ACID-BASE EQUILIBRIUM ERT 207 ANALYTICAL CHEMISTRY SEMESTER 1, ACADEMIC SESSION 2015/16.

ACID-BASE EQUILIBRIUM

ERT 207 ANALYTICAL CHEMISTRY

SEMESTER 1, ACADEMIC SESSION 2015/16

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Overview

ACID-BASE THEORIES ACID DISSOCIATION CONSTANT pH SCALE METHODS OF MEASURING pH POLYPROTIC ACIDS WEAK BASES BASE DISSOCIATION CONSTANT RELATIONSHIP BETWEEN Kw, Ka

AND Kb

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Overview

BEHAVIOR OF SALTS IN WATER SALT SOLUTIONS ACIDS-BASE REACTIONS BUFFER SOLUTIONS HENDERSON-HASSELBALCH

EQUATION PREPARING A BUFFER

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ACID-BASE THEORIES

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Arrhenius (or Classical) Acid-Base Definition An acid is a substance that contains

hydrogen and dissociates in water to yield a hydronium ion : H3O+

A base is a substance that contains the hydroxyl group and dissociates in water to yield : OH –

Neutralization is the reaction of an H+ (H3O+) ion from the acid and the OH - ion from the base to form water, H2O.

Arrhenius 1903

Nobel Prize

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ACID-BASE THEORIES

bblee@unimapH3O+ = H+(aq) = proton in water

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ACID-BASE THEORIES

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ACID-BASE THEORIES

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The neutralization reaction is exothermic and releases approximately 56 kJ per mole of acid and base.

Brønsted-Lowry Acid-Base DefinitionAn acid is a proton donor, any species that donates an H+ ion.

An acid must contain H in its formula; HNO3 and H2PO4

- are two examples, all Arrhenius acids are Brønsted-Lowry acids.

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ACID-BASE THEORIES

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A base is a proton acceptor, any species that accepts an H+ ion. A base must contain a lone pair of electrons to

bind the H+ ion; a few examples are NH3, CO32-,

F -, as well as OH -. Brønsted-Lowry bases are not Arrhenius bases,

but all Arrhenius bases contain the Brønsted-Lowry base OH-.

Therefore in the Brønsted-Lowry perspective, one species donates a proton and another species accepts it: an acid-base reaction is a proton transfer process.

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ACID-BASE THEORIES

bblee@unimapFigure 1: Acid-Base Theories

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ACID-BASE THEORIES

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ACID DISSOCIATION CONSTANT

ACID DISSOCIATION CONSTANT

Strong Acids: 100% dissociation good H+ donor equilibrium lies far to right (HNO3) generates weak base (NO3

-)

Weak Acids: <100% dissociation not-as-good H+ donor equilibrium lies far to left

(CH3COOH) generates strong base (CH3COO-)

ACID DISSOCIATION CONSTANT

ACID DISSOCIATION CONSTANT

Strength vs.

Ka

ACID DISSOCIATION

CONSTANT

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ACID DISSOCIATION CONSTANT

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[H3O+] 1x100 to 1x10-14 in water [OH-] 1x10-14 to 1x100 in water

[H3O+] and [OH-]:

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ACID DISSOCIATION CONSTANT

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Finding [H3O+] and [OH-]:

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pH SCALE

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pH: pH is defined as the negative base-10 logarithm of the hydronium ion concentration.

pH = –log [H3O+] The “p” in pH tells us to take the

negative log of the quantity (in this case, hydronium ions).

Some similar examples are : pOH = –log [OH-] pKw = –log Kw

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pH SCALE

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pH SCALE

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pH and pOH: As [H3O+] rises, [OH-] falls

As pH falls, pOH rises

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pH SCALE

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pH SCALE

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Some pH values:

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METHODS OF MEASURING pH

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pH indicator

pH meter It measures the

voltage in the solution.

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METHODS OF MEASURING pH

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Figure 1: pH indicators

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METHODS OF MEASURING pH

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Relationship between Ka and pKa:

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METHODS OF MEASURING pH

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EXAMPLE 1: Calculating Ka from pH The pH of a 0.10 M solution of formic

acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this

temperature.

We know that

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METHODS OF MEASURING pH

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EXAMPLE 2: Calculating pH from Ka

Calculate the pH of a 0.30 M solution of acetic acid, C2H3O2H, at 25°C.

Ka for acetic acid at 25°C is 1.8 10-5.

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POLYPROTIC ACIDS

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Polyprotic acids have more than one acidic proton.

If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.

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POLYPROTIC ACIDS

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WEAK BASES

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Strength of Bases: Strong:

100% dissociation OH- supplied to solution NaOH(s) Na+

(aq) + OH-(aq)

Weak: <100% dissociation OH- by reaction with water CH3NH2(aq) + H2O(l) CH3NH2(aq) + OH-

(aq)

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WEAK BASES

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Sustainable Sustainability.

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BASE DISSOCIATION CONSTANT

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Bases react with water to produce hydroxide ion.

Sustainability.

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BASE DISSOCIATION CONSTANT

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Kb can be used to find [OH–] and, through it, pH.

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BASE DISSOCIATION CONSTANT

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BASE DISSOCIATION CONSTANT

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EXAMPLE 3: Calculating pH of Basic Solutions

What is the pH of a 0.15 M solution of NH3?

[NH4+] [OH−]

[NH3]Kb = = 1.8 10-5

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RELATIONSHIP BETWEEN Kw, Ka AND Kb

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BEHAVIOR OF SALTS IN WATER

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BEHAVIOR OF SALTS IN WATER

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Properties of salt solutions: The spectator ions in acid-base

reactions form salts. Salts completely ionize in their aqueous solutions.

A salt NH4A, the ionizes

NH4A ↔ NH4+ + A-

The ions of salts interact with water (called hydration),

A- + H2O ↔ HA + OH-

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BEHAVIOR OF SALTS IN WATER

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If the anions are stronger base than H2O, the solution is basic.

NH4+ + H2O = NH3 + H3O+

If the cations are stronger acid than H2O, the solution is acidic.

These competitive reactions make the solution acidic or basic depending on the strength of the acids and bases.

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SALT SOLUTIONS

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Neutral Salt Solutions: Salts consisting of the Anion of a Strong Acid and the Cation of a Strong Base yield a neutral solutions because the ions do not react with water.

Nitrate (NO3-) is a weaker base than

water (H2O) ; reaction goes to completion as the NO3

- becomes fully hydrated; does not react with water

)aq(OH)aq(NOIOH)I(HNO 3323

aqOHaqNa)s(NaNO OH 23

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SALT SOLUTIONS

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Na+ & NO3- do not react with water,

leaving just the “autoionization” of water, i.e., a neutral solution.

Salts that produce Acidic Solutions:A salt consisting of the anion of a

strong acid and the cation of a weak base yields an acidic solution.The cation acts as a weak acid

OHOHOH 322

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SALT SOLUTIONS

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The anion does not react with water. In a solution of NH4Cl, the NH4

+ ion that forms from the weak base, NH3, is a weak acid.

The Chloride ion, the anion from a strong acid does not react with water ClaqNHOH)s(ClNH OH

4242

)Acidic(OHaqNHOHaqNH 3324

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SALT SOLUTIONS

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Salts that produce Basic Solutions: A salt consisting of the anion of a weak

aid and the cation of a strong base yields a basic solution.

The anion acts as a weak base. The cation does not react with water. The anion of the weak acid accepts a

proton from water to yield OH- ion, producing a “Basic” solution.

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SALT SOLUTIONS

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Salts of Weakly Acidic Cations and Weakly Basic Anions: Overall acidity of solution depends on

relative acid strength (Ka) or base strength (Kb) of the separated ions.

Eg. NH4CN - Acidic or Basic? Write equations for any reactions that

occur between the separated ions and water OH)aq(NHIOH)aq(NH 3324

OHaqHCNIOH)aq(CN 2

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SALT SOLUTIONS

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Compare Ka of NH4+ & Kb of CN-

Magnitude of Kb (Kb > Ka) = (1.6 x 10-5

/ 5.7 x 10-10 = 3 x 104),Kb of CN- >> Ka of NH4

+

(Solution is Basic) Acceptance of proton from H2O by CN-

proceeds much further than the donation of a proton to H2O by NH4

+.

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SALTS OF WEAKLY ACIDIC CATIONS AND WEAKLY BASIC ANIONS

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ACIDS-BASE REACTIONS

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There are four classifications or types of reactions: i. strong acid with strong base, ii. strong acid with weak base, iii. weak acid with strong base, and iv. weak acid with weak base.

NOTE: For all four reaction types the limiting reactant problem is carried out first.

Once this is accomplished, one must determine which reactants and products remain and write an appropriate equilibrium equation for the remaining mixture.

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ACIDS-BASE REACTIONS

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(i) STRONG ACID WITH STRONG BASE:

The net reaction is:

The product, water, is neutral.

(ii) STRONG ACID WITH WEAK BASE: The net reaction is:

The product is HB+ and the solution is acidic.

+ -2H + OH H O

14net

w

1K = = 1.0 10

K

+ +3 2H O + B HB + H O

+

14bBnet net

waHB

K1K = = , with 1 < K < 1.0 10

K K

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ACIDS-BASE REACTIONS

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(iii) WEAK ACID WITH STRONG BASE: The net reaction is:

The product is A- and the solution is basic.

(iv) WEAK ACID WITH WEAK BASE: The net reaction is:

Notice that Knet may even be less than one.

This will occur when Ka HB+ > Ka HA.

- -2HA + OH H O + A

-

14aHAnet net

W bA

K 1K = = , with 1 < K < 1.0 10

K K

-

14aHAnet net

W bA

K 1K = = , with 1 < K < 1.0 10

K K

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ACIDS-BASE REACTIONS

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Example 4: You titrate 100 mL of a 0.025 M solution

of benzoic acid with 0.100 M NaOH to the equivalence point (mol HBz = mol NaOH).

What is the pH of (a) the final solution (b) half way point?Note: HBz and NaOH are used up!

HBz + NaOH ---> Na+ + Bz- + H2O

Ka = 6.3 × 10-5 Kb = 1.6 × 10-10

C6H5CO2

H = HBz

Benzoate ion =

Bz-

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BUFFER SOLUTIONS

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HCl is added to pure water.

HCl is added to a solution of a

weak acid H2PO4-

and its conjugate base

HPO42-.

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BUFFER SOLUTIONS

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The function of a buffer is to resist changes in the pH of a solution. Buffers are just a special case of the

common ion effect. Buffer CompositionWeak Acid + Conj. BaseHC2H3O2 + C2H3O2

-

H2PO4- + HPO4

2-

Weak Base+ Conj. AcidNH3 + NH4

+

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BUFFER SOLUTIONS

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Consider HOAc/OAc- to see how buffers work. ACID USES UP ADDED OH-. We know that

OAc- + H2O HOAc + OH-

has Kb = 5.6 x 10-10

Therefore, the reverse reaction of the WEAK ACID with added OH- has Kreverse = 1/ Kb =

1.8 x 109

Kreverse is VERY LARGE, so HOAc completely uses up the OH- !!!!

Acetic acid (HOAc) & a salt of the

acetate ion (OAc)

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BUFFER SOLUTIONS

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Consider HOAc/OAc- to see how buffers work. CONJUGATE BASE USES UP ADDED H+

HOAc + H2O OAc- + H3O+

has Ka = 1.8 x 10-5. Therefore, the reverse reaction of the WEAK

BASE with added H+has

Kreverse = 1/ Ka = 5.6 x 104

Kreverse is VERY LARGE, so OAc- completely

uses up the H+ !

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BUFFER SOLUTIONS

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Example 5 What is the pH of a buffer that has

[HOAc] = 0.700 M and [OAc-] = 0.600 M?

HOAc + H2O OAc- + H3O+

Ka = 1.8 x 10-5

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BUFFER SOLUTIONS Notice that the expression for calculating

the H+ concentration of the buffer is

This leads to a general equation for finding the H+ or OH- concentration of a buffer.

Notice that the H+ or OH- concentrations depend on K and the ratio of acid and base concentrations.

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[H3O+]= Orig. conc. of HOAc

Orig. conc. of OAc• K a

[OH- ]= [Base]

[Conj. acid]•Kb[H3O+ ]=

[Acid][Conj. base]

•Ka

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HENDERSON-HASSELBALCH EQUATION

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Take the negative log of both sides of this equation:

or

This is called the Henderson-Hasselbalch equation.

[H3O+

] = [Acid]

[Conj. base]• K a

pH= pKa + log[Conj. base]

[Acid]

pH= pKa - log[Acid]

[Conj. base]

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HENDERSON-HASSELBALCH EQUATION

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This shows that the pH is determined largely by the pKa of the acid and then adjusted by the ratio of acid and conjugate base.

Sustainability.

pH= pKa + log[Conj. base]

[Acid]

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HENDERSON-HASSELBALCH EQUATION

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EXAMPLE 6 What is the new pH when 1.00 mL

of 1.00 M HCl is added to:a) 1.00 L of pure water (before

HCl, pH = 7.00)

b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)

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PREPARING A BUFFER

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You want to buffer a solution at pH = 4.30. This means [H3O+] = 10-pH = 5.0 x 10-5 M It is best to choose an acid such that

[H3O+] is about equal to Ka (or pH pKa). You get the exact [H3O+] by adjusting the

ratio of acid to conjugate base. Buffer is prepared from:

HCO3- weak acid

CO32- conjugate base

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PREPARING A BUFFER

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You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M

POSSIBLE ACIDS Ka

HSO4- / SO4

2- 1.2 x 10-2

HOAc / OAc- 1.8 x 10-5

HCN / CN- 4.0 x 10-10

Best choice is acetic acid / acetate.

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PREPARING A BUFFER

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You want to buffer a solution at pH = 4.30 or

[H3O+] = 5.0 x 10-5 M

Solve for [HOAc]/[OAc-] ratio = 2.78/1 Therefore, if you use 0.100 mol of

NaOAc and 0.278 mol of HOAc, you will have pH = 4.30.

[H3O+

] = 5.0 x 10 -5 = [HOAc]

[OAc - ] (1.8 x 10 -5 )

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PREPARING A BUFFER

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The concentration of the acid and conjugate base are not important.

It is the RATIO OF THE NUMBER OF MOLES of each.

This simplifying approximation will be correct for all buffers with 3<pH<11, since the [H]+ will be small compared to the acid and conjugate base.

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PREPARING A BUFFER

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Preparing Buffers1)Solid/Solid: mix two solids. 2)Solid/Solution: mix one solid and one

solution. 3)Solution/Solution: mix two solutions.4)Neutralization: Mix weak acid with

strong base or weak base with strong acid.

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PREPARING A BUFFER

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Example 7: Preparing buffer: solid/solid Calculate the pH of a solution made by

mixing 1.5 moles of phthalic acid and 1.2 moles of sodium hydrogen phthalate in 500. mL of solution.

It is given that Ka= 3.0 x 10-4

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PREPARING A BUFFER

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Example 8: Preparing buffer: Solid/solution How many moles of sodium acetate

must be added to 500. mL of 0.25 M acetic acid to produce a solution with a pH of 5.50?

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EXAMPLE 1To calculate Ka, we need all equilibrium

concentrations.We can find [H3O+], which is the same

as [HCOO−], from the pH.

pH = –log [H3O+]– 2.38 = log [H3O+]

10-2.38 = 10log [H3O+] = [H3O+]4.2 10-3 = [H3O+] = [HCOO–]

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EXAMPLE 1

In table form:

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EXAMPLE 2The equilibrium constant expression is:

Use the ICE (Initial Change and Equilibrium) table:

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EXAMPLE 2

Simplify: x is relatively same,

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EXAMPLE 2

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(1.8 10-5) (0.30) = x2

5.4 10-6 = x2

2.3 10-3 = x

pH = –log [H3O+]pH = – log (2.3 10−3)pH = 2.64

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EXAMPLE 3Tabulate the data.

Simplify:

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EXAMPLE 3

Therefore,[OH–] = 1.6 10-3 M pOH = –log (1.6 10-3) pOH = 2.80 pH = 14.00 – 2.80 pH = 11.20

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(1.8 10-5) (0.15) = x2

2.7 10-6 = x2

1.6 10-3 = x

(x)2

(0.15)1.8 10-5 =

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EXAMPLE 4 (a)The product of the titration of benzoic

acid, the benzoate ion, Bz-, is the conjugate base of a weak acid. The final solution is basic.

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++

Kb = 1.6 x 10-10

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EXAMPLE 4 (a)This is a two-step problem:

1st:stoichiometry of acid-base reaction2nd:equilibrium calculation

1. Calculate moles of NaOH required.(0.100L HBz)(0.025M) = 0.0025mol HBz

This requires 0.0025 mol NaOH2. Calculate volume of NaOH required.0.0025 mol (1 L / 0.100 mol) = 0.025 L

25 mL of NaOH required

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EXAMPLE 4 (a)3.Moles of Bz- produced = moles HBz = 0.0025 mol Bz- 4.Calculate concentration of Bz-. There are 0.0025 mol of Bz- in a TOTAL

SOLUTION VOLUME of 125 mL [Bz-] = 0.0025 mol / 0.125 L = 0.020 M

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EXAMPLE 4 (a)

Solving:x = [OH-] = 1.8 x 10-6, pOH = 5.75, pH = 8.25

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Kb = 1.6 x 10-10 = x2

0.020 - x

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EXAMPLE 4 (b)

[H3O+] = { [HBz] / [Bz-] } Ka

At the half-way point, [HBz] = [Bz-], so

[H3O+] = Ka = 6.3 x 10-5

pH = 4.20

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EXAMPLE 5

Assuming that x << 0.700 and 0.600, we have

[H3O+] = 2.1 x 10-5 and pH = 4.68

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EXAMPLE 6(a) Calculate [HCl] after adding 1.00 mL

of HCl to 1.00 L of water M1 • V1 = M2 • V2

M2 = 1.00 x 10-3 M

pH = 3.00

(b) Step 1 — do the stoichiometryH3O+ (from HCl) + OAc- (from buffer) ---> HOAc (from buffer)

The reaction occurs completely because K is very large. bblee@unimap

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EXAMPLE 6

Step 2—Equilibrium.

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EXAMPLE 6Because [H3O+] = 2.1 x 10-5 M

BEFORE adding HCl, we again neglect x relative to 0.701 and 0.599.

[H3O+] = 2.1 x 10-5 M ------>

pH = 4.68

The pH has not changed significantly upon adding HCl to the buffer!

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[H3O+] = [HOAc]

[OAc-]•Ka=

0.701

0.599•(1.8 x 10-5)

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EXAMPLE 7

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Conjugates do not react!!

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EXAMPLE 8Let X = moles NaC2H3O2,

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Conjugates do not react!!