Unit 2 Acids and Bases

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Unit 2 Acids and Bases. http://www.cdli.ca/courses/ http://www.cbhs.k12.nf.ca/adrianyoung/. Topics Properties / Operational Definitions Acid-Base Theories pH & pOH calculations Equilibria ( K w , K a , K b ) Indicators Titrations STSE: Acids Around Us. Operational Definitions. - PowerPoint PPT Presentation

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Unit 2Acids and Bases

http://www.cdli.ca/courses/http://www.cbhs.k12.nf.ca/adrianyoung/

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Topics Properties / Operational Definitions Acid-Base Theories pH & pOH calculations Equilibria (Kw, Ka, Kb) Indicators Titrations STSE: Acids Around Us

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An Operational Definition is a list of properties, or operations that can be performed, to identify a substance.

See p. 550 for operational definitions of acids and bases

Operational Definitions

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Operational Definitions (Properties – see p. 550)

Acids pH < 7 taste sour react with

active metals (Mg, Zn) to produce hydrogen gas

Bases pH > 7 taste bitter no reaction with

active metals feel slippery

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Operational Definitions

Acids blue litmus turns

red react with

carbonates to produce CO2 gas

Bases red litmus turns

blue no reaction with

carbonates

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Operational Definitions

Acids conduct electric

current neutralize bases to

produce water and a “salt”

Bases conduct electric

current neutralize acids to

produce water and a “salt”

any ionic compound

any ionic compound6

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Acid-Base Theories

1. Arrhenius Theory (p. 549 )acid – any substance that dissociates or IONIZES in water to produce H+ ions

ie. an acid must contain H+ ions

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Arrhenius Theoryeg.

HCl(aq) →

H2SO4(aq) →

HSO4-(aq) ⇌

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H+(aq) + Cl-(aq)

H+(aq) + HSO4

-(aq)

H+(aq) + SO4

2-(aq)

Arrhenius Theory

base – any substance that dissociates in water to produce OH- ions

ie. a base must contain OH- ions

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Arrhenius Theory

eg. NaOH(aq) →

Ca(OH)2(aq) →

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Na+(aq) + OH-

(aq)

Ca+(aq) + 2 OH-

(aq)

Arrhenius Theory

Which is an Arrhenius acid?

a) KOH c) CH4

b) HCN d) CH3OH

Which is a Arrhenius base?

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Limitations of Arrhenius theory (p.551)

1. H+ cannot exist as an ion in water.

The positive H+ ions are attracted to the polar water molecules forming

HYDRONIUM ions or H3O+(aq)

H+(aq) + H2O(l)

→ H3O+(aq)

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Limitations of Arrhenius theory2. CO2 dissolves in water to produce

an acid.

NH3 dissolves in water to produce a base.

Neither of these observations can be explained by Arrhenius theory

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Limitations of Arrhenius theory3. Some acid-base reactions can occur

in solvents other than water.

Arrhenius theory can explain only aqueous acids or bases.

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Limitations of Arrhenius theory

4. Arrhenius theory is not able to predict whether certain species are acids or bases.

eg. NaHSO4 H2PO4- HCO3

-

Arrhenius theory needs some work

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Acid-Base Theories

2. Modified Arrhenius Theory (p. 552)

acid – any substance that reacts with water to produce H3O+ ions

eg.

HCl(g) + H2O(l) → H3O+(aq) + Cl-(aq)

To be used when Arrhenius is inadequate

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Modified Arrhenius Theory

base – any substance that reacts with water to produce OH- ions

eg.

NH3(aq) + H2O(l) )→ NH4+

(aq) + OH-(aq)

pp. 558, 559 #’s 1, 3, 8, & 919

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Acid-Base Theories

3. Brønsted-Lowry Theory (p. 553)

acid – any substance from which a proton (H+) may be removed

ie. an acid is a substance that loses a proton (H+)

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Brønsted-Lowry Theory

base – any substance that can remove a proton (H+) from an acid.

ie. a base is a substance that gains a proton (H+)

In BLT , an acid-base reaction requires the transfer of a proton

(H+) from an acid to a base.23

Brønsted-Lowry Theory

eg.

HCN(aq) + NH3(aq) →

acid

base

conjugate base

conjugate acid

← CN-(aq) + NH4

+(aq)

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Brønsted-Lowry Theory

What is a conjugate acid-base pair?? (p. 554)

Two particles (molecules or ions) that differ by one proton are called a conjugate acid-base pair.

The conjugate base forms when an acid loses a proton.

The conjugate acid forms when a base gains a proton (H+).

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Brønsted-Lowry Theory

acid

base

conjugate acid

conjugate base 26

Brønsted-Lowry Theory

eg.

NH3(aq) + H2O(l) → NH4+

(aq) + OH-(aq)

H2O(l) + H2O(l) →

acid

base

conjugate base

conjugate acid

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Brønsted-Lowry Theory

- an amphoteric substance can be either an acid or a base

- these include WATER and negative ions that contain at least one hydrogen atom

eg. H2O, HCO3-(aq), H2PO4

-(aq)

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Brønsted-Lowry Theory

p.557 #’s 1 – 7

p. 558 #’s 8, 9

p. 559 #’s 2, 4-7, 10,11

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Strength of Acids and Bases A strong acid is an acid that ionizes

or dissociates 100% in water

eg. HCl(aq)→

Strong acids react 100% with water (BLT)

eg. HCl(aq) + H2O(l) →

Strong acids produce more H+ ionsOR more H3O+ ions than weak acids

with the same molar concentration

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H+(aq) + Cl-(aq)

H3O+(aq) + Cl-(aq)

Strength of Acids and Bases

NOTE: The equilibrium symbol, , is NOT used for strong acids because there is NO REVERSE REACTION.

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Strength of Acids and BasesA weak acid is an acid that ionizes

or dissociates LESS THAN 100%

eg. HF(aq)

Weak acids react less than 100% with water

eg. HF(aq) + H2O(l)

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Strength of Acids and Bases

For weak acids, an equilibrium is established between the original acid molecule and the ions formed.

DO NOT confuse the terms strong and weak with

concentrated and dilute.

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Strength of Acids and Bases

eg. Classify the following acids: 0.00100 mol/L HCl(aq)

strong and dilute 12.4 mol/L HCl(aq)

strong and concentrated 10.5 mol/L CH3COOH(aq)

weak and concentrated34

Strength of Acids and Basesmonoprotic – acids that contain or lose

one proton

diprotic – acids that contain or lose two protons

polyprotic – acids that have more than one proton

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Strength of Acids and Bases A strong base is a base that dissociates

100% in water, or reacts 100% with water, to produce OH- ion.

The only strong bases are hydroxide compounds of most Group 1 and Group 2 elements

eg. NaOH(s) →

Ca(OH)2(s) →36

Strength of Acids and Bases A weak base is a base that reacts less

than 100% in water to produce OH- ion.

eg. S2-(aq) + H2O(l) HS-

(aq) + OH-(aq)

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Writing Acid-Base Equations (BLT)

Step 1: List all the molecules/ions present in the solution

ionic compounds form cations and anions strong acids exist as hydronium ion and

the anion (conjugate base) for weak acids use full formula of the

compound (i.e. un-ionized molecule) always include water in the list.

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Writing Acid-Base Equations (BLT)

Step 2: Identify the strongEST acid and the strongEST base from Step 1.

Step 3: Write the equation for the reaction by transferring a proton from the strongest acid to the strongest base.

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Writing Acid-Base Equations (BLT)

Step 4: Determine the type of reaction arrow to use in the equation.

Stoichiometric (100%) reactions occur between: Hydronium (H3O+) and bases stronger

than nitrite (NO2-)

hydroxide (OH-)and acids stronger than hypochlorous acid (HOCl)

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Writing Acid-Base Equations (BLT)

Step 5: Determine the position of the equilibrium by comparing the strengths of both acids in the equation.

The favoured side is the side with the weaker acid! 

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Writing Acid-Base Equations (BLT)

Sample problems: Write the net ionic equation for the acid-

base reaction between:

- aqueous sodium hydroxide (NaOH(aq)) and hydrochloric acid (HCl(aq)).

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species present

Na+(aq) OH-

(aq) H3O+(aq) Cl-(aq) H2O(l)

strongest acidstrongest base

H3O+(aq) + OH-

(aq) H2O(l) + H2O(l)

H3O+(aq) + OH-

(aq) → 2 H2O(l)

OR

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Writing Acid-Base Equations (BLT)

Sample problems: Write an equation for the acid-base

reaction between nitrous acid (HNO2(aq)) and aqueous sodium sulfite (Na2SO3(aq)).

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species present

Na+(aq) SO3

2-(aq)HNO2(aq) H2O(l)

SA SB

HNO2(aq) + SO32-

(aq) NO2-(aq) + HSO3

- (aq)

Weaker AcidStronger Acid Products favored

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Write the Net Ionic Equation for each aqueous reaction below:

1. Na2CO3(aq) and CH3COOH(aq)

2. NH3(aq) and HNO2(aq)

3. HNO3(aq) and RbOH

4. H2SO4(aq) and K3PO4(aq)

5. HF(aq) and NH4CH3COO(aq)

6. CaCl2(aq) and PbSO4(aq)

p. 564 #’s 10 &11 46

species present

Na+(aq) CO3

2-(aq)

CH3COOH(aq) H2O(l)

SASB

CH3COOH(aq) + CO32-

(aq) CH3COO-(aq) + HCO3

-(aq)

Weaker Acid

Products Favoured

Stronger Acid

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species present

HNO2(aq)NH3(aq) H2O(l)

SASB

HNO2(aq) + NH3(aq) NO2-(aq) + NH4

+ (aq)

Weaker AcidProducts favoredStronger Acid

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species present

Rb+(aq) OH-

(aq)H3O+(aq) NO3

-(aq) H2O(l)

strongest acid strongest base

H3O+(aq) + OH-

(aq) H2O(l) + H2O(l)

H3O+(aq) + OH-

(aq) → 2 H2O(l)

OR

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species presentK+

(aq)HSO4-(aq) PO4

3-(aq) H2O(l)

SA SB

H3O+(aq) + PO4

3-(aq) H2O(l) + HPO4

2-(aq)

H3O+(aq)

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species present

HF(aq) H2O(l)

SA SB

HF(aq) + CH3COO-(aq) F-

(aq) + CH3COOH(aq)

Weaker AcidProducts favoredStronger Acid

NH4+

(aq) CH3COO-(aq)

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species present

Pb2+(aq) SO4

2-(aq)Ca2+

(aq) Cl-(aq) H2O(l)

strongest acid

strongest base

H2O(l) + SO42-

(aq) HSO4-(aq) + OH-

(l)Weaker Acid

Reactants favored

Stronger Acid

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NO!! Products are NOT always favoured

Try these:

CH3COOH(aq) + NH4F(aq)

HCN(aq) + NaHS(aq)

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Acid-Base Calculations

Kw

Ka

Kb

[H3O+] [OH-]

pH pOH54

Kw (Ionization Constant for water)

With very sensitive conductivity testers, pure water shows slight electrical conductivity.

PURE WATER MUST HAVE

A SMALL CONCENTRATION OF

DISSOLVED IONS

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Kw

H2O(l) + H2O(l) H3O+(aq) + OH-

(aq)

K = [H3O+] [OH-]

[H2O] [H2O]Kw =

Auto-Ionization of water

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Kw

In pure water at 25 °C;

[H3O+] = 1.00 x 10-7 mol/L

[OH-] = 1.00 x 10-7 mol/L

Calculate Kw at 25 °C.

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LCP: What happens if we add OH- ions (NaOH(aq)) to water?

shift to the left [H3O+] ? [OH-] ?

Does Kw change?

GET REAL!!

H2O(l) + H2O(l) H3O+(aq) + OH-

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[H3O+] [OH-]

0.00357 M

4.89 x 10-3 mol/L

12.5 M

1.50 mol/L

Kw = [H3O+] [OH-]

1.00 x 10-14 = [H3O+] [OH-]

2.80 x 10-12

2.04 x 10-12

8.00 x 10-16

6.67 x 10-15

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Calculations with Kw (p. 564 – 566)

For strong acids and strong bases, the [H3O+] and [OH-] may be calculated using the solute concentration.

eg. What is the [H3O+] in a 2.00 mol/L solution of HNO3(aq)?

Ans: 2.00 mol/L

[OH-] = ???60

Calculations with Kw

eg. What is the [OH-] in a 2.00 mol/L solution of NaOH(aq)?

Ans: 2.00 mol/L

eg. What is the [OH-] in a 2.00 mol/L solution of Ca(OH)2(aq)?

Ans: 4.00 mol/L[H3O+] = ???

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Calculations with Kw

eg. What molar concentration of Al(OH)3(aq) is needed to obtain a [OH-] = 0.450 mol/L?

Ans: 0.150 mol/L

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What is the [H3O+] and [OH-] in:

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[H3O+] mol/L

[OH-]mol/L

1.0 x 10-8 1.0 x 10-6

5.00 x 10-14 0.200

1.50 6.67 x 10-15

1.0 x 10-2 1.0 x 10-12

solute [H3O+] [OH-]

0.680 mol/L HCl(aq)

1.50 M NaOH

0.0500 M Ca(OH)2(aq)

_____ mol/L HClO4(aq) 0.450 M

____ mol/L Mg(OH)2(aq) 0.500 mol/L

p. 566 #’s 12 - 15

1.47 x 10-14

6.67 x 10-15

1.00 x 10-13

0.450

0.250

2.22 x 10-14

2.00 x 10-14

0.680

1.50

0.100

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pH and pOH (See p. 568)By what factor does the [H3O+] change when the pH value changes by 1?

by 2?

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pH and pOH (See p. 568)The [H3O+] changes by a factor of 10 (10X) for each

pH changes of 1.

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pH = -log [H3O+]

[H3O+] = 10-pH

pOH = -log [OH-]

[OH-] = 10-pOH

pH and pOH FORMULAS

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pH and pOHeg. What is the pH of a 0.0250 mol/L

solution of HCl(aq)?

What is the pOH of a 0.00087 mol/L solution of NaOH(aq)?

What is the pH of a 1.25 mol/L solution of KOH(aq)?

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[H3O+] = 0.0250 mol/LpH = 1.602

[H3O+] = 8.00 x 10-15 mol/L

[OH-] = 0.00087 mol/LpOH = 3.06

[OH-] = 1.25 mol/L

pH = 14.097

Significant digits in pH values?

The number of significant digits in a concentration should be the same as the number of digits to the right of the decimal point in the pH value.eg. In a sample of OJ the

[H3O+] = 2.5 × 10−4 mol/L

pH = 3.60 (See p. 568)69

[H3O+] [OH-] pH pOH

0.0035

1.2 x 10-5

4.68

9.15

8.33 x 10-15

-1.10

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[H3O+] [OH-] pH pOH

0.0035

1.2 x 10-5

4.68

9.15

8.33 x 10-15

-1.10

2.9 x 10-12 2.46 11.54

8.3 x 10-10 9.08 4.92

2.1 x 10-5 4.8 x 10-10 9.32

4.851.4 x 10-5 7.1 x 10-10

1.20 -0.079 14.079

15.107.9 x 10-1613

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pH, pOH and Kw

p. 569 #’s 16 – 19

p. 572 #’s 20 – 25

Examine #23. Where is the energy term in this equation?

H2O(l) + H2O(l) H3O+(aq) + OH-

(aq)

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Dilutions

When a solution is diluted the number of moles does not change.

OR ninitial = nfinal

CiVi = CfVf

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eg. 400.0 mL of water was added to 25.0 mL of HCl(aq) that had a pH of 3.563. Calculate the pH of the resulting solution.calculate [H3O+]dilution formulacalculate pH

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Before dilution:[H3O+] = 10-3.563

= 2.753 x 10-4 After dilution:

(2.753 x 10-4) (25.0 mL) = (Cf)(425.0 mL)

[H3O+] = 1.609 x 10-5

pH = -log (1.609 x 10-5)

= 4.793p.574 #’s 26 - 29

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