54
Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq ) = K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants] [A][B] 3 2 A florence flask was getting dressed for the opera. All of a sudden she screamed: "Erlenmeyer, my joules! Somebody has stolen my joules!". The husband replied: "Take it easy honey, do not overreact. We'll find a solution".

Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

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Page 1: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Background

[C] [D]

[Products]

A(g) + 2 B(g) 3 C(g) + D(g)

Equilibrium constant (Keq) =

Keq = LeChatelier’s Principle (lu-SHAT-el-YAY’s)

[Reactants]

[A][B]

3

2

A florence flask was getting dressed for the opera. All of a sudden she screamed: "Erlenmeyer, my joules! Somebody has stolen my joules!". The husband replied: "Take it easy honey, do not overreact. We'll find a solution".

Page 2: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Acids, Bases, and Salts

Ch. 19

Page 3: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Acids

• Properties– Taste sour or tart– Change the color of an acid-base indicator– Can be strong or weak electrolytes in

aqueous solution

Page 4: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Bases

• Properties– Taste bitter– Feel slippery– Will change the color of an acid-base indicator– Can be strong or weak electrolytes in aqueous

solution

Q: Why do chemistry professors like to teach about ammonia? A: Because it's basic stuff.

Page 5: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

During class, the chemistry professor was demonstrating the properties of various acids. “Now I’m going to drop this silver coin into this glass of acid. Will it dissolve?”“No sir,” one student called out.“No?” queried the professor. “Perhaps you can explain why the silver won’t dissolve in this particular acid.”“Because if it would, you wouldn’t have dropped it in!”

Page 6: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Acid vs. Base

Acid

pH > 7

bitter taste

does notreact with

metals

pH < 7

sour taste

react withmetals

Alike Different

Related toH+ (proton)

concentration

pH + pOH = 14

Affects pHand

litmus paper

Base

Different

Topic Topic

Page 7: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Properties

electrolytes

turn litmus red

sour taste

react with metals to form H2 gas

slippery feel

turn litmus blue

bitter taste

ChemASAP

vinegar, milk, soda, apples, citrus fruits

ammonia, lye, antacid, baking soda

electrolytes

Page 8: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]
Page 9: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Common Acids and Bases

Strong Acids (strong electrolytes)

HCl hydrochloric acidHNO3 nitric acidHClO4 perchloric acidH2SO4 sulfuric acid

Weak Acids (weak electrolytes)

CH3COOH acetic acidH2CO3 carbonic

Strong Bases (strong electrolytes)

NaOH sodium hydroxideKOH potassium hydroxideCa(OH)2 calcium hydroxide

Weak Base (weak electrolyte)

NH3 ammonia

NH3 + H2O NH4OH

Page 10: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Common AcidsFormula Name of Acid Name of

Negative Ion of Salt

HF hydrofluoric fluorideHBr hydrobromic bromideHI hydroiodic iodideHCl hydrochloric chlorideHClO hypochlorous hypochloriteHClO2 chlorous chloriteHClO3 chloric chlorateHClO4 perchloric perchlorateH2S hydrosulfuric sulfideH2SO3 sulfurous sulfiteH2SO4 sulfuric sulfateHNO2 nitrous nitriteHNO3 nitric nitrateH2CO3 carbonic carbonateH3PO3 phosphorous phosphiteH3PO4 phosphoric phosphate

Page 11: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Common Bases

Sodium hydroxide NaOH lye or caustic soda

Potassium hydroxide KOH lye or caustic potash

Magnesium hydroxide Mg(OH)2 milk of magnesia

Calcium hydroxide Ca(OH) 2 slaked lime

Ammonia water NH3 H2O household ammonia

Have you heard the one about a chemist who was reading a book about heliumand just couldn't put it down?

Page 12: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Arrhenius Acids and Bases• Arrhenius said that acids are hydrogen-containing compounds that

ionize to yield hydrogen ions (H+) in aqueous solution• Bases are compounds that ionize to yield hydroxide ions (OH-) in

aqueous solution• Monoprotic acids = acids that contain 1 ionizable hydrogen like nitric

acid (HNO3)• Diprotic acids = acids that contain 2 ionizable hydrogens like sulfuric

acid (H2SO4)• Triprotic acids = acids that contain 3 ionizable hydrogens like

phosphoric acid (H3PO4)• Hydroxides of group I metals are very soluble in water and caustic to

skin, hydroxide of group II metals are not very soluble in water and very dilute even when saturated (can be taken internally)

Q: if both a bear in Yosemite and one in Alaska fall into the waterwhich one disolves faster?A: The one in Alaska because it is Polar.

Page 13: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Bronsted-Lowry Acids and Bases

• But what about bases like sodium carbonate (Na2CO3) and ammonia (NH3)???– The Bronsted-Lowry theory defines an acid as a

hydrogen-ion donor, and a base as a hydrogen-ion acceptor

– More complete definition

Page 14: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Conjugate Acids and Bases• Conjugate acid = the particle formed when a base

gains a hydrogen ion• Conjugate base = the particle that remains when an

acid has donated a hydrogen ion– Conjugate acids and bases are always paired w/ a base

or an acid, respectively• Conjugate acid-base pair = consists of 2 substances

related by the loss or gain of a single hydrogen ion.

Page 15: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Cont…

• A water molecule that gains a hydrogen ion becomes a positively charged Hydronium ion (H3O+)

• Amphoteric = a substance that can act as both an acid and a base – EX: Water

Page 16: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Lewis Acids and Bases• Lewis proposed that an acid accepts a pair of

electrons during a reaction while a base donates a pair of electrons– More general than either of the other 2 theories

• Lewis acid = a substance that can accept a pair of electrons to form a covalent bond

• Lewis base = a substance that can donate a pair of electrons to form a covalent bond

Page 17: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Acid – Base Systems

Type Acid Base

Arrhenius H+ or H3O + producer

OH - producer

Brønsted-Lowry

Proton (H +) donor

Proton (H +) acceptor

Lewis Electron-pair acceptor

Electron-pair donor

Page 18: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

• Many Lewis acids are also Bronsted-Lowry acids and vice versa but not all

• *PP 1-2, 19.1 sect. assessment #8 pg. 593

Copper leaves Copper Sulfate and says see you: he answers CuSO4!!!!!

Page 19: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Ion Product Constant for Water• Self-ionization of water = the reaction in which water molecules

produce ions

• For aqueous solutions, the product of the hydrogen-ion concentration and the hydroxide-ion concentration equals 1.0 x 10-14

– [H+] x [OH-] = 1.0 x 10-14

• ion-product constant for water (KW) = the product of the concentrations of the hydrogen ions and hydroxide ions in water

• Acidic solution = one in which [H+] is greater than [OH-]– *The [H+] is greater than 1 x 10-7 M*

• Basic solution = one in which [H+] is less than [OH-]– *The [H+] is less than 1 x 10-7 M*

• *SP 19.1, PP 9-10 pg. 596

Page 20: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

pH Scale

Acid Base

0

7

14

[H+] pH

10-14 14

10-13 13

10-12 12

10-11 11

10-10 10

10-9 9

10-8 8

10-7 7

10-6 6

10-5 5

10-4 4

10-3 3

10-2 2

10-1 1

100 0

1 M NaOH

Ammonia(householdcleaner)

BloodPure waterMilk

VinegarLemon juiceStomach acid

1 M HCl

Aci

dic

N

eutr

al

Bas

ic

Page 21: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

pH of Common Substances

1.0 MHCl0

gastricjuice1.6

vinegar2.8

carbonated beverage3.0

orange3.5

apple juice3.8

tomato4.2

lemonjuice2.2 coffee

5.0

bread5.5

soil5.5

potato5.8

urine6.0

milk6.4

water (pure)7.0

drinking water7.2

blood7.4

detergents8.0 - 9.0

bile8.0

seawater8.5

milk of magnesia10.5

ammonia11.0

bleach12.0

1.0 MNaOH(lye)14.0

8 9 10 11 12 14133 4 5 621 70

acidic neutral basic[H+] = [OH-]

Page 22: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

pH of Common Substance

14 1 x 10-14 1 x 10-0 0 13 1 x 10-13 1 x 10-1 1 12 1 x 10-12 1 x 10-2 2 11 1 x 10-11 1 x 10-3 3 10 1 x 10-10 1 x 10-4 4 9 1 x 10-9 1 x 10-5 5 8 1 x 10-8 1 x 10-6 6

6 1 x 10-6 1 x 10-8 8 5 1 x 10-5 1 x 10-9 9 4 1 x 10-4 1 x 10-10 10 3 1 x 10-3 1 x 10-11 11 2 1 x 10-2 1 x 10-12 12 1 1 x 10-1 1 x 10-13 13 0 1 x 100 1 x 10-14 14

NaOH, 0.1 MHousehold bleachHousehold ammonia

Lime waterMilk of magnesia

Borax

Baking sodaEgg white, seawaterHuman blood, tearsMilkSalivaRain

Black coffeeBananaTomatoesWineCola, vinegarLemon juice

Gastric juice

Mor

e ba

sic

Mor

e ac

idic

pH [H1+] [OH1-] pOH

7 1 x 10-7 1 x 10-7 7

Page 23: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]
Page 24: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

pH Concept

• pH = the negative logarithm of the hydrogen-ion concentration of a solution

• A solution in which [H+] if greater than 1 x 10-7 M has a pH less than 7.0 and is acidic. The pH of pure water or a neutral aqueous solution is 7.0. A solution with a pH greater than 7 is basic and has a [H+] of less than 1 x 10-7 M.

Søren Sorensen(1868 - 1939)

Page 25: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Cont…• The pOH of a solution equals the

negative logarithm of the hydroxide-ion concentration

– A solution w/ a pOH less than 7 is basic, greater than 7 is acidic

• For pH calculations, you should express the hydrogen-ion concentration in scientific notation

• *SP 19.2, PP 11 pg. 599• *Given pH = 4.6 determine the

hydronium ion• *SP 19.3-19.4, PP 13-16 pg. 600-

601

Page 26: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]
Page 27: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

pH Calculations

pH

pOH

[H3O+]

[OH-]

pH + pOH = 14

pH = -log[H3O+]

[H3O+] = 10-pH

pOH = -log[OH-]

[OH-] = 10-pOH

[H3O+] [OH-] = 1 x10-14

Page 28: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Strength of Acids and Bases

• Strong Acids = completely ionized in aqueous solution– HCl and H2SO4

• Weak Acids = ionize only slightly in aqueous solution– Ethanoic acid (acetic acid)

Page 29: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Comparison of Strong and Weak Acids

Type of acid, HA Reversibilityof reaction

Ka value Ions existing when acid,HA, dissociates in H2O

StrongNot

reversibleKa value very large

H+ and A-, only.No HA present.

Weak reversible Ka is small H+, A-, and HA

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

The equilibrium expression for the reaction is

Ka = [H3O+] [A-]

[HA]Note: H3O+ = H+

Page 30: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Relative Strengths of Acids and Bases

perchloric HClO4

hydrogen chloride HClnitric HNO3

sulfuric H2SO4

hydronium ion H3O+

hydrogen sulfate ion HSO4-

phosphoric H3PO4

acetic HC2H3O2

carbonic H2CO3

hydrogen sulfide H2Sammonium ion NH4

+

hydrogen carbonate ion HCO3-

water H2Oammonia NH3

hydrogen H2

Dec

reas

ing

Aci

d S

tren

gth

perchlorate ion ClO4-

chloride ion Cl-

nitrate ion NO3-

hydrogen sulfate ion HSO4-

water H2Osulfate ion SO4

2-

dihydrogen phosphate ion H2PO4-

acetate ion C2H3O2-

hydrogen carbonate ion HCO3-

hydro sulfide ion HS-

ammonia NH3

carbonate ion CO32-

hydroxide ion OH-

amide ion NH2-

hydride ion H-

Dec

reas

ing

Bas

e S

tren

gth

Acid Formula Conjugate base Formula

acid conjugate base + H+

Page 31: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]
Page 32: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Acid Dissociation Constant• For dilute solutions, the conc. of water is a constant. It can be

combined w/ Keq to give the acid dissocation constant.• Acid dissociation constant = the ratio of the concentration of the

dissociated (or ionized) form of an acid to the concentration of the undissociated (nonionized) form.

• Weak acids have small Ka values. The stronger an acid is, the larger is its Ka value.

– Nitrous acid (HNO2) has a Ka of 4.4 x 10-4, acetic acid has a Ka of 1.8 x 10-5 so nitrous acid is more ionized and has a higher [H3O+] or [H+] thus is a stronger acid

• Di and triprotic acids lose each H separately so they have multiple dissociation constants

Page 33: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]
Page 34: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

[H3O+]

Equilibrium and pH Calculations

HA + H2O H3O+ + A-

Weak acid

HA + H2O H3O+ + A-

Strong acid

acid-dissociationconstant calculations

Ka = [A-] [H3O+]

[HA]

[HA] = [H3O+]

+

pH0 7 14

antilog(-pH)

-log [H3O+] [OH-]-

1 x 10-14

[OH-]=

1 x 10-14

[H3O+]=

HA H+ + A-

Kw = [H3O+][OH-]

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

Page 35: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]
Page 36: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]
Page 37: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Base Dissociation Constant

• Strong bases = dissociate completely into metal ions and hydroxide ions in aqueous solution– Ex: Ca(OH)2

• Weak bases = react w/ water to form the hydroxide ion and the conjugate acid of the base– Ex: ammonia NH3

• Base dissociation constant (Kb) = the ratio of the concentration of the conjugate acid times the conc. of the hydroxide ion to the conc. of the base

Page 38: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]
Page 39: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Calculating Dissociation Constants

• To find the Ka of a weak acid or the Kb of a weak base, substitute the measured concentration of all the substances present at equilibrium into the expression for Ka or Kb.

• *SP 19.5, PP 22-23 pg. 610

Page 40: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

[C] [D]

[Products]

A(g) + 2 B(g) 3 C(g) + D(g)

Weak Acids (pKa)

Weak Acids – dissociate incompletely (~20%)Strong Acids – dissociate completely (~100%)

Equilibrium constant (Keq) =

Keq = LeChatelier’s Principle (lu-SHAT-el-YAY’s)

[Reactants]

[A][B]

3

2

Page 41: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Formula Name Value of Ka*

Values of Ka for Some Common Monoprotic Acids

HSO4- hydrogen sulfate ion 1.2 x 10-2

HClO2 chlorous acid 1.2 x 10-2

HC2H2ClO2 monochloracetic acid 1.35 x 10-3

HF hydrofluoric acid 7.2 x 10-4

HNO2 nitrous acid 4.0 x 10-4

HC2H3O2 acetic acid 1.8 x 10-5

HOCl hypochlorous acid 3.5 x 10-8

HCN hydrocyanic acid 6.2 x 10-10

NH4+ ammonium ion 5.6 x 10-10

HOC6H5 phenol 1.6 x 10-10

*The units of Ka are mol/L but are customarily omitted.

Incr

easi

ng a

cid

stre

ngth

Page 42: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

H2SO4 2 H+ + SO42- in dilute solutions...occurs ~100%

H2SO4 H+ + HSO4

1- & HSO41- H+ +

SO42-

One gram of concentrated sulfuric acid (H2SO4) is diluted to a 1.0 dm3 volume

with water. What is the molar concentration of the hydrogen ion in this solution? What is the pH?

x mol H2SO4 = 1 g H2SO4

Solution)First determine the number of moles of H2SO4

Sample 1)

= 0.010 mol H2SO4

OVERALL:

pH = - log [H+]

pH = 1.69

0.010 M 0.020 M

substitute into equation pH = - log [0.020 M]

98 g H2SO4

1 mol H2SO4

Page 43: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

A volume of 5.71 cm3 of pure acetic acid, HC2H3O2, is diluted with water at

25 oC to form a solution with a volume of 1.0 dm3.

Step 2) Find the number of moles of acid.

x mol acetic acid = 6.00 g HC2H3O2 = 0.10 mol acetic acid (in 1 L)

M = 0.1 molar HC2H3O2

Step 3) Find the [H+]Ka =

Step 1) Find the mass of the acidMass of acid = density of acid x volume of acid

= 1.05 g/cm3 x 5.71 cm3

= 6.00 g

Molarity: M = mol / LSubstitute into equation M = 0.10 mol / 1 L

What is the molar concentration of the hydrogen ion, H+, in this solution? (The density of pure acetic acid is 1.05 g/cm3.)

(From the formula of acetic acid, you can calculate that the molar mass of acetic acid is 60 g / mol).

60 g HC2H3O2

1 mol HC2H3O2

Page 44: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Step 3) Find the [H+]

H C H O

HC H O

12 3 2

2 3 2

[ ][ ]

[ ]

1.8 x 10-5 =

Ka = 1.8 x 10-5 @ 25 oC for acetic acid

H C H O

HC H O

12 3 2

2 3 2

[ ][ ]

[ ]

Ka =

Substitute into equation: ]OH[HC

[x][x] 10 x 1.8

232

5-

]M [0.10x

10 x 1.82

5-

x2 = 1.8 x 10-6 M

x = 1.3 x 10-3 molar = [H+]

HC2H3O2 H+ + C2H3O2

1-

0.1 M

pH = - log[H+]

pH = - log [1.3 x10-3 M]

pH = 2.9

?0.1 M

weak acid

How do the concentrations of H+ and C2H3O2

1- compare?

Page 45: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

1a) What is the molar hydrogen ion concentration in a 2.00 dm3 solution

of hydrogen chloride in which 3.65 g of HCl is dissolved? 1b) pH

2a) What is the molar concentration of hydrogen ions in a solution containing 3.20 g of HNO3 in 250 cm3 of solution?

2b) pH

3a) An acetic acid solution is 0.25 M. What is its molar concentration of

hydrogen ions?3b) pH

4) A solution of acetic acid contains 12.0 g of HC2H3O2 in 500 cm3 of solution. What is the molar concentration of hydrogen ions?

1a) 0.0500 M 2a) 0.203 M 3a) 2.1 x 10-3 M 4) 2.7 x 10-3 M1b) pH = 1.3 2b) pH = 0.7 3b) pH = 2.7

Practice Problems:

Page 46: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Weak Acids

Cyanic acid is a weak monoprotic acid. If the pH of 0.150 M cyanic acid is 2.32.calculate Ka for cyanic acid.

HCN(aq) H+(aq) + CN1-(aq)

H3O+(aq)

0.150 M 4.8 x 10-3 M

Ka = [Products]

[Reactants]Ka =

[H3O+]

[HCN]

[CN1-]

Ka = [4.8 x 10-3 M]

[0.150 M]

[CN1-]

[4.8 x 10-3 M]

Ka = 1.54 x 10-4

4.8 x 10-3 M

pH = -log[H3O+]

10-pH = [H3O+]

10-2.32 = [H3O+]

4.8 x10-3 M = [H3O+]

Page 47: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Titration• Neutralization Reaction = a reaction b/w an acid and a

base in aqueous solution to produce salt and water• Equivalence point = when the # of moles of hydrogen

ions equals the number of moles of hydroxide ions• Titration = the process of adding a known amt. of

solution of known conc. to determine the conc. of another solution– Standard solution = the solution of known conc.– End point = the point at which the indicator changes color

• The point of neutralization is the end pt. of the titration• *SP 19.7, PP 32-33 pg. 616

Q: How do you get lean molecules?A: Feed them titrations. Q: What did one titration say to the other? A: Let's meet at the endpoint!

Q: How did the chemist survive the famine?A: By subsisting on titrations.

Page 48: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]
Page 49: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Buffers

• Buffer = a solution in which the pH remains relatively constant when small amts. of acid or base are added

• A buffer is a solution of weak acids and one of its salts or a solution of a weak base and one of its salts

• A buffer solution is better able to resist drastic changes in pH than is pure water

• Buffer capacity = the amt. of acid or base that can be added to a buffer solution before a significant change in pH occurs

Page 50: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Naming Acids

_________ ide(chloride, Cl1-)

_________ite(chlorite, ClO2

-)(hypochlorite, ClO-)

_________ ate(chlorate, ClO3

-)(perchlorate, ClO4

-)

Hydro____ ic acid(hydrochloric acid, HCl)

_________ic acid(chloric acid, HClO3)

(perchloric acid, HClO4)

______ous acid(chlorous acid, HClO2)

(hypochlorous acid, HClO)

Anion Acid

add H+

add H+

add H+

ions

ions

ions

Page 51: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Review

What does one do with a dead body? Barium in a krypt-onMaybe he was killed oxydentally.They should have seen the doctor first, he'd Curium.Ah, barium anyway, just to see how he reacts.better though to have helium.Perhaps with a houseplant, a Germanium.And if they stole it, the police would Cesium.Locked up for life, in Irons.They would go crazy in jail, a Silicon.

A physicist, biologist and a chemist were going to the ocean for the first time.

The physicist saw the ocean and was fascinated by the waves. He said he wanted to do some research on the fluid dynamics of the waves and walked into the ocean. Obviously he was drowned and never returned.

The biologist said he wanted to do research on the flora and fauna inside the ocean and walked inside the ocean. He too, never returned.

The chemist waited for a long time and afterwards, wrote the observation, "The physicist and the biologist are soluble in ocean water".

Page 52: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

pH scale0

7

14

acid

neutral

base

[H+] = [OH-]

Soren Sorenson developed pH scale

pOH = -log [OH-]

kW = [H+] [OH-]

pH = -log [H+]

pH + pOH = 14

(alkalinity)

Arnold Beckman invented the pH meter

H+ + H2O H3O+

proton hydronium ion

kw = 1 x 10-14

Page 53: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Strong / Weak Acid

Strong HA H+ + A- (~100% dissociation)

Weak HA H+ + A- (~20% dissociation)

Ka = [Product][Reactant]

acid dissociation constant

Ka

0.8 H3PO4 3H+ + PO43-

0.0021 HF H+ + F-

H2A 2 H+ + A-

Ka = [H+]2 [A-]

[H2A]

Page 54: Background [C] [D] [Products] A (g) + 2 B (g) 3 C (g) + D (g) Equilibrium constant (K eq )= K eq = LeChatelier’s Principle (lu-SHAT-el-YAY’s) [Reactants]

Acid + Base Salt + Water

How would you make calcium sulfate in the lab?

+ CaSO4

ACID

Sour taste, litmus red

Arrhenius – H+ as only ion in water

Brønsted-Lowry – proton donor

BASE

bitter taste, litmus blue

Arrhenius – OH- as only ion in water

Brønsted-Lowry – proton acceptor

H2SO4 Ca(OH)2 + 2 H2O

? ?