibrium Constant (K)

a reaction, the change in product concentration (or

reactant concentration) per unit time

reaction, the concentration of product that results given

unlimited reaction time

• At equilibrium state: rate forward = rate reverse

• A system at equilibrium is dynamic on the molecular

level; no further net change in reactant and product

concentrations is observed because changes in one

direction are balanced by changes in the opposite

direction.

If rateforward = ratereverse

→ kforward[reactants]m = kreverse[products]n

• This expression is also known as the Law of Mass Action

• The exponents (m and n) are equal to the coefficients in

the balanced chemical equation

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Example 1:

What is the K expression for?

17.2 The Reaction Quotient (Q) and the Equilibrium Constant (K)

For a general reaction: aA + bB ⇌ cC + dD

•At the equilibrium state:

•At any time state:

Q is called the Reaction Quotient

•Multiply with n: naA + nbB ⇌ ncC + ndD

•Overall reaction that is equal the sum of many reaction:

• Pure solids and liquids are not included in the equilibrium constant expression

Rules calculating with K

Example 2:

In the cylinders of a car engine, at the very high temperatures,

form nitrogen dioxide, a toxic pollutant that contributes to

photochemical smog

N2(g) + O2(g) ⇌ 2NO(g) (1) K1= 4.3 x 10-25

2NO(g) + O2(g) ⇌ 2NO2(g) (2) K2= 6.4 x 109

a) Show the overall reaction

b) Calculate the K for the overall reaction

d) The acid dissociation reaction

Expressing Equilibria: Kcand Kp

For a general reaction: aA(g) + bB(g) ⇌ cC(g) + dD(g)

• At the equilibrium state, if the substances is gas:

flue gas of a coal-burning power plant for form lime (CaO)

atmospheres

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Example 3:

Which direction does the reaction shift to in each of these

molecular scenes?

1) 2) 3) 4)

Exercise 3:

The scenes below represent different mixtures of the reaction:

In which direction does the reaction shift (if at all) for each mixture

to reach equilibrium?

17.5 How to Solve Equilibrium Problems

3 Forms of Equilibrium Exercises

1 Using Quantities to Determine the Equilibrium Constant (Calculating K from Concentration Data)

2 Using the Equilibrium Constant to Determine Quantities (Determining Equilibrium Concentrations from K)

3 Mixtures of Reactants and Products: Determining Reaction Direction (Predicting Reaction Direction and Calculating Equilibrium Concentrations)

Steps to solve the equilibrium problems

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Example 5:

In a study concerning the conversion of methane to other fuels,

flask at 1200 K At equilibrium the flask contains 0.26 mol of CO,

Conc (M) CH 4(g) + H 2O(g) ⇌ CO(g) + 3H2(g)

flask at 900 K

CO(g) + H2O(g) ⇌ CO2(g) + H2(g)

What is the composition of the equilibrium mixture? At this

Conc (M) CO(g) + H 2O(g) ⇌ CO 2(g) + H2(g)

The research and development unit of a chemical company is

gas:

CH4(g) + 2H2S(g) ⇌ CS2(g)+ 4H2(g)

“When any system at equilibrium is subjected to change in

concentration, temperature, volume, or pressure, then the

system readjusts itself to (partially) counteract the effect of

the applied change and a new equilibrium is established.”

The Effect of a Change in

Temperature Pressure

(Volume) Concentration

2H2S(g) + O2(g) ⇌ 2S(s) + 2H2O(g) ∆H < 0Which direction does the reaction shift to, if?

a)O2is added b) H2S is removed c) sulfur is added

+ When the equilibrium is established:

-a)If O2is added → Qc< Kc→ the equilibrium shift to the right

b)If H2S is removed → Qc> Kc→ the equilibrium shift to the left

c)Sulfur is not part of the Q and K expression because it is a solid Therefore, as long as some sulfur is present the reaction is unaffected

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Example 9:

To improve air quality and obtain a useful product, chemists often

remove sulfur from coal and natural gas by treating the fuel

contaminant hydrogen sulfide with O2:

2H2S(g) + O2(g) ⇌ 2S(s) + 2H2O(g) ∆H < 0

CaCO3(s) ⇌ CaO(s) + CO2(g) ∆Ho= 178 kJ

Which direction does the reactions equilibrium shift to, if?

a)increase in temperature b) increase in pressure

-a)With the case ∆H < 0, so if increase in temperature → decrease

K for a system → the equilibrium shift to the left

The case ∆H > 0 → the equilibrium shift to the right

a)With the case ∆n = -1 < 0, so if increase in pressure → the

equilibrium shift to the right

The case ∆n = +1 > 0 → the equilibrium shift to the left

R = 8.314 J/mol*K

K = the equilibrium constant

Van’t Hoff equation (Chapter 17)

-R = 8.314 J/mol*K

k = the rate constant

Arrhenius equation (Chapter 16)

-R = 8.314 J/mol*K

P = the vapor pressure

Clausius-Clapeyron equation (Chapter 12)

The Van’t Hoff Equation = the Effect of T on K

If we add 0.075 M of Cl2to the above equilibrium mixture, what is the

composition of the new equilibrium mixture? At this temperature, Kcis

24 for the equation

For the reaction: X(g) + Y2(g) ⇌ XY(g) + Y(g) H > 0

the following molecular scenes depict different reaction mixtures

(X = green, Y = purple)

(a) If K = 2 at the temperature of the reaction, which scene

represents the mixture at equilibrium?

(b) Will the reaction mixtures in the other two scenes proceed toward reactant or toward products to reach equilibrium?

(c) For the mixture at equilibrium, how will a rise in temperature affect [Y2]?

Lecture PowerPoint

Chemistry

The Molecular Nature of

Matter and Change

18.1 Acids and Bases in Water

18.2 Autoionization of Water and the pH Scale

18.3 Proton Transfer and the Brønsted-Lowry Acid-Base Definition

18.4 Solving Problems Involving Weak-Acid Equilibria

18.5 Weak Bases and Their Relation to Weak Acids

18.6 Molecular Properties and Acid Strength

18.7 Acid-Base Properties of Salt Solutions

18.8 Generalizing the Brønsted-Lowry Concept: The Leveling Effect

18.9 Electron-Pair Donation and the Lewis Acid-Base Definition

3

18.1 Acids and Bases in Water

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• An acid is a substance that has H in its formula and

dissociates in water to yield H3O+

• A base is a substance that has OH in its formula and

• When an acid reacts with a base, they undergo

6

The Base Constant, Kb

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7

Perform the dissociations of Strong and Weak Acids – Bases

• A strong acid dissociates completely into ions in water:

HA(l) + H2O(l) → H3O+(aq) + A-(aq)

• A weak acid dissociates slightly to form ions in water:

HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)

• Similarly for strong and weak bases

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Classifying of Strong Bases - Acids

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• Strong acids include

 The hydrohalic acids (HCl, HBr, and HI)

 Oxoacids in which the number of O atoms exceeds

the number of ionizable H atoms by two or more

(HNO3, H2SO4, HClO4…)

• Strong bases include

 The cations are usually those of the most active

where M = (Ca, Sr, Ba)

9

Classifying of Weak Bases - Acids

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• Weak acids include

 The acid HF and acids in which H is not bonded to O or

to a halogen (HCN…)

 Oxoacids in which the number of O atoms equals or exceeds the number of ionizable H atoms by one (HClO,

 Carboxylic acids, which have the general formula

• Weak bases include

 The common structural feature is an N atom with a lone electron pair

10

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Exercise 1

Classify each of the following compounds as a strong acid,

weak acid, strong base, or weak base:

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18.2 Autoionization of Water,

pH scale

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• Autoionization of water is a reaction that has form

2H2O(l) ⇌ H3O+(aq) + OH-(aq)

pH = -log[H 3 O + ] & pOH = -log[OH - ]

The pH of a solution indicates its relative acidity:

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• An acid is a proton donor, any species that donates an

H + ion An acid must contain H in its formula.

• A base is a proton acceptor, any species that accepts an

H + ion A base must contain a lone pair of electrons to

Arrhenius Acids- Bases

20

Conjugate Acid-Base Pairs Definition

Conjugate Pair

Conjugate Pair

Acid 1 & Base 1 = a conjugate acid-base pair

Acid 2 & Base 2 = a conjugate acid-base pair

 Base 1 = the conjugate base of the Acid 1.

 Acid 2 = the conjugate acid of the Base 2.

e) N2H5+ + H2SO4 ⇌ HSO4-+ N2H62+

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Rule to determine the net

Direction of Reaction

A reaction will favor the formation of the weaker acid and base:

Stronger Acid + Stronger Base ⇌ Weaker Base + Weaker Acid

Example 2: Show the direction of the below reaction:

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3 Forms of Weak-Acid Equilibria Exercises

H3PO4(aq) + H2O(l) ⇌ H2PO4-(aq) + H3O+(aq) K a1 = 7.2x10-3

H2PO4-(aq) + H2O(l) ⇌ HPO42-(aq) + H3O+(aq) K a2= 6.3x10-8

HPO42-(aq) + H2O(l) ⇌ PO43-(aq) + H3O+(aq) K a3 = 4.2x10-13

We can see that K a1 >> K a2 >> K a3

Polyprotic Acids

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B(aq) + H2O(l) ⇌ BH+(aq) + OH-(aq)

• The Brønsted-Lowry base must have a lone electron pair at

least

• Any anions of weak acids will be Weak Bases

Example 7:

Because (3) = (1) + (2)

 Kw = Kax KbThis relationship is true for any conjugate acid-base pair.

What is the pH of 0.25 M sodium acetate (CH3COONa,

or NaAc)? Kaof acetic acid (HAc) is 1.8x10-5.

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18.6 Molecular Properties and Acid Strength

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• Trends in Acid Strength of Nonmetal Hydrides

o Across a period, nonmetal hydride acid strength increases

HCl > H2S

o Down a group, nonmetal hydride acid strength increases

HF << HCl < HBr < HI

• Trends in Acid Strength of Oxoacids

o For oxoacids with the same number of oxygens around E,

acid strength increases with the electronegativity of E

HOCl > HOBr > HOI

o For oxoacids with different numbers of oxygens around a

given E, acid strength increases with number of O atoms

HOCl < HOClO < HOClO2< HOClO3

HNO2< HNO3

H2SO3< H2SO4

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18.7 Acid-Base Properties of Salt Solutions

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Salts → Neutral Solutions

A salt that consists of the anion of a strong acid and the cation

of a strong base yields a neutral solution

Example 8:

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A salt that consists of the anion of a strong acid and the cation

of a weak base yields an acidic solution

Example 9:

to produce H3O+:

NH4(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)

Salts → Acidic Solutions

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A salt that consists of the anion of a weak acid and the cation

of a strong base yields a basic solution

Example 10:

Salts → Basic Solutions

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If a salt that consists of the anion of a weak acid and the cation

of a weak base, the pH of the solution will depend on the relative acid strength or base strength of the ions

Example 11:

NH4(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)

CN-(1.6x10-5)  since Kbof CN-> Kaof NH4

Salts → Neutral, Acidic, Basic Solution

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Exercise 6

Predict whether aqueous solutions of the following are acidic,

basic, or neutral Write an equation for the reaction of any ion

with water (Kaof Zn(H2O)62+= 1x10-9; Kbof HCOO-= 5.6x10-11)

(a) Chromium(III) nitrate, Cr(NO3)3

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18.7

Generalizing the Brønsted-Lowry Concept:

The Leveling Effect

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The Leveling Effect

• All strong acids and bases are equally strong in water

Because:

o All strong acids dissociate completely to form H3O+

o All strong bases dissociate completely to form OH-

The Lewis Acid-Base Definition

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• A Lewis acid is a specie that accepts an electron pair to form

a bond A Lewis base is a specie that donates an electron pair

to form a bond

o A Lewis acid must have a vacant orbital (or be able to rearrange its bonds to form one) to accept a lone pair and form a new bond

o A Lewis base must have a lone pair of electrons to donate

• The Lewis definition views an acid-base reaction as the donation and acceptance of an electron pair to form a covalent bond

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Lewis Acid-Base Definition

Lewis Acids

• Atom B and Al often form electron-deficient molecules, and

these atoms have an unoccupied p orbital that can accept a

pair of electrons

• Molecules that contain a polar multiple bond often function

as Lewis acids

• A metal cation acts as a Lewis acid when it dissolves in

water to form a hydrated ion

pair to K+

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Lecture PowerPoint

Chemistry

The Molecular Nature of

Matter and Change

Ionic Equilibria in Aqueous Systems

19.1 Equilibria of acid-base buffer systems

19.2 Acid-base titration curves

19.3 Equilibria of slightly soluble ionic compounds

19.4 Equilibria involving complex ions

An acid-base buffer is a solution that lessens the impact of pH

from the addition of acid or base

An acid-base buffer usually:

 a conjugate acid-base pair with appreciable quantities.

 a solution of a weak acid and its conjugate base,

 a solution of weak base and its conjugate acid.

equilibrium position.

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Example 2: The effect of addition of acid or base to …

an unbuffered solution.

a buffered solution.

Example 3:

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3 O +(aq)

H3O+(aq) + OH-(aq) → H2O(l)

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3 O +(aq)

How a Buffer Works: The Common-Ion Effect

A buffer consists of a weak acid and its conjugate base:

The Henderson-Hasselbalch Equation (2)

pOH = pKb+ log[acid]

[base]

A buffer consists of a weak base and its conjugate acid:

(buffer-component ratio)

Buffer Capacity

The buffer capacity is a measure of the “strength” of the buffer,

its ability to maintain the pH following addition of strong acid or

base

The buffer capacity is high when:

 The concentrations of all the buffer components are high

 The concentrations are close to each other

 A buffer has the highest capacity when the component

concentrations are equal

Buffer Range

The buffer range is the pH range over which the buffer acts effectively, and it is related to the relative component concentrations

 The buffers usually have a usable range within ±1 pH unit

buffering action is poor (in practice)

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Example 4:

b) After adding 0.020 mol of solid NaOH to 1.0 L of the buffer

solution in part (a)

c) After adding 0.020 mol of HCl to 1.0 L of the buffer solution

b) The closer 1 the buffer-component ratio is, the more capacity the buffer has → sample 3 Why?

Preparing a Buffer

acid component should be close to the desired pH

 Calculate the ratio of buffer component concentrations

 Determine the buffer concentration, and calculate the required

volume of stock solutions and/or masses of components

 Mix the solution and correct the pH

Example 5:

We need a carbonate buffer of pH 10.00 How many grams of

make the buffer? Kaof HCO3-is 4.7x10-11

 nNa2CO3= nCO32-= 0.14 mol

 mNa2CO3= n.M = 0.14×106 = 15 gram