chapter 15 acids and bases chemistry ii. stomach acid & heartburn the cells that line your...

Chapter 15Acids and Bases

Chemistry II

Properties of Acids

• sour taste

• react with “active” metals i.e., Al, Zn, Fe, but not Cu, Ag, or Au

2 Al + 6 HCl → 2 AlCl3 + 3 H2

Corrosive

• react with carbonates, producing CO2

marble, 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 Acids

Chemical 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 HC2H3O2 plastics & 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

Structure of Acids

• binary acids have H-atoms attached to a nonmetal atom

e.g. HCl, HF

Structure of Acids

• oxy acids have H-atoms attached to an O-atom:

Structure of Acids

• carboxylic acids have COOH group:

e.g. Acetic acid

CH3COOH

Properties of Bases

• also known as alkalis

• taste bitter alkaloids = plant product that is alkaline

often poisonous

• solutions feel slippery

• change color of vegetable dyes different color than acid red litmus turns blue

• react with acids to form ionic salts neutralization

http://wps.prenhall.com/wps/media/objects/477/489045/st1404.html

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

CO32− + 2H2O → HCO3

− + H2O + OH−

HCO3− + H2O + OH− → H2CO3 + 2OH−

Definitions of Acids and Bases

• What are the main characteristics of molecules and ions that exhibit acid and base behavior?

• 3 different definitions:ArrheniusBrønsted-LowryLewis

Arrhenius Theory (1880s)

• 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)

CH3COOH(aq) → H+(aq) + CH3COO–(aq)

• bases dissociate in water to produce OH- ions and cations ionic substances dissociate in water

NaOH(aq) → Na+(aq) + OH–(aq)

Hydronium Ion

• the H+ ions produced by the acid are so reactive they cannot exist in water H+ ions are protons!!

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

H+ + H2O H3O+

Chemists use H+(aq) and H3O+(aq) interchangeably

Arrhenius Acid-Base Reactions

• the H+ from the acid combines with the OH- from the base to make a molecule of H2O

H+ + OH- → H2O

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

Na+ + Cl- → NaCl

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

acid + base → salt + water

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 (1923)

• in a Brønsted-Lowry Acid-Base reaction, an H+ is transferred

Acid: proton (H+) donorBase: proton (H+) acceptor

• base structure must contain an atom with unshared pair of e-

• 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+ donorsAnything that has H+ can potentially be Brønsted-Lowry

acid

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

HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)

acid base

Brønsted-Lowry Bases

• Brønsted-Lowry bases are H+ acceptorsany material that has atoms with lone pairs can

potentially be a Brønsted-Lowry base

• NH3(aq) is basic because NH3 accepts an H+ from H2O (acid, proton donor), forming OH–(aq)

NH3(aq) + H2O(l) NH⇌ 4+(aq) + OH–(aq)

base acid

Amphoteric Substances

• amphoteric substances can act as either an acid or a base have both transferable H and atom with lone pair

• water acts as base, accepting H+ from HCl

HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)

• water acts as acid, donating H+ to NH3

NH3(aq) + H2O(l) NH⇌ 4+(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 e- 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 conjugate base acid

HCHO2 + H2O ⇌ CHO2– + H3O+


H2O + NH3 ⇌ HO– + NH4+


Conjugate (joined) Pairs

In the reaction H2O + NH3 OH⇌ – + NH4+

H2O and OH– 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

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

H2O

NH3

CO32−

H2PO4−

H3O+

NH4+

HCO3−

H3PO4

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

H2O

NH3

CO32−

H2PO4−

OH−

NH2−

No H+, it cannot be an acid

HPO42−

Strong or Weak• a strong acid is a strong electrolyte

practically all the acid molecules ionize, →

• a strong base is a strong electrolytepractically 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 H+

use water as the standard base

• strong acids donate practically all their H+’s 100% ionized in water strong 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 water

much 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 acids the ionizable H+’s may have different acid strengths or be equal 1 H+ = monoprotic, 2 H + = diprotic, 3 H + = triprotic

HCl = monoprotic, H2SO4 = diprotic, H3PO4 = triprotic

• polyprotic acids ionize in steps each 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 Bases HClO4 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

HA + H2O A⇌ - + H3O+

B: + H2O HB⇌ + + OH-

• 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+ > H2O > OH-; NH4

+ > NH3 > NH2-

base trend opposite

Acid Ionization Constant, Ka

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

HA + H2O A⇌ - + H3O+

• acid ionization constant, Ka

larger Ka = stronger acid

[HA]

]O[H][A 3a

K

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 H⇌ 3O+ + 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 H3O+ and OH– concentrations is always the same n

• the number is called the ion product of water - symbol, Kw

Kw = [H3O+] [OH–] = 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[ 3wK

]OH[]-OH[

]-OH][OH[

3

w

3w

K

K

M 107.6105.1

100.1]-[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

Question

Complete the following table of pH values

[H+] pH Significant Figures

1 x 10-7 7.0 1

1.0 x 10-7

6.80

4.39

1.78 x 10-11

1.6 x 10-7 2

3

2

4.1 x 10-52

7.00

10.750

pH

• the lower the pH, the more acidic the solution; the higher the pH, the more basic the solution 1 pH unit corresponds to a factor of 10 difference in acidity

• normal range 0 to 14 pH can be negative (very acidic) or larger than 14 (very

alkaline)

52

pH of Common SubstancesSubstance pH

1.0 M HCl 0.0

0.1 M HCl 1.0

stomach acid 1.0 to 3.0

lemons 2.2 to 2.4

soft drinks 2.0 to 4.0

plums 2.8 to 3.0

apples 2.9 to 3.3

cherries 3.2 to 4.0

unpolluted rainwater 5.6

human blood 7.3 to 7.4

egg whites 7.6 to 8.0

milk of magnesia (sat’d Mg(OH)2) 10.5

household ammonia 10.5 to 11.5

1.0 M NaOH 14

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

acidic, basic, or neutral

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[ 3wK

2

14

3

3w

103.1

100.1]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 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 pOH Complete 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 = 14.00at 25°Ccan use pOH to find pH of a solution

14.00pOHpH

00.14]-[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 pKa

larger 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 negligible

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

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

• for a monoprotic strong acid [H3O+] = [HA] 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

HA + H2O ⇌ A + H3O+

[HNO2] [NO2-] [H3O+]

initial

change

equilibrium

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

Write the reaction for the acid with water

Construct an ICE table for the reaction

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

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

HNO2 + H2O NO⇌ 2 + H3O+

[HNO2] [NO2-] [H3O+]

initial 0.200 0 ≈ 0

change

equilibrium

[HNO2] [NO2-] [H3O+]

initial 0.200 0 0

change

equilibrium


represent the change in the concentrations in terms of x

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

substitute into the equilibrium constant expression

+x+xx

0.200 x x x

x

xxK

12

3-2

a 1000.2HNO

]OH][[NO


x

xxK

12

3-2

a 1000.2HNO

OHNO

determine the value of Ka from Table 15.5

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

1

2

3-2

a 1000.2HNO

OHNO

xxK

1

24

1000.2106.4

x

3

14

106.9

1000.2106.4

x

x

Ka for HNO2 = 4.6 x 10-4

[HNO2] [NO2-] [H3O+]

initial 0.200 0 ≈ 0

change -x +x +x

equilibrium 0.200 x x0.200 x


Ka for HNO2 = 4.6 x 10-4

[HNO2] [NO2-] [H3O+]

initial 0.200 0 ≈ 0

change -x +x +x

equilibrium 0.200 x x

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

[HNO2] [NO2-] [H3O+]

initial 0.200 0 ≈ 0

change -x +x +x

equilibrium 0.200-x x x

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

[HNO2] [NO2-] [H3O+]

initial 0.200 0 ≈ 0

change -x +x +x

equilibrium 0.190 0.0096 0.0096


Ka for HNO2 = 4.6 x 10-4

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

[HNO2] [NO2-] [H3O+]

initial 0.200 0 ≈ 0

change -x +x +x

equilibrium 0.190 0.0096 0.0096

02.2106.9log

OH-logpH3

3


Ka for HNO2 = 4.6 x 10-4

[HNO2] [NO2-] [H3O+]

initial 0.200 0 ≈ 0

change -x +x +x

equilibrium 0.190 0.0096 0.0096

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

109.4190.0

106.9

HNO

OHNO

Kthough not exact, the answer is reasonably close

Tro, Chemistry: A Molecular Approach

88

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 initial

acid ionized ofmolarity IonizationPercent

%100[HA]

]O[HIonizationPercent

init

equil3

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

Write the reaction for the acid with water


Enter the Initial Concentrations

Define the Change in Concentration in terms of x

Sum the columns to define the Equilibrium Concentrations

HNO2 + H2O NO⇌ 2 + H3O+

[HNO2] [NO2-] [H3O+]

initial

change

equilibrium

[HNO2] [NO2-] [H3O+]

initial 2.5 0 ≈ 0

change

equilibrium

+x+xx2.5 x x x

determine the value of Ka from Table 15.5

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

5.2HNO

OHNO

2

3-2

a

xxK

5.2106.4

24 x

2

4

104.3

5.2106.4

x

x

Ka for HNO2 = 4.6 x 10-4

[HNO2] [NO2-] [H3O+]

initial 2.5 0 ≈ 0

change -x +x +x

equilibrium 2.5-x ≈2.5 x x



HNO2 + H2O NO⇌ 2 + H3O+

[HNO2] [NO2-] [H3O+]

initial 2.5 0 ≈ 0

change -x +x +x

equilibrium 2.5 0.034 0.0342.5 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 NO⇌ 2 + H3O+

[HNO2] [NO2-] [H3O+]

initial 2.5 0 ≈ 0

change -x +x +x

equilibrium 2.5 0.034 0.034

Apply the Definition and Compute the Percent Ionization

%4.1%1005.2

104.3

%100][HNO

]O[HIonizationPercent

2

init2

equil3

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

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)

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

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

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

HF + H2O F⇌ + H3O+ Ka = 3.5 x 10-4

[HF] [F-] [H3O+]

initial 0.150 0 ≈ 0

change

equilibrium

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

H2O + H2O OH⇌ + H3O+ Kw = 1.0 x 10-14

[HF] [F-] [H3O+]

initial 0.150 0 0

change

equilibrium





+x+xx

0.150 x x x

x

xxK

13

-

a 1050.1HF

]OH][[F


x

xxK

150.0HF

OHF 3-

a

determine the value of Ka for HF

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

150.0HF

OHF 3-

a

xxK

1

24

1050.1105.3

x

3

14

102.7

1050.1105.3

x

x

Ka for HF = 3.5 x 10-4

[HF] [F-] [H3O+]

initial 0.150 0 ≈ 0

change -x +x +x


Tro, Chemistry: A Molecular Approach

99


Ka for HF = 3.5 x 10-4

[HF] [F-] [H3O+]

initial 0.150 0 ≈ 0

change -x +x +x


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

%5%8.4%1001050.1

102.71

3


x = 7.2 x 10-3


Ka for HF = 3.5 x 10-4

[HF] [F-] [H3O+]

initial 0.150 0 ≈ 0

change -x +x +x

equilibrium 0.150-x x x

x = 7.2 x 10-3


M 143.0102.7150.0150.0HF 3 x

M 102.7OHF 33

- x

[HF] [F-] [H3O+]

initial 0.150 0 ≈ 0

change -x +x +x

equilibrium 0.143 0.0072 0.0072


Ka for HF = 3.5 x 10-4


14.2102.7log

OH-logpH3

3

[HF] [F-] [H3O+]

initial 0.150 0 ≈ 0

change -x +x +x

equilibrium 0.143 0.0072 0.0072


Ka for HF = 3.5 x 10-4

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

4

23

3-

a

106.3143.0

102.7

HF

OHF


[HF] [F-] [H3O+]

initial 0.150 0 ≈ 0

change -x +x +x

equilibrium 0.143 0.0072 0.0072

NaOH → Na+ + OH-

Strong Bases

• the stronger the base, the more willing it is to accept H use water as the standard acid

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

dissociated

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

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

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

pH

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

]-][OHOH[ 3wK

3

14

3

3w

100.3

100.1]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

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

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

[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’s weak electrolyte most of the weak base molecules do not

take H from water much 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-

[NH3] [NH4+] [OH]

initial

change

equilibrium

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

Write the reaction for the base with water


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


NH3 + H2O NH⇌ 4+ + OH

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0

change

equilibrium

[NH3] [NH4+] [OH]

initial 0.100 0 0

change

equilibrium





+x+xx

0.100 x x x

x

xxK

13

4b 1000.1NH

]OH][[NH


x

xxK

13

4b 1000.1NH

OHNH

determine the value of Kb from Table 15.8

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

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

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0

change -x +x +x



Kb for NH3 = 1.76 x 10-5

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0

change -x +x +x


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

%5%33.1%1001000.1

1033.11

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

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.100 x x x

x = 1.33 x 10-3

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.099 1.33E-3 1.33E-3


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


124.111052.7log

OH-logpH12

3

Kb for NH3 = 1.76 x 10-5

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.099 1.33E-3 1.33E-3

12-3

3-

14-

3

-3w

107.52]O[H

101.33

101.00]O[H

]][OHO[H

K


Kb for NH3 = 1.76 x 10-5

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.099 1.33E-3 1.33E-3

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

5

23

3

4b

108.1099.0

1033.1

NH

OHNH


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

[B] [BH+] [OH]

initial 0.0015 0 ≈ 0

change

equilibrium

[B] [BH+] [OH]

initial 0.0015 0 0

change

equilibrium





+x+xx

0.0015 x x x

x

xxK

3b 105.1B

]OH][[BH


x

xxK

3b 105.1B

OHBH

determine the value of Kb

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

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

[B] [BH+] [OH]

initial 0.0015 0 ≈ 0

change -x +x +x



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

%5%3.3%100105.1

109.43

5


x = 4.9 x 10-5


[B] [BH+] [OH]

initial 0.0015 0 ≈ 0

change -x +x +x




M 0015.0109.40015.00015.0Morphine 5 x

M 109.4][OH]BH[ 5- x

x = 4.9 x 10-5


[B] [BH+] [OH]

initial 0.0015 0 ≈ 0

change -x +x +x

equilibrium 0.0015 x x x

[B] [BH+] [OH]

initial 0.0015 0 ≈ 0

change -x +x +x

equilibrium 0.0015 4.9E-5 4.9E-5




69.9100.2log

OH-logpH10

3

10-3

5-

14-

3

-3w

100.2]O[H

109.4

101.00]O[H

]][OHO[H

K


[B] [BH+] [OH]

initial 0.0015 0 ≈ 0

change -x +x +x




6

25

b

106.10015.0

109.4

B

OHBH

Kthe answer matches the given Kb


[B] [BH+] [OH]

initial 0.0015 0 ≈ 0

change -x +x +x


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 basic

Na+ is the cation of the strong base NaOH HCO3

− 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 acidic

NH4+ is the conjugate acid of the weak base NH3

Cl− is the anion of the strong acid HCl

Acid-Base Properties of SaltsAnions 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 neutral

Cl−(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 basic

F−(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

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) solutionNa+ is the cation of a

strong base – pH neutral. The CHO2

− is the anion of a weak acid – pH basic

Write the reaction for the anion with water



CHO2− + H2O HCHO⇌ 2 + OH

[CHO2−] [HCHO2] [OH]

initial 0.100 0 ≈ 0

change

equilibrium

0.100 x

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



Calculate the value of Kb from the value of Ka of the weak acid from Table 15.5


+x+xxx x

x

xxK

12

2b 1000.1CHO

]OH][[HCHO


initial 0.100 0 ≈ 0

change

equilibrium

114

14

b

wba

106.5108.1

100.1

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

104.2

1000.1106.5

x

x

Kb for CHO2− = 5.6 x 10-11


initial 0.100 0 ≈ 0

change -x +x +x


x

xxK

12

2b 1000.1CHO

]OH][[HCHO

12

2b 1000.1CHO

]OH][[HCHO

xxK


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

−]init

%5%0024.0%1001000.1

104.21

6


x = 2.4 x 10-6

Kb for CHO2− = 5.6 x 10-11


initial 0.100 0 ≈ 0

change -x +x +x




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


initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.100 −x x x


initial 0.100 0 ≈ 0

change -x +x +x





38.8102.4log

OH-logpH9

3

9-3

6-

14-

3

-3w

102.4]O[H

104.2

101.00]O[H

]][OHO[H

K

Kb for CHO2− = 5.6 x 10-11


initial 0.100 0 ≈ 0

change -x +x +x




11

26

2

2b

108.5100.0

104.2

CHO

OHHCHO


Kb for CHO2− = 5.6 x 10-11


initial 0.100 0 ≈ 0

change -x +x +x


Polyatomic Cations as Weak Acids

• some cations can be thought of as the conjugate acid of a base others are the counterions of a strong base

• therefore, some cation can potentially be an acid MH+(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 neutral a cation that is the conjugate acid of a weak base is acidic

NH4+(aq) + H2O(l) NH⇌ 3(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 acidic alkali metal cations and alkali earth metal cations pH neutral cations are hydrated

Al(H2O)63+(aq) + H2O(l) Al(H⇌ 2O)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 neutral

c) 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

• Ion with higher K value dominates

NH4F = NH4+, F-

NH4+ is conjugate acid of weak base (NH3) = acidic

F- is conjugate base of weak acid = basic Ka of NH4

+ (5.68 x 10-10) is larger than Kb of the F− (2.9 x 10-11); 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) AlBr3

Al3+ is a small, highly charged metal ion, weak acidCl− is the conjugate base of a strong acid, pH neutralsolution will be acidic

c) CH3NH3NO3

CH3NH3+ is the conjugate acid of a weak base, acidic

NO3− is the conjugate base of a strong acid, pH neutral

solution 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 neutral

CHO2− is the conjugate base of a weak acid, basic

solution will be basic

e) NH4F

NH4+ is the conjugate acid of a weak base, acidic

F− is the conjugate base of a weak acid, basic

Ka(NH4+) > Kb(F−); solution will be acidic

Practice - Lewis Acid-Base ReactionsLabel the Nucleophile and Electrophile

• BF3 + HF H⇌ +BF4-

• CaO + SO3 Ca⇌ +2SO4-2

• KI + I2 KI⇌ 3

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 I K+1 I -1••

••

•• •• +ElecNuc K+1 I I I -1

chapter 15 acids and bases chemistry ii. stomach acid & heartburn the cells that line your...

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