Acids, Bases, and Salts

Chapters 4.3, 16 and 17

Classification of Acids and Bases is one of the oldest in History. The medieval Alchemists first used the terms “acid”, “alkali”, and “salt”.Acids were probably the most easily recognized chemical because of their sour taste.

Properties of Acids

1. Have a sour taste2. Change the color of some plant pigments3. Dissolve or corrode certain metals and

minerals; damages skin -“Corrosive”4. Produce gas bubbles when combined

with minerals.5. Dissolve in water to form aqueous

solutions.6. Ionize in water to produce H+ ions

Examples of Acids

HCl hydrochloric acid (stomach acid) H2SO4 sulfuric acid (battery acid)

HNO3 nitric acid (used to make fertilizer) HBr hydrobromic acid HI hydroiodic acid HClO3 chloric acid

HClO4 perchloric acid (all other acids are weak. . .)(all other acids are weak. . .)

HCHC22HH33OO22 acetic acid acetic acid

HH22COCO33 carbonic acid carbonic acid

There are 7 strongAcids.

Properties of Bases

1. Have a bitter, chalky taste

2. Have a soapy, slippery feeling

3. Change the color of some plant pigments

4. Can burn and irritate skin - “caustic”

5. Bases can destroy the properties of acids when mixed in proper proportions.

6. Dissolve in water to form aqueous solution; accept H+ ions and produce OH- ions

Examples of Bases• NaOH sodium hydroxide (lye)

• LiOH

• KOH

• RbOH

• CsOH

• Ca(OH)2

• Sr(OH)2

• Ba(OH)2

Group IA metal hydroxides

Heavy Group IIA metal hydroxides

ALL other bases are weak: NH3

Electrolytes and Nonelectrolytes:

Strong acids and bases completely ionize in solution making them strong electrolytes.

Weak acids and bases only partly ionize making them weak electrolytes in solution.

Acids and Bases are the only molecular compounds that can act as electrolytes.

Arrhenius Definition• Acids produce hydrogen ions in aqueous

solution.

HCl + H2O ---> H+(aq) + Cl- (aq)

• Bases produce hydroxide ions when dissolved in water.

NaOH + H2O ---> Na+ (aq) + OH- (aq)

• Limited to aqueous solutions.• Arrhenius limits us to one kind of base.

• NH3 ammonia could not be an Arrhenius base.

Bronsted-Lowry Definitions

• An acid is a H+ (proton) donor and a base is a H+ (proton) acceptor.

• Acids and bases always come in pairs.• Acid + Base Conjugate + Conjugate

acid base• HCl is an acid.• When it dissolves in water it gives its proton to

water.

• HCl(g) + H2O(l) H3O+ + Cl-

H+ donor H+ acceptor conjugate conjugate

acid base

Bronsted-Lowry expands the list

• When ammonia is dissolved in water -

NH3 + H2O NH4+ + OH-

H+ acceptor H+ donor conjugate conjugate

acid base

The conjugate acid is what results when the base accepts the H+

The conjugate base is what remains after the acid has donated the H+

BASE ACID

Competing Pairs

NH3 + H2O NH4+ + OH-

H+ acceptor H+ donor conjugate conjugate base acid acid base

• This is an equilibrium.• Competition for H+ between NH3 and OH

• The stronger base controls direction.• If OH is a stronger base it takes the H+, forms water.• Equilibrium moves to left.• If NH3 is stronger, it takes the H+

, forms NH4

+

• Equilibrium moves to right.

Relative strength of Acids and Bases:

The stronger the acid or base, the weaker its conjugate pair.

1. Strong acids completely transfer their protons to water; therefore, the conjugate bases have a tendency to be protonated.

2. Weak acids only partly dissociate; therefore, the conjugate bases show a slight ability to remove protons from water.

3. Substances with negligible acidity, contain hydrogen, but do not demonstrate any acidic behavior; their conjugate bases are strong bases.

• Most general definition. Gives us many things which act as acid-bases.

• Acids are electron pair acceptors.• Bases are electron pair donors.

B FF

F

:NH

H

H

Lewis Acids and Bases

Lewis Acids involve molecules which have an incomplete octet. Lewis Bases have an unshared pair of electrons.

Lewis Acids and Bases

B FF

F

:NH

H

H

• Boron triflouride wants more electrons.

• It acts as a Lewis Acid.

Ammonia, NHAmmonia, NH33 donates a pair of electrons. donates a pair of electrons.It acts as a Lewis base.It acts as a Lewis base.

Lewis Acids and Bases

BF

F

F

N

H

H

H

• BF3 is Lewis acid

• NH3 is a Lewis base.

Lewis Acid Lewis Base

Water H+ + OH- ---> H2O

[ O - H ]-• •

••• •

H+

Donates a pair of electrons

Lewis Acid

Lewis Base

A New View of Water:Autoionization

Water is amphoteric: it behaves as both an acid and a base

In any sample of water, to a very small degree, water self-ionizes:

H2O ---> H+ + OH

actually,

2 H2O H3O+ + OH

Kw

Since there is an equilibrium established

2 H2O H3O+ + OH

we can write an equilibrium expression:

[H3O+] [OH]

Keq =

[H2O] 2 We don’t include water in the equilibrium expression

because the amount is huge and constant

KKww = [H = [H33OO++] [OH] [OH]] KKw w - “Ion Product Constant for water”- “Ion Product Constant for water”

Kw• Kw = [H3O+] [OH]

For any aqueous solution at 25C, Kw = 1.0 x 10 14 (a very small amt. of ionic activity!)

This means:

Kw = [H3O+] [OH]

1.0 x 10 14 = [H3O+] [OH]

[H3O+] = [OH] = 1.0x 10 7

So, any Acidic Solution: [H+] > [OH]And any Basic Solution: [H+] < [OH]

[H[H33OO++] is also [H] is also [H++]]

Acid vs. Basic Solutions• Remember water behaves as both an acid

and a base.• 2H2O(l) H3O+(aq) + OH-(aq)• KW= [H3O+][OH-] = [H+][OH-]• At 25ºC, KW = 1.0 x10-14 • In EVERY aqueous solution. Therefore,• Neutral solution [H+] = [OH-]= 1.0 x10-7 • Acidic solution [H+] > [OH-]• Basic solution [H+] < [OH-]

pH• pH= -log[H+]

• Used because [H+] is usually very small,

always less than 1 M

• As pH decreases, [H+] increases exponentially

• Helpful to see a graph of y = log [x]

pH = log [H+]y = log [x]

[H+]

(1,0)

pHThis is wherepH is concerned.

pH• [H+] = 1.0 x 10-8 M [H+] = 1.0 x 103 M

pH= 8.00 pH = 3.00

Sig figs - only the digits after the Sig figs - only the digits after the

decimal place of a pH are decimal place of a pH are significant -significant -

only 2 sig. figs in pH = 8.00only 2 sig. figs in pH = 8.00p stands for “p stands for “ log” log”

Also. . . Also. . .

pOH = -log[OHpOH = -log[OH--]]

pKpKww = -log K = -log Kww

Basic

Acidic Neutral

100

10-

1

10-

3

10-

5

10-

7

10-

9

10-

11

10-

13

10-

14

[H+]

0 1 3 5 7 9 11

13

14

pH

Basic

100

10-

1

10-

3

10-

5

10-

7

10-

9

10-

11

10-

13

10-

14

[OH-]

01357911

13

14

pOH

Relationships

KW = [H+][OH-]

-log KW = -log([H+][OH-])

-log KW = -log[H+]+ -log[OH-] So, pKW = pH + pOH

and since, KW = 1.0 x10-14

Then, 14.00 = pH + pOH [H+], [OH-], pH and pOH; given any one of these we can find the other

three.

Calculations with pH

Remember every solution has both a pH and a pOH.

Remember the chemistry, but don’t try to memorize there is no one way to do this.

Example #1

A solution has [H+] = 1.5 x106 M Determine the pH, pOH, and [OH].

pH = log[H+] = log [1.5 x 106 M] = 5.82 14.00 = pH + pOH

14.00 = 5.82 = pOH pOH = 8.18

Kw = [H+] [OH] 1.0 x 1014 = 1.5 x 106 M [OH] [OH] = 6.6 x 10 9 M

Or, take the antilogof 8.18.

Example #2

A solution has a pOH = 4.92 Calculate the pH, [H+], and [OH].

pH = 14.00 pOHpH = 14.00 4.92pH = 9.08

Take the anti-log of 9.08[H+] = 8.3 x 1010 M

Kw = [H+] [OH]1 x 10 14 = [8.3 x 1010 M] [OH][OH] = 1.2 x 105 M

Acid and Base dissociation constants

Ka and Kb

Ka - “Acid dissociation

constant” General equation for an acid:

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

Ka = [H3O+][A-]

[HA]

H3O+ is often written H+ ignoring the water in

equation (it is implied).

Ka -Acid dissociation constant

Ignoring the water:HA(aq) H+(aq) + A-(aq) Ka = [H+][A-]

[HA]We can write the expression for

any acid.

Strong Acids HCl, HNO3, H2SO4, HBr, HI, HClO4, HClO3

Strong acids dissociate completely. HCl ---> H+ (aq) + Cl (aq)

Ka = [H+][Cl-]

[HCl] Strong acids favor products. Equilibrium far to right. Ka is very large. Conjugate base must be weak.

Weak Acids

Weak acids only partly dissociate. Weak acids favors reactants. HF (aq) H+ (aq) + F (aq)

Ka = [H+][F-]

[HF] Ka will be small. ALWAYS WRITE THE MAJOR SPECIES. It will be an equilibrium problem from the start. Rest is just like last chapter - an equilibrium

problem!

Summary

Strong acids

Ka is very large

[H+] is equal to [HA] A- is a weaker base

than water

Weak acids

Ka is small

[H+] <<< [HA] A- is a stronger

base than water

Types of Acids

Oxyacids - Proton is attached to the oxygen of an ion.

Example: nitrous acid, HNO2

Organic acids contain the Carboxyl group -COOH with the H attached to O

example: Benzoic acid, C6H5COOH Polyprotic Acids- more than 1 acidic

hydrogen (diprotic, triprotic).

Polyprotic acids• Always dissociate stepwise.• The first H+ comes off much easier than the

second.• Ka for the first step is much bigger than Ka for the

second.• Denoted Ka1, Ka2, Ka3

• Example: Carbonic Acid

• H2CO3 H+ + HCO3-

Ka1= 4.3 x 10-7

• HCO3- H+ + CO3

-2 Ka2= 4.3 x 10-10

Almost all of the HAlmost all of the H++ comes from the first step, so successive steps comes from the first step, so successive steps are often ignored except for sulfuric acid.are often ignored except for sulfuric acid.

Sulfuric acid is special

H2SO4 (aq) ----> H+ (aq) + HSO4 (aq)

In first step it is a strong acid. HSO4

(aq) H+ (aq) + SO42 (aq)

2nd step it’s weak: Ka2 = 1.2 x 10-2

However, 2nd step too large Ka to ignore.

Ka Problems• Problem: What is the pH of a 0.0658M HCl

solution?• Always write the major species: HCl --> H+ + Cl

• Strong acids completely dissociate!

• [H+] = [HCl] HCl ---> H+ (aq) + Cl (aq) [H+] = [HCl]

[H+] = 0.0658 MpH = log [0.0658 M] = 1.18

Try this.• Calculate the pH of 2.0 M acetic acid

HC2H3O2. Then find the pOH and[OH-].

KKaa = [H = [H++][C][C22HH33OO22--]]

[HC [HC22HH33OO22] ]

Weak acids only partly Weak acids only partly dissociate.dissociate.

HCHC22HH33OO2 2 H H++ + C + C22HH33OO22-- Look up Look up

Ka = 1.8 x 10Ka = 1.8 x 1055

1.8 x 101.8 x 1055 = [ x ][ x ] = [ x ][ x ] [ 2.0 M ][ 2.0 M ]

XX22 = 1.8 x 10 = 1.8 x 10-5-5 x x 2.02.0[H[H++] = X = 0.0060 M] = X = 0.0060 M

pH = 2.2pOH = 11.8[OH] = 1.6 x 1012 M

Bases• Strong Bases: LiOH, NaOH, KOH, RbOH,

CsOH, Ca(OH)2, Sr(OH)2 and Ba(OH)2.• The OH-

is a strong base. • Hydroxides of the alkali metals are strong

bases because they dissociate completely when dissolved.

• The hydroxides of alkaline earths Ca(OH)2 etc. are strong dibasic bases, but they don’t dissolve well in water.

• Again, strong bases favor products. Kb is very large, [OH-] = [B]

• Calculations just like strong acids.

Bases without OH-

Bases are proton acceptors.

NH3 + H2O NH4+ + OH-

It is the lone pair on nitrogen that accepts the proton.

Many weak bases contain N

Weak Bases

NH3 + H2O NH4+ + OH-

General Equation: B(aq) + H2O(l) BH+(aq) + OH- (aq)

Kb = [BH+][OH- ]

[B]

Kb “base dissociation constant”

Strength of Bases

Hydroxides are strong. Others are weak. Smaller the Kb weaker the base.

Kb problem

• Calculate the pH of a 0.25 M NH3.• Always write the equation and the species!

NHNH33 + H + H22O NHO NH44++ + OH + OH--

KKbb = [ = [NHNH44++][][OHOH- - ]]

[ [NHNH33]]

Look up Look up KKbb = 1.8 x 10 = 1.8 x 1055

1.8 x 101.8 x 1055 = [ = [ x x ][][ x x ]] [[NHNH33]](1.8 x 10(1.8 x 105 5 ) (0.25 M) = x) (0.25 M) = x22

x = 0.0021 M = [OHx = 0.0021 M = [OH--]] 14.00 = pH + pOHpH = 11.33pOH = 2.67pOH = 2.67

Percent Dissociation

The amount of the acid (or base) [HA] that has dissociated [x] divided by the acid’s initial conentration, [HA0], then multiplied by 100%.

Percent Dissociation = (x/[HA0]) x 100% When making assumptions in equilibrium

concentrations, it is best to test the assumption by making sure that the percent dissociation is less than or equal to 5%.

Example

The percent dissociation of an acid, HA, which is 0.100M is 2.5%. Calculate the Ka of the acid.

(x/.100) x 100% = 2.5% x = 2.5 x 10-3M Ka = (2.5x10-3)2 / (.100 – 2.5x10-3) Ka = 6.4 x 10-5

A mixture of Weak Acids/Bases

The process is the same. Determine the major species. The stronger will predominate. If one acid has a relatively highier Ka value, it will

be the focus of the solution. Calculate the pH of a mixture:

1.20 M HF (Ka = 7.2 x 10-4) and 3.4 M HOC6H5 (Ka = 1.6 x 10-10)

HF dominates so the pH of the mixture will be based on it.

Relationship of Ka and Kb

NH3 + H2O NH4+ + OH- NH4

+ + H2O NH3 + H3O+

Kb = [NH4+] [OH-] Ka = [NH3] [H+]

[NH3] [NH4+]

if added together, NH3 + H2O NH4+ + OH

+ NH4+ + H2O NH3 + H3O+

= 2 H2O H3O+ + OH-

we get the autoionization of water equation

Relationship of Ka and Kb

Ka x Kb = ( [NH3] [H+] ) ( [NH4+] [OH-] )

[NH4+] [NH3]

Ka x Kb = [H+] [OH-] = Kw = 1.0 x 10-14

So, Ka x Kb = Kw

Relationship of Ka and Kb

Ka x Kb = Kw = 1.0 x 1014

As the strength of an acid increases (larger Ka), the strength of it’s conjugate base must decrease (smaller Kb).

We can now calculate Kb for any base if we know Ka for conjugate acid and vice versa.

Ka x Kb = Kw = 1.0 x 1014

Taking the – log of both sides and simplifying . . . pKa + pKb = pKw = 14.00

Adding Salts to Water

When certain salts are dissolved in water they sometimes create an acidic or basic solution.

Many ions can react with water to produce H+ or OH- called a hydrolysis reaction.

In general, Anions which react with water produce basic solutions.

Cations which react with water produce acidic solutions. However, each salt must be examined carefully.

1. Salts that consist of cations of strong bases and the anions of strong acids have no effect on pH when dissolved in water.

2. Cations of strong bases (Na+, K+, group 1A)

3. Anions of strong acids (Cl-, NO3-)

Salts that produceNeutral Solutions

Salts that produceBasic Solutions

1. For any salt whose cation has neutral properties and whose anion is the conjugate base of a weak acid, the aqueous solution will be basic:

C2H3O2- + H2O HC2H3O2 + OH-

base acid acid base

Salts the produceAcidic Solutions

1. Salts in which the anion is not a base and the cation is the conjugate acid of a weak base produce acidic solutions

NH4+ NH3 + H+

1. Salts that possess a highly charged metallic ion, such as Al+3

Aluminum ion in water is hydrated Al(H2O)6+3

High metallic charge polarizes O-H bond in water

Hydrogens in water become acidic.

Examples

Adding NaC2H3O2 to water NaC2H3O2 Na+ + C2H3O2

-

C2H3O2- + H2O HC2H3O2 + OH-

the solution is basic

Example: NH4Cl NH4+ + Cl-

NH4+ + H2O NH3 + H3O+

the solution is acidic

In Summary

1. Cl-, Br-, I-, NO3-, ClO3

-, and ClO4-

Do not react with water = neutral solutions2. Anions from a weak acid (conjugate base)

= basic solution3. Cations from a weak base (conjugate acid)

= acidic solution4. Metal ions (except group IA metals and

Ca+2, Sr+2, Ba+2) + water = acidic solution

In Summary

5. Ions which are amphoteric, the biggest K value determines results

HPO4-2 can be acid or base in water,

HPO4-2 + OH- PO4

-3 + H3O+

HPO4-2 + H2O H2PO4

- + OH-

Ka = 6.2 x 10-8 and Kb = 1.6 x 10-7

Kb wins, solution is basic

WHY?

Structure and Acid/Base Properties Why are some acids weak, others strong, and

some neither? Molecular structure Any molecule with an H in it is a potential acid. The stronger the X-H bond the less acidic

(compare bond dissociation energies). The more polar the X-H bond the stronger the

acid (use electronegativities). The more polar H-O-X bond -stronger acid.

Strength of Oxyacids

The more oxygen hooked to the central atom, the more acidic the hydrogen.

HClO4 > HClO3 > HClO2 > HClO

Remember that the H is attached to an oxygen atom.

The oxygens are electronegative Pull electrons away from hydrogen

Strength of oxyacidsStrength of oxyacids

Electron Density

Cl O H

Hypochlorous acid

Strength of oxyacidsStrength of oxyacids

Electron Density

Cl O HO

Chlorous acid

Strength of oxyacidsStrength of oxyacids

Cl O H

O

O

Electron Density

Chloric acid

Strength of oxyacids

Cl O H

O

O

O

Electron Density

Perchloric acid

Hydrated metals

• Highly charged metal ions pull the electrons of surrounding water molecules toward them.

• Make it easier for H+ to come off.

• Metals + H2O = acidic (except Group IA metals and

Ca+2, Sr +2, and Ba+2)

Al+3 OH

H

Lewis Acids and Bases

Al+3 ( )H

HO

Al ( )6

H

HO

+ 6

+3

Don’t confuse this with Oxides + Water

• Metal oxides dissolve in water to produce bases.

• CaO(s) + H2O(l) Ca(OH)2(aq)

• Non-metal oxides dissolved in water can make acids.

• SO3 (g) + H2O(l) H2SO4(aq)

Strategy for solving Acid-Base Problems (In summary)

1. List the major species in solution

2. Look for reaction that can be assumed to go to completion, for a strong acid dissociating or H+ reacting with OH-

3. For a reaction that can be assumed to go to completion:

Determine the concentration of the products Write down the major species in solution after the

reaction

4. Look at each major component of the solution and decide if it is an acid or a base.

5. Pick the Equilibrium that will control pH. Use known value of the dissociation constants for the various species to help decide on the dominant equilibrium

Looking into step 5

Write the equation for the reaction and the equilibrium expression

Compute the initial concentrations (assuming dominant equilibrium hasn’t occurred yet)

Define x Compute equilbrium concentrations in terms of x Substitute the concentrations into the equilibrium

expression Check validitiy of approximation Calculate the pH and other concentrations required

Calculating the pH of Salts

First you must decide if the salt is acidic, basic, or neutral.

If it is basic: Anion is conjugate base of a weak acid (ion will undergo hydrolysis)

Determine the pH of a 0.100 M aqueous solution of NaCN. The Ka for HCN is 5.8x10-10.

First write the equilibrium expression for the solution. It is basic, so Kb expression

Use Ka and Kw to calculate Kb Construct an ice chart to determine expected

concentrations. Make assumptions and solve for x. Use the pOH to calculate the pH to finalize

question.

General Acid/Base Question:

The Kw for water at 25oC is 1.0x10-14, but is 1.0x10-13 at 60oC.

Give the chemical equation for the autoionizatoin of water.

Determine the [OH-] for water at 60oC. Determine the pH of water at 60oC. Is the reaction endothermic or exothermic?

Support your answer.

Phosphoric acid, H3PO4 is a triprotic acid:

Show the three equations involved in the dissociation of this substance.

Illustrate how these three equations might be combined to show the complete dissociation of phosphoric acid.

If a 7.0 M H3PO4 solution is dissociated, calculate the pH of the solution. Ka1=7.5x10-3; Ka2=6.2x10-8; Ka3=4.8x10-13

Determine the concentration for the ions: H2PO4-1 ;

HPO4-2 ; PO4

-3 ; Determine the pOH for the same 7.0M H3PO4 solution

Chapter 17

More Equilibrium ApplicationsMore Equilibrium ApplicationsBuffers, Titrations, and Ksp

The Common Ion Effect

When the salt with the anion of a weak acid is added to that acid, it reverses the dissociation of the acid.

HCN H+ + CN-

Adding NaCN to the solution shifts the equilibrium to the left. (“Common Ion Effect” is really LeChatelier’s Principle)

Lowers the percent ionization of the acid. The same principle applies to salts with the

cation of a weak base.

Common Ion Problem• What is the pH of a solution where 0.30 moles

of HC2H3O2 and 0.20 moles of NaC2H3O2 are added to make a 1.00 L solution?

First, identify the species and write the equation.

HC2H3O2 H+ + C2H3O2

Recognize that adding more C2H3O2 shifts to left,

and the [H+] and [C2H3O2] are not equal.

HC2H3O2 H+ + C2H3O2

I 0.30 0 0.20 C - x +x +x E 0.30 - x +x 0.20 + x

What is the pH of a solution where 0.30 moles of HC2H3O2 and 0.20 moles of NaC2H3O2 are

added to make a 1.00 L solution? HC2H3O2 H+

+ C2H3O2

I 0.30 0 0.20 C - x +x +x E 0.30 - x +x 0.20 + x

[H+] [C2H3O2] [x] [0.20 + x]

Ka = [HC2H3O2]

= [0.30 -x]

= 1.8 x 10 -5

X = [H+] = 2.7 x 10-5, pH = 4.57The resulting [H+] is very small, which justifies ignoring x.

Without the addition of 0.20 moles C2H3O2, the problem

works out to [H+] = 7.7 x 10-3, pH = 2.11

Buffers• Buffers are solutions that resist a change in pH.• Human blood is a complex aqueous mixture with

a pH buffered at 7.4• Seawater is an aqueous mixture buffered at 8.1 near

the surface.• A buffer is a weak acid and its salt, OR

a weak base and its salt. • Buffers contain both acidic species to neutralize OH-

and a basic species to neutralize H+ ions.

Buffers work to maintain pH levels because of the Common Ion Effect

Buffers• A buffer is a weak acid and its salt, OR a weak base and its salt.

HC2H3O2 and NaC2H3O2

NH3 and NH4Cl

• Example: buffer with equal concentrations of weak acid and its conjugate base

HC2H3O2/C2H3O2

Buffers• Example: buffer with equal concentrations of weak acid and its conjugate base HC2H3O2/C2H3O2

<---------->add OH, neutralizes the acid to form water

HC2H3O2 + OH ---> H2O + C2H3O2

after adding OH-, more C2H3O2 than before

add H+, they react with the base component of the buffer

H+ + C2H3O2 ---> HC2H3O2

after adding H+, more acid than beforeAdding OH- reduces [HC2H3O2] and increases [C2H3O2

]Adding H+ increases [HC2H3O2] and reduces [C2H3O2

]

Buffer Problem (common ion problem)

• Calculate the pH of a buffer that is

0.75 M lactic acid (HC3H5O3) and

0.25 M sodium lactate (Ka = 1.4 x 10-4)HC3H5O3 <-----> H+ + C3H5O3

I 0.75 0 0.25C -x +x +xE 0.75 - x x 0.25 +x

Ka = 1.4 x 10-4 = [H+] [C3H5O3] = [x] [0.25 +x]

[HC3H5O3] [0.75 -x]X = [H+] = 4.2 x 104 M pH = 3.37

Buffers• Buffers are most effective when

concentrations of weak acid and salt are the same.

• 2 Important characteristics:

1. Buffer capacity - the amount of acid or

base the buffer can neutralize before the

pH begins to change substantially

depends on amount of acid/base it’s made from

2. Buffer pH - depends on the Ka for the acid

Why?

General Weak Acid Equation

• Ka = [H+] [A-][HA]

• so [H+] = Ka [HA] [A-]

• The [H+] depends on the ratio [HA]/[A-]• taking the negative log of both sides• pH = -log(Ka [HA]/[A-])• pH = -log(Ka)-log([HA]/[A-])

• pH = pKa + log([A-]/[HA])

This is called the Henderson-Hasselbach Equation

pH = pKa + log([A-]/[HA])

pH = pKa + log(base/acid)

• Calculate the pH of the following mixtures:

a.) 0.75 M lactic acid (HC3H5O3) and 0.25 M

sodium lactate (Ka = 1.4 x 10-4)

b.) 0.25 M NH3 and 0.40 M NH4Cl

(Kb = 1.8 x 10-5)

pH =3.37

pH = 9.05

Buffer capacity

• The pH of a buffered solution is determined by the ratio [A-]/[HA].

• As long as this doesn’t change much, the pH won’t change much.

• The more concentrated these two are, the more H+ and OH- the solution will be able to absorb.

• Larger concentrations bigger buffer capacity.

Buffer capacity

• Henderson-Hasselbach Equation:

pH = pKa + log([A-]/[HA])• The best buffers have a ratio: [A-]/[HA] = 1• This is most resistant to change• True when [A-] = [HA]• Make pH = pKa (since log1=0)

Addition of Strong Acids or Bases to buffers:

A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol of NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.

First do the stoichiometry (Create a Before and after reaction chart) with the

neutralization reaction

• B: 0.3 mol 0.02 mol 0.3 mol

• HC2H3O2 + OH- H2O + C2H3O2-

• A: 0.28 mol 0 mol 0.32 mol

• NaOH was completely consumed by the weak acid component. Since OH- is limiting, all that was added would be consumed.

Equilibrium: Now we focus on the equilibrium that will determine the pH

of the buffer (i.e. the ionization of the acetic acid)

• HC2H3O2 H+ + C2H3O2-

• We then use the new quantities from the stoichiometry in the H.H. equation:

• pH = 4.74 + log (0.320 mol / 0.280 mol)

• pH = 4.80

Note: we can use mole amounts in place of concentrations in the H.H. equation.

Prove they’re buffers:

Determine the pH of a solution made by adding 0.020 mol of NaOH to 1.00 L of pure water.

pH = 14 – ( -log 0.020) pH = 12.30 Quite a BIG difference from the

previous problem

Another practice

A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol of NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of HCl is added.

Do the stoichiometry Use equilibrium info with H.H. pH = 4.68

One more…

Calculate the change in pH that occurs when 0.010 mol of HCl is added to 1.0 L of each of the following:

a.) 5.00 M HC2H3O2 and 5.00 M NaC2H3O2

b.) 0.050 M HC2H3O2 and 0.050 M NaC2H3O2

Ka= 1.8x10-5

Remember: Make pH = pKa to find the original pH

Last one…

What would the pH be if 0.050 mol of solid NaOH is added to a 1.0L mixture of:

1.40 M HC2H3O2 and 1.40 M NaC2H3O2

Ka= 1.8x10-5

pH = 4.74 + log (1.45/1.35)pH = 4.77

Titrations

• Adding a solution of known concentration until the substance being tested is consumed.

• This is called the equivalence point.

• The end point is a visible sign that the equivalence point has been reached.

• Graph of pH vs. mL is a titration curve.

Helpful to Remember: #moles # molesM = # moles = M x L L = L M

Strong acid with Strong Base• Do the stoichiometry. Use a BCA table.

• There is no equilibrium .

• They both dissociate completely.

• The titration of 50.0 mL of 0.200 M HNO3

with 0.300 M NaOH requires x ml NaOH?HNO3 + NaOH -----> H2O + NaNO3

B 0.01 mol 0.01 mol xs 0

C -0.01 -0.01 + 0.01 +0.01

A 0 0 xs 0.01

The titration of 50.0 mL of 0.200 M HNO3 with 0.300 M NaOH requires x ml NaOH?

# moles 0.01 mol NaOH

L = L = = 0.0333 L

M 0.300 M NaOH= 33.3 ml NaOH

Lets look at a titration curve of this . . .

Because they are both strongwe could also use:

MAVA = MBVB

pH

mL of Base added

7

• Strong acid with strong Base• Equivalence at pH 7

Weak acid with Strong base• There is an equilibrium.• Do stoichiometry. Then do equilibrium.• Titrate 50.0 mL of 0.10 M HF (Ka = 7.2 x 10-4) with 0.10 M

NaOH. How many ml of NaOH are needed?

HF + NaOH ---> H2O + NaF .0050 mol .0050 mol .0050 mol before after

Notice the product F- is a conjugate base.Some F- will react with water to produce OH-

How much? Do the equilibrium problem

It requires .0050 mol NaOH to neutralize the acid; L = .0050 mol / 0.10 M NaOH = .050 L = 50 ml NaOH The titration is at the equivalence point.

Titrate 50.0 mL of 0.10 M HF (Ka = 7.2 x 10-4) with 0.10 M NaOH. How many ml of NaOH are needed?

F- + H2O HF + OH-

I 0.050 0 0

C -x +x +x

E 0.050 -x x x [HF] [OH-]Kb = [ F- ]

x2

1.39 x 10-11 = .050 -x

X = 8.33 x 107 M OH-

pOH = 6.08 pH = 7.92

So, at the equivalencepoint, the mixture is basic

pH

mL of Base added

>7

Weak acid with strong Base Equivalence at pH >7

7

pH

mL of Base added

7

Strong base with strong acid Equivalence at pH 7

pH

mL of Base added

<7

Weak base with strong acid Equivalence at pH <7

7

• 75 mL of 1.5 M HF is titrated with 2.25 M KOH. Calculate the pH at the equivalence point.

• pH = 8.55

Summary

For a titration, any acid-base will neutralize and follow rules for stoichiometry. (BCA table)

If asked to find the pH of resulting weak acid - strong base titration:

– Stoichiometry first– Then Henderson-Hasselbach

Think about the chemistry involved. Don’t try to memorize steps.

Indicators• Weak acids that change color when they

become bases.• weak acid written HIn• Weak base• HIn H+ + In-

clear red• Equilibrium is controlled by pH• End point - when the indicator changes

color.

Indicators• Since it is an equilibrium the color change

is gradual.

• It is noticeable when the ratio of [In-]/[HI] or [HI]/[In-] is 1/10

• Since the Indicator is a weak acid, it has a Ka.

• pH the indicator changes at is.

• pH=pKa +log([In-]/[HI]) = pKa +log(1/10)

• pH=pKa - 1 on the way up

Indicators

• pH=pKa + log([HI]/[In-]) = pKa + log(10)

• pH=pKa+1 on the way down

• Choose the indicator with a pKa 1 less than the pH at equivalence point if you are titrating with base.

• Choose the indicator with a pKa 1 greater than the pH at equivalence point if you are titrating with acid.

Solubility Equilibria

Will it all dissolve, and if not, how much?

• All dissolving is an equilibrium.

• If there is not much solid it will all dissolve.

• As more solid is added the solution will become saturated.

• Solid dissolved particles

• Equilibrium: the rate that the solid dissolves is equal to the rate that the dissolved particles rejoin the solid.

Ksp• Metal ion bonded to nonmetal, then it dissociates.• AgCl (s) Ag+(aq) + Cl (aq)

[Ag+]1[Cl-]1

Keq =

[AgCl]• But the concentration of a solid doesn’t change.

So we leave it out.

• Ksp = [Ag+] [Cl-]

• Called the solubility product expression

• Ksp called the solubility product constant

Watch out

• Solubility is not the same as

solubility product.

• Solubility product is an equilibrium constant, (Ksp).

• It doesn’t change except by temperature.• Solubility is how much it can dissolve

Units: g/L (set by the equilibrium position)

• A common ion can change this.

Calculating KspThe solubility of Li2CO3 is 4.35 g/L

What is the Ksp?First, Always Write the Equation.Then write the Ksp expression.

change g/L into moles/L (M)

Li2CO3 2Li+ + CO3-2

Ksp = [Li+] 2 [CO3-2]

4.35 g/L ( 1 mol / 73.8 g ) = 0.05894 M

Ksp = [ 2 x 0.05894 M] 2 [0.05894 M]Ksp = 8.2 x 104

Calculating Solubility• The solubility is determined by equilibrium.

• Its an equilibrium problem.

• Calculate the solubility of AgCl.

Ksp = 1.8 x 10-10

Write the equation and the Ksp expression.

AgCl (s) Ag+(aq) + Cl (aq) Ksp = [Ag+] [ Cl ]

1.8 x 1010 = [ x ] [ x ]1.8 x 1010 = x2

x = [Ag+] [ Cl ] =1.34 x 105 M

1.34 x 105 mol/L x Molar Mass = 0.00192 g/L solubility

Relative solubilities

The bigger the Ksp the more soluble.

1 x 103

The smaller the Ksp, the less soluble.

1 x 1015

Ksp describes precipitates -- substances which dissolve less than 0.1 M

pKsp = log Ksp

Common Ion Effect

AgCl (s) Ag+(aq) + Cl (aq)

• If we add a common ion to a saturated solution, equilibrium shifts left, more solid results.

• If we try to dissolve the solid in a solution with either the cation or anion already present less will dissolve. (more solid results)

pH and solubility• OH- can be a common ion.

Zn(OH)2 Zn+2 + 2OH

• Substance is more soluble in acid.• For other anions if they come from a

weak acid they are more soluble in a acidic solution than in water.

• CaC2O4 Ca+2 + C2O4-2

• H+ + C2O4-2 HC2O4

-

• Reduces C2O4-2 in acidic solution.

• More solid CaC2O4 dissolves.

Predicting Precipitation• Ion Product, Q = [Ag+] [Cl-] • If Q > Ksp a precipitate forms.• If Q < Ksp No precipitate.• If Q = Ksp equilibrium.

Will a precipitate form when 100 ml of 0.050 M AgNO3 is added to 100 ml of 0.040 M NaCl?

Equation: AgNO3 + NaCl ---> AgCl (ppt) + NaNO3

Will a precipitate form when 100 ml of 0.050 M AgNO3 is added to 100 ml of 0.040 M NaCl?

• Equation: AgNO3 + NaCl ---> AgCl (ppt) + NaNO3

Find the M [ ] of each ion. New volume = 200 ml

.1L x .05 M = .005 mol Ag+ and .1 x .04 M = .004 mol Cl

0.200 L 0.200 L

= 0.025 M Ag+ = 0.02 M Cl

Ksp = [Ag+] [Cl]

1.8 x 1010 = Ksp

Ion product, Q = [0.025 M] [0.02 M] Q = .0005

Since Q > Ksp, a precipitate forms

Selective Precipitations

• Used to separate mixtures of metal ions in solutions.

• Add anions that will only precipitate certain metals at a time.

• Used to purify mixtures.

• Often use H2S because in acidic solution

Hg+2, Cd+2, Bi+3, Cu+2, Sn+4 will precipitate.

Selective Precipitation

• In Basic adding OH-solution S-2 will increase so more soluble sulfides will precipitate.

• Co+2, Zn+2, Mn+2, Ni+2, Fe+2, Cr(OH)3,

Al(OH)3

Selective precipitation

• Follow the steps first with insoluble chlorides (Ag, Pb, Ba)

• Then sulfides in Acid.

• Then sulfides in base.

• Then insoluble carbonate (Ca, Ba, Mg)

• Alkali metals and NH4+ remain in solution.

Complex ion Equilibria

• A charged ion surrounded by ligands.

• Ligands are Lewis bases using their lone pair to stabilize the charged metal ions.

• Common ligands are NH3, H2O, Cl-,CN-

• Coordination number is the number of attached ligands.

• Cu(NH3)4+2 has a coordination # of 4

The addition of each ligand has its own equilibrium

• Usually the ligand is in large excess.

• And the individual K’s will be large so we can treat them as if they go to equilibrium.

• The complex ion will be the biggest ion in solution.

• Calculate the concentrations of Ag+, Ag(S2O3)-, and Ag(S2O3)-3 in a solution

made by mixing 150.0 mL of AgNO3 with 200.0 mL of 5.00 M Na2S2O3

• Ag+ + S2O3-2 Ag(S2O3)- K1=7.4 x 108

• Ag(S2O3)- + S2O3-2 Ag(S2O3)-3

K2=3.9 x 104