chapter 15 acids and bases - center for nonlinear...

Chapter 15Acids and

Bases

2008, Prentice Hall

Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro

Roy KennedyMassachusetts Bay Community College

Wellesley Hills, MA

Tro, Chemistry: A Molecular Approach 2

Stomach Acid & Heartburn• the cells that line your stomach produce

hydrochloric acidto kill unwanted bacteriato help break down foodto activate enzymes that break down food

• if the stomach acid backs up into your esophagus, it irritates those tissues, resulting in heartburn

acid refluxGERD = gastroesophageal reflux disease = chronic leaking of stomach acid into the esophagus


Curing Heartburn

• mild cases of heartburn can be cured by neutralizing the acid in the esophagus

swallowing saliva which contains bicarbonate iontaking antacids that contain hydroxide ions and/or carbonate ions


Properties of Acids• sour taste• react with “active” metals

i.e., Al, Zn, Fe, but not Cu, Ag, or Au2 Al + 6 HCl → 2 AlCl3 + 3 H2

corrosive

• react with carbonates, producing CO2marble, baking soda, chalk, limestone

CaCO3 + 2 HCl → CaCl2 + CO2 + H2O• change color of vegetable dyes

blue litmus turns red

• react with bases to form ionic salts


Common AcidsChemical Name Formula Uses Strength

Nitric Acid HNO3 explosive, fertilizer, dye, glue Strong

Sulfuric Acid H2SO4 explosive, fertilizer, dye, glue,

batteries Strong

Hydrochloric Acid HCl metal cleaning, food prep, ore refining, stomach acid Strong

Phosphoric Acid H3PO4 fertilizer, plastics & rubber,

food preservation Moderate

Acetic Acid HC2H3O2plastics & rubber, food preservation, Vinegar Weak

Hydrofluoric Acid HF metal cleaning, glass etching Weak

Carbonic Acid H2CO3 soda water Weak

Boric Acid H3BO3 eye wash Weak


Structures of Acids

• binary acids have acid hydrogens attached to a nonmetal atom

HCl, HF


Structure of Acids• oxy acids have acid hydrogens attached to

an oxygen atomH2SO4, HNO3


Structure of Acids• carboxylic acids have

COOH groupHC2H3O2, H3C6H5O7

• only the first H in the formula is acidic

the H is on the COOH


Properties of Bases• also known as alkalis• taste bitter

alkaloids = plant product that is alkalineoften poisonous

• solutions feel slippery• change color of vegetable dyes

different color than acidred litmus turns blue

• react with acids to form ionic saltsneutralization


Common BasesChemical

Name Formula Common Name Uses Strength

sodium hydroxide NaOH lye,

caustic soda soap, plastic,

petrol refining Strong

potassium hydroxide KOH caustic potash soap, cotton,

electroplating Strong

calcium hydroxide Ca(OH)2 slaked lime cement Strong

sodium bicarbonate NaHCO3 baking soda cooking, antacid Weak

magnesium hydroxide Mg(OH)2

milk of magnesia antacid Weak

ammonium hydroxide

NH4OH, {NH3(aq)}

ammonia water

detergent, fertilizer,

explosives, fibers Weak


Structure of Bases

• most ionic bases contain OH ionsNaOH, Ca(OH)2

• some contain CO32- ions

CaCO3 NaHCO3

• molecular bases contain structures that react with H+

mostly amine groups


Indicators• chemicals which change color depending on

the acidity/basicity• many vegetable dyes are indicators

anthocyanins• litmus

from Spanish mossred in acid, blue in base

• phenolphthaleinfound in laxativesred in base, colorless in acid


Arrhenius Theory• bases dissociate in water to produce OH- ions and

cationsionic substances dissociate in water

NaOH(aq) → Na+(aq) + OH–(aq)• acids ionize in water to produce H+ ions and anions

because molecular acids are not made of ions, they cannot dissociate they must be pulled apart, or ionized, by the water

HCl(aq) → H+(aq) + Cl–(aq)in formula, ionizable H written in front

HC2H3O2(aq) → H+(aq) + C2H3O2–(aq)


Arrhenius Theory

HCl ionizes in water,producing H+ and Cl– ions

NaOH dissociates in water,producing Na+ and OH– ions


Hydronium Ion• the H+ ions produced by the acid are so reactive they

cannot exist in waterH+ ions are protons!!

• instead, they react with a water molecule(s) to produce complex ions, mainly hydronium ion, H3O+

H+ + H2O → H3O+

there are also minor amounts of H+ with multiple water molecules, H(H2O)n

+


Arrhenius Acid-Base Reactions

• the H+ from the acid combines with the OH-

from the base to make a molecule of H2Oit is often helpful to think of H2O as H-OH

• the cation from the base combines with the anion from the acid to make a salt

acid + base → salt + waterHCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)


Problems with Arrhenius Theory• does not explain why molecular substances, like

NH3, dissolve in water to form basic solutions –even though they do not contain OH– ions

• does not explain how some ionic compounds, like Na2CO3 or Na2O, dissolve in water to form basic solutions – even though they do not contain OH–

ions• does not explain why molecular substances, like

CO2, dissolve in water to form acidic solutions –even though they do not contain H+ ions

• does not explain acid-base reactions that take place outside aqueous solution


Brønsted-Lowry Theory• in a Brønsted-Lowry Acid-Base reaction, an

H+ is transferreddoes not have to take place in aqueous solutionbroader definition than Arrhenius

• acid is H donor, base is H acceptorbase structure must contain an atom with an unshared pair of electrons

• in an acid-base reaction, the acid molecule gives an H+ to the base molecule

H–A + :B ⇔ :A– + H–B+


Brønsted-Lowry Acids• Brønsted-Lowry acids are H+ donors

any material that has H can potentially be a Brønsted-Lowry acidbecause of the molecular structure, often one H in the molecule is easier to transfer than others

• HCl(aq) is acidic because HCl transfers an H+ to H2O, forming H3O+ ions

water acts as base, accepting H+

HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)acid base


Brønsted-Lowry Bases• Brønsted-Lowry bases are H+ acceptors

any material that has atoms with lone pairs can potentially be a Brønsted-Lowry basebecause of the molecular structure, often one atom in the molecule is more willing to accept H+ transfer than others

• NH3(aq) is basic because NH3 accepts an H+

from H2O, forming OH–(aq)water acts as acid, donating H+

NH3(aq) + H2O(l) ⇔ NH4+(aq) + OH–(aq)

base acid


Amphoteric Substances• amphoteric substances can act as either an

acid or a basehave both transferable H and atom with lone pair

• water acts as base, accepting H+ from HClHCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)

• water acts as acid, donating H+ to NH3

NH3(aq) + H2O(l) ⇔ NH4+(aq) + OH–(aq)


Brønsted-Lowry Acid-Base Reactions

• one of the advantages of Brønsted-Lowry theory is that it allows reactions to be reversible

H–A + :B ⇔ :A– + H–B+

• the original base has an extra H+ after the reaction – so it will act as an acid in the reverse process

• and the original acid has a lone pair of electrons after the reaction – so it will act as a base in the reverse process

:A– + H–B+ ⇔ H–A + :B


Conjugate Pairs• In a Brønsted-Lowry Acid-Base reaction, the

original base becomes an acid in the reverse reaction, and the original acid becomes a base in the reverse process

• each reactant and the product it becomes is called a conjugate pair

• the original base becomes the conjugate acid; and the original acid becomes the conjugate base


Brønsted-Lowry Acid-Base Reactions

H–A + :B ⇔ :A– + H–B+

acid base conjugate conjugatebase acid

HCHO2 + H2O ⇔ CHO2– + H3O+


H2O + NH3 ⇔ HO– + NH4+



Conjugate PairsIn the reaction H2O + NH3 ⇔ HO– + NH4

+

H2O and HO– constitute an Acid/Conjugate Base pair

NH3 and NH4+ constitute a

Base/Conjugate Acid pair


Ex 15.1a – Identify the Brønsted-Lowry Acids and Bases and Their Conjugates in the Reaction

H2SO4 + H2O ⇔ HSO4– + H3O+


H2SO4 + H2O ⇔ HSO4– + H3O+

When the H2SO4 becomes HSO4−, it lost an H+ − so

H2SO4 must be the acid and HSO4− its conjugate base

When the H2O becomes H3O+, it accepted an H+ − so H2O must be the base and H3O+ its conjugate acid


Ex 15.1b – Identify the Brønsted-Lowry Acids and Bases and Their Conjugates in the Reaction

HCO3– + H2O ⇔ H2CO3 + HO–

base acid conjugate conjugateacid base

HCO3– + H2O ⇔ H2CO3 + HO–

When the HCO3− becomes H2CO3, it accepted an H+ −

so HCO3− must be the base and H2CO3 its conjugate acid

When the H2O becomes OH−, it donated an H+ − so H2O must be the acid and OH− its conjugate base


Practice – Write the formula for the conjugate acid of the following

H2O

NH3

CO32−

H2PO41−


Practice – Write the formula for the conjugate acid of the following

H2O H3O+

NH3 NH4+

CO32− HCO3

−

H2PO41− H3PO4


Practice – Write the formula for the conjugate base of the following

H2O

NH3

CO32−

H2PO41−


Practice – Write the formula for the conjugate base of the following

H2O HO−

NH3 NH2−

CO32− since CO3

2− does not have an H, it cannot be an acid

H2PO41− HPO4

2−


Arrow Conventions• chemists commonly use two kinds

of arrows in reactions to indicate the degree of completion of the reactions

• a single arrow indicates all the reactant molecules are converted to product molecules at the end

• a double arrow indicates the reaction stops when only some of the reactant molecules have been converted into products

⇔ in these notes


Strong or Weak• a strong acid is a strong electrolyte

practically all the acid molecules ionize, →• a strong base is a strong electrolyte

practically all the base molecules form OH– ions, either through dissociation or reaction with water, →

• a weak acid is a weak electrolyteonly a small percentage of the molecules ionize, ⇔

• a weak base is a weak electrolyteonly a small percentage of the base molecules form OH– ions, either through dissociation or reaction with water, ⇔


Strong Acids• The stronger the acid, the

more willing it is to donate Huse water as the standard base

• strong acids donate practically all their H’s

100% ionized in waterstrong electrolyte

• [H3O+] = [strong acid]

HCl → H+ + Cl-

HCl + H2O→ H3O+ + Cl-


Weak Acids• weak acids donate a small

fraction of their H’smost of the weak acid molecules do not donate H to watermuch less than 1% ionized in water

• [H3O+] << [weak acid]

HF ⇔ H+ + F-

HF + H2O ⇔ H3O+ + F-


Polyprotic Acids• often acid molecules have more than one ionizable H –

these are called polyprotic acidsthe ionizable H’s may have different acid strengths or be equal1 H = monoprotic, 2 H = diprotic, 3 H = triprotic

HCl = monoprotic, H2SO4 = diprotic, H3PO4 = triprotic

• polyprotic acids ionize in stepseach ionizable H removed sequentially

• removing of the first H automatically makes removal of the second H harder

H2SO4 is a stronger acid than HSO4−


Acids Conjugate BasesHClO4 ClO4

-1 H2SO4 HSO4

-1 HI I-1

HBr Br-1 HCl Cl-1

HNO3 NO3-1

H3O+1 H2O HSO4

-1 SO4-2

H2SO3 HSO3-1

H3PO4 H2PO4-1

HNO2 NO2-1

HF F-1 HC2H3O2 C2H3O2

-1 H2CO3 HCO3

-1 H2S HS-1

NH4+1 NH3

HCN CN-1 HCO3

-1 CO3-2

HS-1 S-2 H2O OH-1

CH3-C(O)-CH3 CH3-C(O)-CH2-1

NH3 NH2-1

CH4 CH3-1

OH-1 O-2

Incr

easi

ng A

cidi

ty

Increasing Basicity


Strengths of Acids & Bases• commonly, acid or base strength is measured by

determining the equilibrium constant of a substance’s reaction with water

HAcid + H2O ⇔ Acid-1 + H3O+1

Base: + H2O ⇔ HBase+1 + OH-1

• the farther the equilibrium position lies to the products, the stronger the acid or base

• the position of equilibrium depends on the strength of attraction between the base form and the H+

stronger attraction means stronger base or weaker acid


General Trends in Acidity• the stronger an acid is at donating H, the

weaker the conjugate base is at accepting H• higher oxidation number = stronger oxyacid

H2SO4 > H2SO3; HNO3 > HNO2

• cation stronger acid than neutral molecule; neutral stronger acid than anion

H3O+1 > H2O > OH-1; NH4+1 > NH3 > NH2

-1

base trend opposite


Acid Ionization Constant, Ka

• acid strength measured by the size of the equilibrium constant when react with H2O

HAcid + H2O ⇔ Acid-1 + H3O+1

• the equilibrium constant is called the acid ionization constant, Ka

larger Ka = stronger acid

[HAcid]]O[H][Acid 1

31

a

+− ×=K

Name Formula Ka1 Ka2 Ka3 Benzoic C6H5COOH 6.14 x 10-5

Propanoic CH3CH2COOH 1.34 x 10-5 Formic HCOOH 1.77 x 10-5 Acetic CH3COOH 1.75 x 10-5

Chloroacetic ClCH2COOH 1.36 x 10-5 Trichloroacetic Cl3C-COOH 1.29 x 10-4

Oxalic HOOC-COOH 5.90 x 10-2 6.40 x 10-5 Nitric HNO3 strong

Nitrous HNO2 4.6 x 10-4 Phosphoric H3PO4 7.52 x 10-3 6.23 x 10-8 2.2 x 10-13

Phosphorous H3PO3 1.00 x 10-2 2.6 x 10-7 Arsenic H3AsO4 6.0 x 10-3 1.05 x 10-7 3.0 x 10-12

Arsenious H3AsO3 6.0 x 10-10 3.0 x 10-14 very small Perchloric HClO4 > 108

Chloric HClO3 5 x 102 Chlorous HClO2 1.1 x 10-2

Hypochlorous HClO 3.0 x 10-8 Boric H3BO3 5.83 x 10-10

Carbonic H2CO3 4.45 x 10-7 4.7 x 10-11


Autoionization of Water• Water is actually an extremely weak electrolyte

therefore there must be a few ions present• about 1 out of every 10 million water molecules

form ions through a process called autoionization

H2O ⇔ H+ + OH–

H2O + H2O ⇔ H3O+ + OH–

• all aqueous solutions contain both H3O+ and OH–

the concentration of H3O+ and OH– are equal in water[H3O+] = [OH–] = 10-7M @ 25°C


Ion Product of Water• the product of the H3O+ and OH–

concentrations is always the same number• the number is called the ion product of

water and has the symbol Kw• [H3O+] x [OH–] = Kw = 1 x 10-14 @ 25°C

if you measure one of the concentrations, you can calculate the other

• as [H3O+] increases the [OH–] must decrease so the product stays constant

inversely proportional


Acidic and Basic Solutions• all aqueous solutions contain both H3O+ and

OH– ions• neutral solutions have equal [H3O+] and [OH–]

[H3O+] = [OH–] = 1 x 10-7

• acidic solutions have a larger [H3O+] than [OH–][H3O+] > 1 x 10-7; [OH–] < 1 x 10-7

• basic solutions have a larger [OH–] than [H3O+][H3O+] < 1 x 10-7; [OH–] > 1 x 10-7

Example 15.2b – Calculate the [OH−] at 25°C when the [H3O+] = 1.5 x 10-9 M, and determine if the solution is

acidic, basic, or neutral

The units are correct. The fact that the [H3O+] < [OH−] means the solution is basic

[H3O+] = 1.5 x 10-9 M

[OH−]

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

[H3O+] [OH−]

]-][OHOH[ 3w+=K

]OH[]-OH[

]-OH][OH[

3

w

3w

+

+

=

=K

KM 107.6

105.1100.1]-[OH 6

9

14−

−

−×=

××

=


Complete the Table[H+] vs. [OH-]

OH-H+ H+ H+ H+ H+

OH-OH-OH-OH-

[OH-]

[H+] 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14


Complete the Table[H+] vs. [OH-]

OH-H+ H+ H+ H+ H+

OH-OH-OH-OH-

[OH-]10-14 10-13 10-11 10-9 10-7 10-5 10-3 10-1 100

[H+] 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14

even though it may look like it, neither H+ nor OH- will ever be 0the sizes of the H+ and OH- are not to scale

because the divisions are powers of 10 rather than units

Acid Base


pH• the acidity/basicity of a solution is often

expressed as pH• pH = -log[H3O+], [H3O+] = 10-pH

exponent on 10 with a positive signpHwater = -log[10-7] = 7need to know the [H+] concentration to find pH

• pH < 7 is acidic; pH > 7 is basic, pH = 7 is neutral


Sig. Figs. & Logs• when you take the log of a number written in scientific

notation, the digit(s) before the decimal point come from the exponent on 10, and the digits after the decimal point come from the decimal part of the number

log(2.0 x 106) = log(106) + log(2.0)= 6 + 0.30303… = 6.30303...

• since the part of the scientific notation number that determines the significant figures is the decimal part, the sig figs are the digits after the decimal point in the log

log(2.0 x 106) = 6.30

51

pH• the lower the pH, the more acidic the solution; the

higher the pH, the more basic the solution1 pH unit corresponds to a factor of 10 difference in acidity

• normal range 0 to 14pH 0 is [H+] = 1 M, pH 14 is [OH–] = 1 MpH can be negative (very acidic) or larger than 14 (very alkaline)

52

pH of Common Substances

141.0 M NaOH10.5 to 11.5household ammonia

10.5milk of magnesia (sat’d Mg(OH)2)7.6 to 8.0egg whites7.3 to 7.4human blood

5.6unpolluted rainwater3.2 to 4.0cherries2.9 to 3.3apples2.8 to 3.0plums2.0 to 4.0soft drinks2.2 to 2.4lemons1.0 to 3.0stomach acid

1.00.1 M HCl0.01.0 M HClpHSubstance

Example 15.3b – Calculate the pH at 25°C when the [OH−] = 1.3 x 10-2 M, and determine if the solution is


pH is unitless. The fact that the pH > 7 means the solution is basic

[OH−] = 1.3 x 10-2 M

pH

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

]-][OHOH[ 3w+=K

2

14

3

3w

103.1100.1]OH[

]-OH][OH[

−

−+

+

××

=

=K M 107.7]O[H 133

−+ ×=

[H3O+][OH−] pH

]OH[log pH 3- +=

( )12.11 pH

107.7log- pH 13

=×= −


pOH• another way of expressing the acidity/basicity of

a solution is pOH• pOH = -log[OH−], [OH−] = 10-pOH

pOHwater = -log[10-7] = 7need to know the [OH−] concentration to find pOH

• pOH < 7 is basic; pOH > 7 is acidic, pOH = 7 is neutral


pH and pOHComplete the Table

OH-H+ H+ H+ H+ H+

OH-OH-OH-OH-

[OH-]10-14 10-13 10-11 10-9 10-7 10-5 10-3 10-1 100

[H+] 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14

pH

pOH


pH and pOHComplete the Table

OH-H+ H+ H+ H+ H+

OH-OH-OH-OH-

[OH-]10-14 10-13 10-11 10-9 10-7 10-5 10-3 10-1 100

[H+] 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14

pH 0 1 3 5 7 9 11 13 14

pOH 14 13 11 9 7 5 3 1 0


Relationship between pH and pOH• the sum of the pH and pOH of a solution = 14.00

at 25°Ccan use pOH to find pH of a solution

( )( ) ( )( )( ) ( )( )

14.00pOHpH00.14]-[OHlog]OH[log

100.1log]-][OHOH[log

100.1]-][OHOH[

3

143

14w3

=+=−+−

×−=−

×==

+

−+

−+ K


pK• a way of expressing the strength of an acid or

base is pK• pKa = -log(Ka), Ka = 10-pKa

• pKb = -log(Kb), Kb = 10-pKb

• the stronger the acid, the smaller the pKalarger Ka = smaller pKa

because it is the –log


Finding the pH of a Strong Acid• there are two sources of H3O+ in an aqueous

solution of a strong acid – the acid and the water• for the strong acid, the contribution of the water

to the total [H3O+] is negligibleshifts the Kw equilibrium to the left so far that [H3O+]water is too small to be significant

except in very dilute solutions, generally < 1 x 10-4 M

• for a monoprotic strong acid [H3O+] = [HAcid]for polyprotic acids, the other ionizations can generally be ignored

• 0.10 M HCl has [H3O+] = 0.10 M and pH = 1.00


Finding the pH of a Weak Acid• there are also two sources of H3O+ in and

aqueous solution of a weak acid – the acid and the water

• however, finding the [H3O+] is complicated by the fact that the acid only undergoes partial ionization

• calculating the [H3O+] requires solving an equilibrium problem for the reaction that defines the acidity of the acid

HAcid + H2O ⇔ Acid− + H3O+


[NO2-]

equilibriumchangeinitial

[H3O+][HNO2]

Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution @ 25°C

Enter the initial concentrations –assuming the [H3O+] from water is ≈ 0

Construct an ICE table for the reaction

Write the reaction for the acid with water

since no products initially, Qc = 0, and the reaction is proceeding forward

HNO2 + H2O ⇔ NO2− + H3O+

0[NO2

-]

equilibriumchange

≈ 00.200initial[H3O+][HNO2]


0[NO2

-]

equilibriumchange

00.200initial[H3O+][HNO2]


substitute into the equilibrium constant expression

sum the columns to find the equilibrium concentrations in terms of x

represent the change in the concentrations in terms of x

+x+x−x0.200 −x x x

[ ]( )( )

( )xxxK

−×== −

+

12

3-2

a 1000.2HNO]OH][[NO



[ ][ ][ ]

( )( )( )x

xxK−×

== −

+

12

3-2

a 1000.2HNOOHNO

since Ka is very small, approximate the [HNO2]eq = [HNO2]initand solve for x

determine the value of Ka from Table 15.5

[ ][ ][ ]

( )( )( )1

2

3-2

a 1000.2HNOOHNO

−

+

×==

xxK

1

24

1000.2106.4 −

−

×=×

x

( )( )3

14

106.9

1000.2106.4−

−−

×=

××=

x

x

Ka for HNO2 = 4.6 x 10-4

x+x0

[NO2-]

x0.200equilibrium+x-xchange≈ 00.200initial

[H3O+][HNO2]

0.200 −x



Ka for HNO2 = 4.6 x 10-4

x+x0

[NO2-]


[H3O+][HNO2]check if the approximation is valid by seeing if x < 5% of [HNO2]init

%5%8.4%1001000.2

106.91

3<=×

××

−

−

the approximation is valid

x = 9.6 x 10-3



Ka for HNO2 = 4.6 x 10-4

x+x0

[NO2-]

x0.200-xequilibrium+x-xchange≈ 00.200initial

[H3O+][HNO2]

x = 9.6 x 10-3

substitute x into the equilibrium concentration definitions and solve

[ ] ( ) M 190.0106.9200.0200.0HNO 32 =×−=−= −x

[ ] [ ] M 106.9OHNO 33

-2

−+ ×=== x

0.0096

+x0

[NO2-]

0.00960.190equilibrium+x-xchange≈ 00.200initial

[H3O+][HNO2]



Ka for HNO2 = 4.6 x 10-4

substitute [H3O+] into the formula for pH and solve

0.0096

+x0

[NO2-]


[H3O+][HNO2]

( )( ) 02.2106.9log

OH-logpH3

3

=×−=

=−

+



Ka for HNO2 = 4.6 x 10-4

0.0096

+x0

[NO2-]


[H3O+][HNO2]check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [ ][ ]

[ ]( )

( )4

23

2

3-2

a

109.4190.0106.9

HNOOHNO

−−

+

×=×

=

==Kthough not exact, the answer is reasonably close


Practice - What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2? (Ka = 1.4 x 10-5 @ 25°C)


Practice - What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2?




HC6H4NO2 + H2O ⇔ C6H4NO2− + H3O+

0[A-]

equilibriumchange

≈ 00.012initial[H3O+][HA]


0[A-]

equilibriumchange

00.012initial[H3O+][HA]

Practice - What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2?




+x+x−x0.012 −x x x

[ ]( )( )

( )xxxK

−×== −

+

2246

3-246

a 102.1NOHHC]OH][NOH[C



[ ][ ][ ]

( )( )( )x

xxK−×

== −

+

23

-

a 102.1HAOHA

since Ka is very small, approximate the [HA]eq = [HA]init and solve for x

determine the value of Ka

[ ][ ][ ]

( )( )( )2

3-

a 102.1HAOHA

−

+

×==

xxK

2

25

102.1104.1 −

−

×=×

x

( )( )4

25

101.4

102.1104.1−

−−

×=

××=

x

x

x+x0

[A2-]


[H3O+][HA]

0.012 −x

Practice - What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2? Ka = 1.4 x 10-5 @ 25°C




Ka for HC6H4NO2 = 1.4 x 10-5

check if the approximation is valid by seeing if x < 5% of [HC6H4NO2]init

%5%4.3%100102.1101.4

2

4<=×

××

−

−


x = 4.1 x 10-4

x+x0

[A2-]


[H3O+][HA]



x = 4.1 x 10-4


[ ] ( ) M 012.0101.4012.0012.0NOHHC 4246 =×−=−= −x

[ ] [ ] M 101.4OHNOHC 43

-246

−+ ×=== x

x+x0

[A2-]


[H3O+][HA]




( )( ) 39.3101.4log

OH-logpH4

3

=×−=

=−

+

0.00041

+x0

[A2-]


[H3O+][HA]



check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka

the values match

0.00041

+x0

[A2-]


[H3O+][HA]

[ ]( )( )

52

24

246

3-246

a

104.1102.1101.4

NOHHC]OH][NOH[C

−−

−

+

×=××

=

=K


0[ClO2

-]

equilibriumchange

≈ 00.100initial[H3O+][HClO2]

Ex 15.7 Find the pH of 0.100 M HClO2(aq) solution @ 25°C



Write the reaction for the acid with water HClO2 + H2O ⇔ ClO2

− + H3O+






[ ]( )( )

( )xxxK

−×== −

+

12

3-2

a 1000.1HClO]OH][[ClO

x

+x0

[ClO2-]


[H3O+][HClO2]



since Ka is very small, approximate the [HClO2]eq = [HClO2]initand solve for x


[ ][ ][ ]

( )( )( )1

2

3-2

a 1000.1HClOOHClO

−

+

×==

xxK

1

22

1000.1101.1 −

−

×=×

x

( )( )2

12

103.3

1000.1101.1−

−−

×=

××=

x

x

Ka for HClO2 = 1.1 x 10-2

x

+x0

[ClO2-]


[H3O+][HClO2]




check if the approximation is valid by seeing if x < 5% of [HNO2]init

%5%33%1001000.1

103.31

2>=×

××

−

−

the approximation is invalid

x = 3.3 x 10-2

x

+x0

[ClO2-]


[H3O+][HClO2]



[ ][ ][ ]

( )( )( )x

xxK−×

== −

+

12

3-2

a 1000.1HClOOHClO

( )xx

−×=× −

−1

22

1000.1101.1

( )

0.039-or 028.0)1(2

)0011.0)(1(4011.0011.0

0011.0011.002

2

=

−−±−=

−+=

x

x

xx


if the approximation is invalid, solve for xusing the quadratic formula




x = 0.028


[ ] ( ) M 072.0028.0100.0100.0HClO2 =−=−= x

[ ] [ ] M 028.0OHClO 3-2 === + x

x

+x0

[ClO2-]


[H3O+][HClO2]

0.028

+x0

[ClO2-]


[H3O+][HClO2]





( )( ) 55.1028.0log

OH-logpH 3

=−== +

0.028

+x0

[ClO2-]


[H3O+][HClO2]




check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [ ][ ]

[ ]( )( )

22

2

3-2

a

101.1072.0028.0

HClOOHClO

−

+

×==

==Kthe answer matches

0.028

+x0

[ClO2-]


[H3O+][HClO2]


Ex 15.8 - What is the Ka of a weak acid if a 0.100 M solution has a pH of 4.25?

Use the pH to find the equilibrium [H3O+]

Enter the initial concentrations and [H3O+]equil



HA + H2O ⇔ A− + H3O+

[A-]


[H3O+][HA]

M 106.51010]OH[ 525.4-pH3

−−+ ×===

0[A-]

5.6E-05equilibriumchange

≈ 00.100initial[H3O+][HA]


0[A-]

equilibrium

change00.100initial

[H3O+][HA]

Ex 15.8 - What is the Ka of a weak acid if a 0.100 M solution has a pH of 4.25?

if the difference is insignificant, [HA]equil = [HA]initial

substitute into the Kaexpression and compute Ka

fill in the rest of the table using the [H3O+] as a guide

+5.6E-05+5.6E-05−5.6E-05

0.100 −5.6E-05

5.6E-05 5.6E-05

[ ]( )( )

( )8

a

553

-

a

101.3100.0

106.5106.5HA

]OH][[A

−

−−+

×=

××==

K

K

HA + H2O ⇔ A− + H3O+

0.100


Percent Ionization• another way to measure the strength of an acid is

to determine the percentage of acid molecules that ionize when dissolved in water – this is called the percent ionization

the higher the percent ionization, the stronger the acid

• since [ionized acid]equil = [H3O+]equil

%100acid ofmolarity initialacid ionized ofmolarity IonizationPercent ×=

%100[HA]

]O[HIonizationPercent

init

equil3 ×=+


Ex 15.9 - What is the percent ionization of a 2.5 M HNO2 solution?

Sum the columns to define the Equilibrium Concentrations

Define the Change in Concentration in terms of x

Enter the Initial Concentrations



HNO2 + H2O ⇔ NO2− + H3O+

[NO2-]


[H3O+][HNO2]0

[NO2-]

equilibriumchange

≈ 02.5initial[H3O+][HNO2]

+x+x−x2.5 − x x x


since Ka is very small, approximate the [HNO2]eq = [HNO2]initand solve for x


[ ][ ][ ]

( )( )( )5.2HNO

OHNO

2

3-2

axxK ==

+ 5.2106.4

24 x

=× −

( )( )2

4

104.3

5.2106.4−

−

×=

×=

x

x

Ka for HNO2 = 4.6 x 10-4

x

+x0

[NO2-]

x2.5-x ≈2.5equilibrium+x-xchange≈ 02.5initial

[H3O+][HNO2]




HNO2 + H2O ⇔ NO2− + H3O+

0.034

+x0

[NO2-]


[H3O+][HNO2]

2.5 − x x x

substitute x into the Equilibrium Concentration definitions and solve

[ ] ( ) M 5.2034.05.25.2HNO2 =−=−= x

[ ] [ ] M 034.0OHNO 3-2 === + x

x = 3.4 x 10-2



HNO2 + H2O ⇔ NO2− + H3O+

0.034

+x0

[NO2-]


[H3O+][HNO2]Apply the Definition and Compute the Percent Ionization

%4.1%1005.2104.3

%100][HNO

]O[HIonizationPercent

2init2

equil3

=××

=

×=

−

+since the percent ionization is < 5%, the “x is small” approximation is valid


Relationship Between [H3O+]equilibrium & [HA]initial

• increasing the initial concentration of acid results in increased H3O+

concentration at equilibrium• increasing the initial concentration

of acid results in decreased percent ionization

• this means that the increase in H3O+

concentration is slower than the increase in acid concentration

%100[HA]

]O[H

IonizationPercent

init

equil3 ×=+

93

Why doesn’t the increase in H3O+

keep up with the increase in HA?• the reaction for ionization of a weak acid is:

HA(aq) + H2O(l) ⇔ A−(aq) + H3O+

(aq)

• according to Le Châtelier’s Principle, if we reduce the concentrations of all the (aq) components, the equilibrium should shift to the right to increase the total number of dissolved particles

we can reduce the (aq) concentrations by using a more dilute initial acid concentration

• the result will be a larger [H3O+] in the dilute solution compared to the initial acid concentration

• this will result in a larger percent ionization


Finding the pH of Mixtures of Acids• generally, you can ignore the contribution of the

weaker acid to the [H3O+]equil

• for a mixture of a strong acid with a weak acid, the complete ionization of the strong acid provides more than enough [H3O+] to shift the weak acid equilibrium to the left so far that the weak acid’s added [H3O+] is negligible

• for mixtures of weak acids, generally only need to consider the stronger for the same reasons

as long as one is significantly stronger than the other, and their concentrations are similar


Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)


If the Kas are sufficiently different, use the strongest acid to construct an ICE table for the reaction

Write the reactions for the acids with water and determine their Kas

HF + H2O ⇔ F− + H3O+ Ka = 3.5 x 10-4

0[F-]

equilibriumchange

≈ 00.150initial[H3O+][HF]

HClO + H2O ⇔ ClO− + H3O+ Ka = 2.9 x 10-8

H2O + H2O ⇔ OH− + H3O+ Kw = 1.0 x 10-14


0[F-]

equilibriumchange

00.150initial[H3O+][HF]





+x+x−x0.150 −x x x

[ ]( )( )

( )xxxK

−×== −

+

13

-

a 1050.1HF]OH][[F



[ ][ ][ ]

( )( )( )x

xxK−

==+

150.0HFOHF 3

-

a

since Ka is very small, approximate the [HF]eq = [HF]init and solve for x

determine the value of Ka for HF

[ ][ ][ ]

( )( )( )150.0HF

OHF 3-

axxK ==

+

1

24

1050.1105.3 −

−

×=×

x

( )( )3

14

102.7

1050.1105.3−

−−

×=

××=

x

x

Ka for HF = 3.5 x 10-4

x+x0

[F-]


[H3O+][HF]

0.150 −x



Ka for HF = 3.5 x 10-4

x+x0

[F-]


[H3O+][HF]check if the approximation is valid by seeing if x < 5% of [HF]init

%5%8.4%1001050.1102.7

1

3<=×

××

−

−


x = 7.2 x 10-3



Ka for HF = 3.5 x 10-4

x+x0

[F-]


[H3O+][HF]

x = 7.2 x 10-3


[ ] ( ) M 143.0102.7150.0150.0HF 3 =×−=−= −x

[ ] [ ] M 102.7OHF 33

- −+ ×=== x

0.0072

+x0

[F-]


[H3O+][HF]



Ka for HF = 3.5 x 10-4


( )( ) 14.2102.7log

OH-logpH3

3

=×−=

=−

+

0.0072

+x0

[F-]


[H3O+][HF]



Ka for HF = 3.5 x 10-4


[ ]( )

( )4

23

3-

a

106.3143.0102.7

HFOHF

−−

+

×=×

=

==Kthough not exact, the answer is reasonably close

0.0072

+x0

[F-]


[H3O+][HF]


NaOH → Na+ + OH-

Strong Bases• the stronger the base, the more

willing it is to accept Huse water as the standard acid

• for strong bases, practically all molecules are dissociated into OH– or accept H’s

strong electrolytemulti-OH strong bases completely dissociated

• [HO–] = [strong base] x (# OH)

Example 15.11b – Calculate the pH at 25°C of a 0.0015 M Sr(OH)2 solution and determine if the solution is acidic,

basic, or neutral


[Sr(OH)2] = 1.5 x 10-3 M

pH

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

]-][OHOH[ 3w+=K

3

14

3

3w

100.3100.1]OH[

]-OH][OH[

−

−+

+

××

=

=K M 103.3]O[H 123

−+ ×=

]OH[log pH 3- +=

( )11.48 pH

103.3log- pH 12

=×= −

[H3O+][OH−] pH[Sr(OH)2]

[OH−]=2[Sr(OH)2]

[OH−] = 2(0.0015)= 0.0030 M

Example 15.11b (alternative) – Calculate the pH at 25°C of a 0.0015 M Sr(OH)2 solution and determine if the

solution is acidic, basic, or neutral


[Sr(OH)2] = 1.5 x 10-3 M

pH

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

]--log[OHpOH = 4.001 pOHpH =+

( )2.52 pOH

100.3log- pOH 3

=×= −

pOH[OH−] pH[Sr(OH)2]

[OH−]=2[Sr(OH)2]

[OH−] = 2(0.0015)= 0.0030 M

11.482.52-14.00pHpOH-14.00 pH

===


Practice - Calculate the pH of a 0.0010 M Ba(OH)2 solution and determine if it is



Practice - Calculate the pH of a 0.0010 M Ba(OH)2 solution and determine if it is


[H3O+] = 1.00 x 10-14

2.0 x 10-3 = 5.0 x 10-12M

pH > 7 therefore basic

Ba(OH)2 = Ba2+ + 2 OH- therefore [OH-] = 2 x 0.0010 = 0.0020 = 2.0 x 10-3 M

pH = -log [H3O+] = -log (5.0 x 10-12)pH = 11.30

Kw = [H3O+][OH−]


Weak Bases• in weak bases, only a small

fraction of molecules accept H’sweak electrolytemost of the weak base molecules do not take H from watermuch less than 1% ionization in water

• [HO–] << [weak base]• finding the pH of a weak base

solution is similar to finding the pH of a weak acid

NH3 + H2O ⇔ NH4+ + OH-


Structure of Amines


[NH4+]


[OH−][NH3]

Ex 15.12 Find the pH of 0.100 M NH3(aq) solution

Enter the initial concentrations –assuming the [OH−] from water is ≈ 0


Write the reaction for the base with water


NH3 + H2O ⇔ NH4+ + OH−

0[NH4

+]

equilibriumchange

≈ 00.100initial[OH−][NH3]


0[NH4

+]

equilibriumchange

00.100initial[OH−][NH3]





+x+x−x0.100 −x x x

[ ]( )( )

( )xxxK

−×== −

−+

13

4b 1000.1NH

]OH][[NH



[ ][ ][ ]

( )( )( )x

xxK−×

== −

−+

13

4b 1000.1NH

OHNH

since Kb is very small, approximate the [NH3]eq = [NH3]initand solve for x

determine the value of Kb from Table 15.8

[ ][ ][ ]

( )( )( )1

3

4b 1000.1NH

OHNH−

−+

×==

xxK

1

25

1000.11076.1 −

−

×=×

x

( )( )3

15

1033.1

1000.11076.1−

−−

×=

××=

x

x

Kb for NH3 = 1.76 x 10-5

x+x0

[NH4+]


[OH−][NH3]

0.100 −x



Kb for NH3 = 1.76 x 10-5

x+x0

[NH4+]


[OH−][NH3]check if the approximation is valid by seeing if x < 5% of [NH3]init

%5%33.1%1001000.11033.1

1

3<=×

××

−

−


x = 1.33 x 10-3




[ ] ( ) M 099.01033.1100.0100.0NH 33 =×−=−= −x

M 1033.1][OH]NH[ 3-4

−+ ×=== x

Kb for NH3 = 1.76 x 10-5

x+x0

[NH4+]

x0.100 −xequilibrium+x-xchange≈ 00.100initial

[OH−][NH3]

x = 1.33 x 10-3

1.33E-3

+x0

[NH4+]

1.33E-30.099equilibrium+x-xchange≈ 00.100initial

[OH−][NH3]




use the [OH-] to find the [H3O+] using Kw

( )( ) 124.111052.7log

OH-logpH12

3

=×−=

=−

+

Kb for NH3 = 1.76 x 10-5

1.33E-3

+x0

[NH4+]


[OH−][NH3]

12-3

3-

14-

3

-3w

107.52]O[H101.33101.00]O[H

]][OHO[H

×=××

=

=

+

+

+K



use pOH to find pH

use the [OH-] to find the pOH

( )( )

876.21033.1log

OH-logpOH3

=×−=

=−

−

Kb for NH3 = 1.76 x 10-5

1.33E-3

+x0

[NH4+]


[OH−][NH3]

11.1242.876-14.00

pOH -14.00pH

==

=


Ex 15.12 Find the pH of 0.100 M NH3(aq) solutionKb for NH3 = 1.76 x 10-5

1.33E-3

+x0

[NH4+]


[OH−][NH3]check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb

[ ][ ][ ]

( )( )

523

3

4b

108.1099.0

1033.1

NHOHNH

−−

−+

×=×

=

=Kthough not exact, the answer is reasonably close


Practice – Find the pH of a 0.0015 M morphine solution, Kb = 1.6 x 10-6


Practice – Find the pH of a 0.0015 M morphine solution



Write the reaction for the base with water


B + H2O ⇔ BH+ + OH−

0[BH+]

equilibriumchange

≈ 00.0015initial[OH−][B]


0[BH+]

equilibriumchange

00.0015initial[OH−][B]





+x+x−x0.0015 −x x x

[ ]( )( )

( )xxxK

−×== −

−+

3b 105.1B]OH][[BH



[ ][ ][ ]

( )( )( )x

xxK−×

== −

−+

3b 105.1BOHBH

since Kb is very small, approximate the [B]eq = [B]init and solve for x

determine the value of Kb

[ ][ ][ ]

( )( )( )3b 105.1B

OHBH−

−+

×==

xxK

1

26

105.1106.1 −

−

×=×

x

( )( )5

36

109.4

105.1106.1−

−−

×=

××=

x

x

Kb for Morphine = 1.6 x 10-6

x+x0

[BH+]


[OH−][B]

0.0015 −x



check if the approximation is valid by seeing if x < 5% of [B]init

%5%3.3%100105.1109.4

3

5<=×

××

−

−


x = 4.9 x 10-5


x+x0

[BH+]


[OH−][B]




[ ] ( ) M 0015.0109.40015.00015.0Morphine 5 =×−=−= −x

M 109.4][OH]BH[ 5- −+ ×=== x

x = 4.9 x 10-5


x+x0

[BH+]


[OH−][B]

4.9E-5

+x0

[BH+]


[OH−][B]





( )( ) 69.9100.2log

OH-logpH10

3

=×−=

=−

+

10-3

5-

14-

3

-3w

100.2]O[H109.4101.00]O[H

]][OHO[H

×=××

=

=

+

+

+K


4.9E-5

+x0

[BH+]


[OH−][B]



use pOH to find pH


( )( )

31.4109.4log

OH-logpOH5

=×−=

=−

−

69.94.31-14.00

pOH -14.00pH

==

=


4.9E-5

+x0

[BH+]


[OH−][B]



check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb

[ ][ ][ ]

( )( )

625

b

106.10015.0

109.4

BOHBH

−−

−+

×=×

=

=Kthe answer matches the given Kb


4.9E-5

+x0

[BH+]


[OH−][B]


Acid-Base Properties of Salts• salts are water soluble ionic compounds• salts that contain the cation of a strong base and an

anion that is the conjugate base of a weak acid are basic

NaHCO3 solutions are basicNa+ is the cation of the strong base NaOHHCO3

− is the conjugate base of the weak acid H2CO3

• salts that contain cations that are the conjugate acid of a weak base and an anion of a strong acid are acidic

NH4Cl solutions are acidicNH4

+ is the conjugate acid of the weak base NH3

Cl− is the anion of the strong acid HCl


Anions as Weak Bases• every anion can be thought of as the conjugate base of an

acid• therefore, every anion can potentially be a base

A−(aq) + H2O(l) ⇔ HA(aq) + OH−(aq)• the stronger the acid is, the weaker the conjugate base is

an anion that is the conjugate base of a strong acid is pH neutralCl−(aq) + H2O(l) ← HCl(aq) + OH−(aq)

since HCl is a strong acid, this equilibrium lies practically completely to the left

an anion that is the conjugate base of a weak acid is basicF−(aq) + H2O(l) ⇔ HF(aq) + OH−(aq)

since HF is a weak acid, the position of this equilibrium favors the right


Ex 15.13 - Use the Table to Determine if the Given Anion Is Basic or Neutral

a) NO3−

the conjugate base of a strong acid, therefore neutral

b) NO2−

the conjugate base of a weak acid, therefore basic

130

Relationship between Ka of an Acid and Kb of Its Conjugate Base

• many reference books only give tables of Ka values because Kb values can be found from them

][A]OHA][H[ )(OH ) HA()O( H)(A

[HA]]O][HA[ )(O H )(A )O( H)HA(

3b2

3a32

−

+−−

+−+−

=+⇔+

=+⇔+

Kaqaqlaq

Kaqaqlaqwhen you add equations, you multiply the K’s

)(OH )(OH )O(H 2 32 aqaql −+ +⇔

w3ba

3ba

]OH][OH[]A[

]OH][HA[]HA[

]OH][A[

KKK

KK

==×

×=×

−+

−

−+−


Ex 15.14 Find the pH of 0.100 M NaCHO2(aq) solution

Write the reaction for the anion with water



Na+ is the cation of a strong base – pH neutral. The CHO2

−

is the anion of a weak acid – pH basic

CHO2− + H2O ⇔ HCHO2 + OH−

0

[HCHO2]

equilibriumchange

≈ 00.100initial[OH−][CHO2

−]

132

0.100 −x


Calculate the value of Kb from the value of Ka of the weak acid from Table 15.5substitute into the equilibrium constant expression



+x+x−xx x

[ ]( )( )

( )xxxK

−×== −−

−

12

2b 1000.1CHO

]OH][[HCHO

0

[HCHO2]

equilibriumchange

≈ 00.100initial[OH−][CHO2

−]

114

14

b

wba

106.5108.1100.1 −

−

−

×=××

=

=×

K

KKK



since Kb is very small, approximate the [CHO2

−]eq = [CHO2−]init

and solve for x

1

211

1000.1106.5 −

−

×=×

x

( )( )6

111

104.21000.1106.5

−

−−

×=××=

xx

Kb for CHO2− = 5.6 x 10-11

x+x0

[HCHO2]


[OH−][CHO2−]

0.100 −x

[ ]( )( )

( )xxxK

−×== −−

−

12

2b 1000.1CHO

]OH][[HCHO[ ]

( )( )( )1

2

2b 1000.1CHO

]OH][[HCHO−−

−

×==

xxK



check if the approximation is valid by seeing if x < 5% of [CHO2

−]init

%5%0024.0%1001000.1104.2

1

6<=×

××

−

−


x = 2.4 x 10-6

Kb for CHO2− = 5.6 x 10-11

x+x0

[HCHO2]


[OH−][CHO2−]




[ ] ( ) M 100.0104.2100.0100.0CHO 62 =×−=−= −− x

M 104.2][OH]HCHO[ 6-2

−×=== x

x = 2.4 x 10-6

Kb for CHO2− = 5.6 x 10-11

x+x0

[HCHO2]


[OH−][CHO2−]

2.4E-6

+x0

[HCHO2]


[OH−][CHO2−]





( )( ) 38.8102.4log

OH-logpH9

3

=×−==

−

+

9-3

6-

14-

3

-3w

102.4]O[H104.2101.00]O[H

]][OHO[H

×=××

=

=

+

+

+K

Kb for CHO2− = 5.6 x 10-11

2.4E-6

+x0

[HCHO2]


[OH−][CHO2−]



use pOH to find pH


( )( )

62.5104.2logOH-logpOH

6

=×−=

=−

−

38.85.62-14.00

pOH -14.00pH

==

=

Kb for CHO2− = 5.6 x 10-11

2.4E-6

+x0

[HCHO2]


[OH−][CHO2−]



check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb

[ ][ ][ ]

( )( )

1126

2

2b

108.5100.0104.2

CHOOHHCHO

−−

−

−

×=×

=

=Kthough not exact, the answer is reasonably close

Kb for CHO2− = 5.6 x 10-11

2.4E-6

+x0

[HCHO2]


[OH−][CHO2−]


Polyatomic Cations as Weak Acids• some cations can be thought of as the conjugate acid

of a baseothers are the counterions of a strong base

• therefore, some cation can potentially be an acidMH+(aq) + H2O(l) ⇔ MOH(aq) + H3O+(aq)

• the stronger the base is, the weaker the conjugate acid is

a cation that is the counterion of a strong base is pH neutrala cation that is the conjugate acid of a weak base is acidic

NH4+(aq) + H2O(l) ⇔ NH3(aq) + H3O+(aq)

since NH3 is a weak base, the position of this equilibrium favors the right


Metal Cations as Weak Acids• cations of small, highly charged metals are weakly

acidicalkali metal cations and alkali earth metal cations pH neutralcations are hydrated

Al(H2O)63+(aq) + H2O(l) ⇔ Al(H2O)5(OH)2+ (aq) + H3O+(aq)


Ex 15.15 - Determine if the Given Cation Is Acidic or Neutral

a) C5N5NH2+

the conjugate acid of a weak base, therefore acidic

b) Ca2+

the counterion of a strong base, therefore neutralc) Cr3+

a highly charged metal ion, therefore acidic


Classifying Salt Solutions asAcidic, Basic, or Neutral

• if the salt cation is the counterion of a strong base and the anion is the conjugate base of a strong acid, it will form a neutral solution

NaCl Ca(NO3)2 KBr• if the salt cation is the counterion of a strong

base and the anion is the conjugate base of a weak acid, it will form a basic solution

NaF Ca(C2H3O2)2 KNO2



• if the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a strong acid, it will form an acidic solution

NH4Cl • if the salt cation is a highly charged metal ion

and the anion is the conjugate base of a strong acid, it will form an acidic solution

Al(NO3)3



• if the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a weak acid, the pH of the solution depends on the relative strengths of the acid and base

NH4F since HF is a stronger acid than NH4+, Ka of

NH4+ is larger than Kb of the F−; therefore the

solution will be acidic


Ex 15.16 - Determine whether a solution of the following salts is acidic, basic, or neutral

a) SrCl2Sr2+ is the counterion of a strong base, pH neutralCl− is the conjugate base of a strong acid, pH neutralsolution will be pH neutral

b) AlBr3Al3+ is a small, highly charged metal ion, weak acidCl− is the conjugate base of a strong acid, pH neutralsolution will be acidic

c) CH3NH3NO3CH3NH3

+ is the conjugate acid of a weak base, acidicNO3

− is the conjugate base of a strong acid, pH neutralsolution will be acidic


Ex 15.16 - Determine whether a solution of the following salts is acidic, basic, or neutral

d) NaCHO2

Na+ is the counterion of a strong base, pH neutralCHO2

− is the conjugate base of a weak acid, basicsolution will be basic

e) NH4FNH4

+ is the conjugate acid of a weak base, acidicF− is the conjugate base of a weak acid, basicKa(NH4

+) > Kb(F−); solution will be acidic


Polyprotic Acids• since polyprotic acids ionize in steps, each H has a

separate Ka• Ka1 > Ka2 > Ka3• generally, the difference in Ka values is great enough so

that the second ionization does not happen to a large enough extent to affect the pH

most pH problems just do first ionizationexcept H2SO4 ⇒ use [H2SO4] as the [H3O+] for the second ionization

• [A2-] = Ka2 as long as the second ionization is negligible


0[SO4

2 −]

equilibriumchange

0.01000.0100initial[H3O+][HSO4

−]

Ex 15.18 Find the pH of 0.0100 M H2SO4(aq) solution @ 25°C

Enter the initial concentrations –assuming the [HSO4

−] and [H3O+] is ≈ [H2SO4]


Write the reactions for the acid with water HSO4

− + H2O ⇔ SO42− + H3O+

H2SO4 + H2O ⇔ HSO4− + H3O+






( )( )( )x

xxK−

+==

+

0100.00100.0

]HSO[]OH][[SO

-4

3-2

4a

x

+x0

[SO42 −]

0.0100 −x0.0100 −xequilibrium+x−xchange


−]



( )( )

00012.0022.001000.1102.1102.1

0100.00100.0012.0

2

2224

−+=+×=×−×

−+

=

−−−

xxxxx

xxx

( )

0.0045or 027.0)1(2

)00012.0)(1(4022.0022.0 2

−=

−−±−=

x

x

Ka for HSO4− = 0.012

expand and solve for x using the quadratic formula

( )( )( )x

xxK−

+==

+

0100.00100.0

]HSO[]OH][[SO

-4

3-2

4a



x = 0.0045


( ) M 0055.00045.00100.00100.0]HSO[ 4 =−=−=− x( ) M 0145.00100.0]OH[ 3 =+=+ x

x

+x0

[SO42 −]

0.0100 −x0.0100 −xequilibrium+x−xchange


−]Ka for HSO4

− = 0.012

0.0045

+x0

[SO42 −]

0.01450.0055equilibrium+x−xchange


−]

M 0045.0]SO[ -24 == x




( )( ) 839.10145.0log

OH-logpH 3=−=

= +


0.0045

+x0

[SO42 −]



−]




[ ]( )( )

( )2

-4

3-2

4a

102.10055.0

0145.00045.0HSO

OHSO

−

+

×==

==Kthe answer matches


0.0045

+x0

[SO42 −]



−]


Strengths of Binary Acids

• the more δ+ H-X δ- polarized the bond, the more acidic the bond

• the stronger the H-X bond, the weaker the acid

• binary acid strength increases to the right across a period

H-C < H-N < H-O < H-F• binary acid strength increases down

the columnH-F < H-Cl < H-Br < H-I


Strengths of Oxyacids, H-O-Y

• the more electronegative the Y atom, the stronger the acid

helps weakens the H-O bond• the more oxygens attached to Y, the stronger the

acidfurther weakens and polarizes the H-O bond


Lewis Acid - Base Theory• electron sharing• electron donor = Lewis Base = nucleophile

must have a lone pair of electrons• electron acceptor = Lewis Acid = electrophile

electron deficient• when Lewis Base gives electrons from lone

pair to Lewis Acid, a covalent bond forms between the molecules

Nucleophile: + Electrophile ⇔ Nucleophile:Electrophile

• product called an adduct• other acid-base reactions also Lewis


Example - Complete the Following Lewis Acid-Base Reactions

Label the Nucleophile and ElectrophileOH

H C H⊕

+ OH-1 ⇔

OH

H C H

OH

OH

H C H⊕

+ OH-1 ⇔Electrophile Nucleophile

••••

••


Practice - Complete the Following Lewis Acid-Base Reactions

Label the Nucleophile and Electrophile

• BF3 + HF ⇔

• CaO + SO3 ⇔

• KI + I2 ⇔


Practice - Complete the Following Lewis Acid-Base Reactions

Label the Nucleophile and Electrophile

• BF3 + HF ⇔ H+1BF4-1

• CaO + SO3 ⇔ Ca+2SO4-2

• KI + I2 ⇔ KI3

F

B F

F

H F••••

•• +

F

B F

F

F-1

H+1NucElec

O

S O

O

••••

Ca+2 O -2•• •• +O

S O

O

O-2

Ca+2

ElecNuc

I IK+1 I -1••

••

•• •• +ElecNuc K+1 I I I -1


What Is Acid Rain?

• natural rain water has a pH of 5.6naturally slightly acidic due mainly to CO2

• rain water with a pH lower than 5.6 is called acid rain

• acid rain is linked to damage in ecosystems and structures


What Causes Acid Rain?• many natural and pollutant gases dissolved in the air

are nonmetal oxidesCO2, SO2, NO2

• nonmetal oxides are acidicCO2 + H2O ⇔ H2CO3

2 SO2 + O2 + 2 H2O ⇔ 2 H2SO4• processes that produce nonmetal oxide gases as waste

increase the acidity of the rainnatural – volcanoes and some bacterial actionman-made – combustion of fuel

• weather patterns may cause rain to be acidic in regions other than where the nonmetal oxide is produced

pH of Rain in Different Regions

Sources of SO2 from Utilities


Damage from Acid Rain• acids react with metals, and materials that contain

carbonates• acid rain damages bridges, cars, and other metallic

structures• acid rain damages buildings and other structures made

of limestone or cement• acidifying lakes affecting aquatic life • dissolving and leaching more minerals from soil

making it difficult for trees


Acid Rain Legislation

• 1990 Clean Air Act attacks acid rainforce utilities to reduce SO2

• result is acid rain in northeast stabilized and beginning to be reduced

167

Damage from Acid Rain

chapter 15 acids and bases - center for nonlinear...

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