1

Ch 13 Acids & Bases

PropertiesAcid-Base TheoriesAcid-Base Reactions

Most assignments are from the Ch 13 handout.

2

Properties

Both conduct electricity (electrolytes) because they break apart to some degree in water.Acids produce H+ (proton) in water.Bases produce OH- (hydroxide) in water.

Samples:Acids: vinegar(acetic acid), lactic acid in sour milk, citric acid, Bases: ammonia, lye (NaOH), Milk of Magnesia Mg(OH)2.

3

More on Acids

1. Sour taste. NEVER taste acids in lab situations.2. Change color of indicators.3. Some acids react with metals & release H2 gas.4. Acids react with bases to produce salt & water.

When neutralization occurs, #1.-#3 disappear.5. Conduct electric current.

4

Acid Nomenclature

• Binary Acids contain Hydrogen and another element:

• Hydro + root of 2nd element + ic– HF hydrofluoric acid– HCl hydrochloric acid– HBr hdrobromic acid– HI hydroiodic …– H2S hydrosulfuric …

5

Oxyacids• Contain H, O, and a 3rd element. More are

listed in your book.

6

Common Industrial Acids

• Sulfuric• Nitric• Phosphoric• Hydrochloric• Acetic

7

Bases

• Bitter taste (NEVER taste bases in labs).• Change the color of indicators.• Slippery feel (dilute bases, don’t touch

concentrated bases)• React with acids to produce salt & water• Conduct electric current.

8

Neutralization Reaction

Acid + Base --> Salt + WaterHCl + NaOH --> NaCl + H20H2SO4 + Ca(OH)2 --> CaSO4 + 2H20

Assignment from Ch 13 A-B handout:117/1-3 AND 123/1 (naming/formulas)

9

Arrhenius Acids & Bases

• Arrhenius Acid is a chemical compound that increases the concentration of hydrogen ions, H+, in aqueous solutions.

• Arrhenius Base is a chemical compound that increases the concentration of hydroxide ions, OH-, in aqueous solutions.

• When put with water, these compounds dissociate (break apart) forming ions

10

• HNO3 (l) + H20 (l) --> NO3- (aq) + H30+ (aq)

• When put in water, HNO3 , ionizes and the charged particles formed can conduct electricity.

• The amount of H30+ (hydronium) produced is an indication of the acid’s strength.

11

Strong Acids ionize completely in water.

Strong Acids:• HI• HClO4

• HBr• HCl• H2SO4

• HClO3

Weak Acide release few hydrogen ions in water.

Weak Acids:• HSO4

-

• H3PO4

• HF• CH3COOH

• H2CO3

• H2S• HCN• HCO3

-

12

13

For Bases, the strength depends on how it dissociates (ionizes)

Strong Bases ionize completely.

Strong Bases• Ca(OH)2 --> Ca2+ + 2OH-

• Sr(OH)2

• Ba(OH)2

• NaOH• KOH• RbOH• CsOH

Weal Bases ionize slightly.Weak Bases• NH3 + H2O NH4

+ + OH-

• C6H5NH2

“ “ means the reaction is

reversible

14

Assignment

• Book: 437/33,34 (strengths & ionization) Answer these on page 118(bottom) of handout.

• H/out: 117/5,6 (reactions) and 122/5,6ac (neutralization reactions, balancing and mole ratio)

15

Acid-Base Theories

Bronsted-Lowry Acids donate protons (H+)Molecules or ions can donate protons.HCl + NH3 NH4

+ + Cl-

Na+

Na+

-

16

The HCl is a Bronsted-Lowry Acid. It donates a proton to water

Water can act as a Bronsted-Lowry Acid also as in the following reaction:

H2O (l) + NH3 OH- + NH4+

17

Bronsted-Lowry Bases

accept protons. In the equation below, ammonia is the base, because it accepts the proton to become an ammonium ion.

acid baseHCl + NH3 NH4

+ + Cl-

18

Mono- and Polyprotic Acids• Monoprotic acids can only donate one proton

per molecule. Ex.: HCl, HNO3

• Polyprotic acids can donate 2 or more protons per molecule. Ex.: H2SO4, H3PO4

• For polyprotic acids the donations occur in stages, losing one H+ at a time.

19

Lewis Acids and Bases

• Arrhenius and Bronsted-Lowery definitions have some limitations. Lewis classification is based on bonding and structure including substances without hydrogen. The Lewis classification is more complete than the other 2 methods.

20

A Lewis acid

is an atom, ion or molecule that accepts an electron pair to form a covalent bond.

Dot notation

Structural formula – a bar represents what?A pair of shared electrons.

21

A Lewis base

is an atom, ion, or molecule that donates an electron pair to form a covalent.

22

Lewis Acid-Base Reaction• is the formation of one or more covalent

bonds between an electron-pair donor and an electron-pair acceptor.

Pair of donated electrons

23

Assignment:

• 119/1-3 (ionization in stages)• 120/4,5ab (BL a&b, L’LP)

24

Sample: Dilute HCl(aq) and KOH(aq) are mixed in chemically equivalent quantities.

a) Write the formula equation for the reaction.HCl(aq) + KOH(aq) --> KCl(aq) + H2O(l)

b) Write the overall ionic equation.H3O+(aq) + Cl-(aq) + K+(aq) + OH-(aq) -->

K+(aq) + Cl-(aq) + 2H20(l)c) Write the net ionic equation.

H3O+(aq) + OH-(aq) --> 2H20(l)

25

Sample: Write the formula equation and net ionic equation for this reaction.

Formula equation for: Zn(s) + HCl(aq) --> Zn(s) + 2HCl(aq) --> ZnCl2(aq) + H2(g)Overall ionic equation:

Zn(s) + 2H3O+(aq) + 2Cl-(aq) --> Zn2+(aq) + 2Cl-(aq) + H2(g) + 2H20(l)Net ionic equation:

Zn(s) + 2H30+(aq) --> Zn2+(aq) + H2(g) + 2H20(l)

26

Acid-Base Reactions

• Now we are going to use Bronsted-Lowry description to explore acid-base reactions.

• What was the Bronsted-Lowery theory?• B-L acid donates protons• B-L base accepts protons

• Proton: H+ (a hydrogen nucleus)

27

• A conjugate base is the species that remains after a Bronsted-Lowery acid has given up a proton.

• A conjugate acid is the species that forms when a Bronsted-Lowery base gains a proton.

28

29

Acid-Base Reactions

Using Bronsted-Lowry definitions to study Acid-Base reactions, continued:

The species that remains after a Bronsted-Lowry acid has given up a proton is the conjugate base of that acid.

HF + H2O F- + H30+

Acid conjugate base

30

The species that is formed when a Bronsted-Lowry base gains a proton is the conjugate acid of that base.

HF(aq) + H2O(l) F-(aq) + H30+(aq) Base conjugate acid

31

HF(aq) + H2O(l) F-(aq) + H30+(aq) Acid Base conjugate conjugate

base acid acid1 base2 base1 acid2

Conjugate pairs: (1) HF and F- (2) H20 and H30+

32

Assignment

• 121/1,2 (conjugate A&B)• 123/2,3 (same as above)

33

Strength of Conjugate Acids & Bases

• On Page 1 of your handout for this chapter, you have a table which lists and compares the strengths of various acids and their conjugate bases. Get your Ch. 14 handout out now.

34

Determining direction of equilibrium in Acid-Base reactions

The stronger an acid is, the weaker its conjugate base will be.

The stronger a base is, the weaker its conjugate acid will be.

From these concepts, we can predict the outcome of a reaction.

35

36

37

38

39

)

Sample problem on next page:

40

Sample: Identify the proton donor or acid and the proton acceptor or base. Label each acid-

base conjugate pair.CH3COOH + H20 H30+ + CH3COO-

acid base conjugate conjugate acid base

41

Another sample.

Write the formula equation, the overall ionic equation, and the net ionic equation for a neutralization reaction that would form RbClO4.

Formula equation:RbOH(aq) + HClO4(aq) --> RbClO4(aq) + H20(l)

42

Overall Ionic equation:Rb+(aq) + OH-(aq) + H30+(aq) + ClO4

-(aq) --> Rb+(aq) + ClO4

-(aq) + 2H20(l)Net ionic equation:H30+(aq) + OH-(aq) --> 2H20(l)

43

Amphoteric Compounds

These can act as either an acid or a base.Water acts as a base in this reaction:H2SO4(aq) + H20(l) --> H30+(aq) + HSO4

-(aq) acid1 base2 acid2 base1

But, water acts as an acid here:NH3(g) + H20(l) NH4

+(aq) + OH-(aq)Base1 acid2 acid1 base2

44

Assignment:

• 123/5 (amphoteric/stages)• 124/6ac (conduction, strength)

45

Review

Acids BasesArrhenius concentration of: [H+] [OH-]

Bronsted-Lowry H+ donor H+ acceptor

Lewis, e- pair: acceptor donor

46

Chapter 14, cont.pH and Acid-Base Titration

47

pH – What is it?

pH is an indication of the hydronium ion concentration present in a solution.

[H30+] is the symbol for concentration of hydronium ion in moles per liter, mol/L or M

pOH is an indication of the hydroxide ion concentration present in a solution.

[OH-] is the symbol for concentration of hydroxide ion in mol/L or M

48

Water self ionizes

H20(l) + H20(l) H30+(aq) + OH-(aq)

In the above reaction, two water molecules produce a hydronium ion and a hydroxide ion by transfer of a proton. Water is self Ionizing.

At 25oC, the concentrations of H30+ and OH- are each only 1.0x10-7 mol/L of water.

49

Math product of these ions

is a constant, kw, the ionization constant of water. Kw = [H30+ ] [OH-] = 1.0x10-7(1.0x10-7) =1.0x10-14

This occurs at 25oC. If the temperature changes, the ion product, Kw changes.

When both [H30+ ] and[OH-] are 1.0x10-7, the solution is neutral.

If [H30+ ] is greater than 1.0x10-7, the solution isAcidic. (10-6 or 10-4 would be greater)

If [OH-] is greater than 1.0x10-7, the solution is Basic.

50

Calculating without a calculator

Kw = [H30+ ] [OH-] = 1.0x10-7(1.0x10-7) =1.0x10-14

Let’s say that the [H30+ ] is 1.0x 10-6 and you are asked to find the [OH-].

Kw = [H30+ ] [OH-] --> [OH-] = kw = 1.0x10-14

[H30+ ] 1.0x10-6

-14 – (-6) = -14 + 6 = -8 so: [OH-] = 10-8 mol/Liter More practice: 10-14/10-2 = 10-12

and 10-14/10-9 = 10-5

51

Asgn: H/O: 125/1,2

52

Calculating [H30+ ] and [OH-]

Your own scientific calculator is a MUST here!!!Find these keys: 2nd, either EE or EXP, and change

sign (-) or (+/-) on your calculator.Let’s practice putting in numbers in sci. not.1x10-7: Press keys in this sequence:

1 2nd EE (-) 7 on your display you see something similar to this: 1E -7

To get 2 x10-4, press: 2 2nd EE (-) 4 display: 2 E -4

53

For concentration, M means moles/L

The [H30+] is 2.34 x 10-5 M in a solution. Calculate the [OH-] of the solution.

[OH-] = Kw = 1.0x10-14

[H30+ ] 2.34 x 10-5

Key sequence:

1 2nd EE (-) 14 : 2.34 2nd EE (-) 5 enterDisplay: 4.27 E -10 which means: 4.27 x 10-10 M

54

Calculate hydronium and hydroxide ion concentrations in a solution that is 1x10-4 M HCl.

HCl is a strong acid that ionizes completely. So the concentration of H30+ is 1x10-4 M.

Find [OH-]: [OH-] = Kw = 1.0x10-14

[H30+ ] 1x10-4 Answer: [OH-] = 10-10 M

55

The pH Scale is used to show how acidic or basic (alkaline) a solution is.

pH of a solution is the negative of the common logarithm of the hydronium ion concentration.pH = - log [H30+ ] A common logarithm of a number is “the power to which 10 must be raised to equal the number.”

56

“pHinding” pH

The logarithm of 1.0x10-7 is - 7.0The pH = - log [H30+ ] = - log (1.0x10-7) = 7.0

pOH is the negative log of [OH-].pOH = -log [OH-]. In a neutral solution where

[OH-] is 1.0x10-7, the pOH = -log [OH-] = -log 1.0x10-7 = 7.0

57

Kw = [H30+ ] [OH-] = 1.0x10-7(1.0x10-7) =1.0x10-14

From above: pH + pOH = 14.0pOH can also be found by: pOH = 14.0 – pH = 14.0 – 7.0 = 7.0

58

Determine the pH of these solutions:

a. 1x10-3 M HCl. HCl is a strong acid that ionizes completely. So the concentration of H30+ is 1x10-3 M.

pH = - log [H30+ ] = - log (1.0x10-3) = 3.0

b. 1x10-4 M NaOH. NaOH is a strong base that ionizes completely. The concentration of OH- is 1x10-4 M

[H30+ ] = Kw = 10-14 = 10-10

[OH-] 10-4

pH = -log(10-10) = 10.0

59

More pH samples…Find the pH of a solution where [H30+ ] is 2.8 x 10-5 M?pH = -log [H30+ ] = -log 2.8 x 10-5 = 4.55 or 4.6 Key sequence: (-) log 2.8 2nd EE (-) 5 =Find the pH of a 4.7 x 10-2 M NaOH solution.Find the concentration of [H30+] first:[H30+ ] = Kw = 10-14

[OH-] 4.7x10-2

[H30+ ] = 2.1x10-13

pH = -log [H30+ ] = - log 2.1x10-13 pH = 12.7

60

“pHinding” pH Assignment

• Asgn: H/O: 125/4a-d and 129/1-3. Hint for 3a: H2SO4 will produce double [H30+]

61

Calculating [H30+] & [OH-] from whole number pH

pH = -log [H30+]log [H30+] = -pH[H30+] = antilog (-pH)[H30+] = 10-pH

If pH = 7, [H30+] = 1x10-7

If pH = 2, [H30+] = 1x10-2

62

Find the [H30+] of a solution with pH 7.52. (Not a whole number)

pH = -log [H30+] log [H30+] = -pH[H30+] =antilog(-pH) = antilog (-7.52) = 1x10 -7.52 = 3.0 x 10-8 M H30+

Raise 10 to the -7.52 power using thisFlow chart: [2nd] [log] [+/- ] [7.52] = 3.10x10-8 M

Practice

2. If the pH = 12.0 then [H30+] = 1x10-12 M3. The pH of an aqueous solution is measured as

1.50. Calculate the [H30+] and [OH-].[H30+] = [2nd] [Log] [+/-] [1.5] [H30+] = 1x10-1.5 = 3.16x10-2M

[OH-] = ___Kw___ = 1x10-14 = 3.16 x 10-13M 3.16x10-2 3.16x10-2

The pH of an aqueous solution is 3.67. Determine [H30+] .

[H30+] = antilog (-pH) = antilog (-3.67) …OR[2nd] [log] [+/-] [3.67] = 2.14 x10-4 MAsgn from book: 437/30-32 (labeling A+B),42,61-63

(calc pH)

and 439/64-68 a-c only (calc pH + pOH)

65

Solution Concentrations*

Molarity – one mole of solute dissolved in enough solvent (water) to make exactly one liter of solution.

Molality – one mole of solute dissolved in exactly 1,000 grams of solvent.

Normality – one gram equivalent weight (gew) of solute dissolved in enough solvent to make exactly one liter of solution.

These are on your handout titled: “ph/Acid/Base Equations”

66

Molarity (M) Formulae

M = grams of solute given GFW of solute (L of solvent)

Grams of solute needed= M(GFW of solute)(Liters Solvent)

L of solvent needed = g solute/GFW solute (M)

67

A molarity problemYou have 3.50L of solution that contains 90.0g of NaCl.

What is the molarity of the solution?g NaCl x 1mol NaCl = mol NaCl, the solute g NaClmol of Solute = molarity of solutionL of solution90.0g NaCl x 1 mol NaCl = 1.54 mol NaCl

58.44 g NaCl1.54 mol NaCl = 0.440 M NaCl 3.50 L

68

Another molarity problem.

1. What is the molarity of a solution composed of 5.85g KI, dissolved in enough water to make 0.125 L of solution?

5.85g KI x 1 mol KI = 0.0352 mol KI166g KI

0.0352 mol KI = 0.282 M KI 0.125 L

69

molality, m – one mole of solute dissolved in exactly 1,000 g of solvent.

m = g of solute given X 1000 GFW of solute X g of solvent

g of solute needed = m(GFW)(g solvent)1000

g solvent needed = g solute(1000) GFW solute x m

Molality = molarity if water is the solvent (aqueous solutions)

70

Normality, N – one gram equivalent weight of solute dissolved in enough solvent to

make exactly one liter of solution.N = _________g solute_____

GEW solute x L solventg solute = N X GEW solute X L of solventL solvent = ____g of solute___ GEW of solute X NNormality to MolarityM = N(valence of cation)(subscript of cation)GEW = _________GFW of solute_______ charge X subscript of solute cation

GEW – gram equivalent weight

71

EquivalentsOne equivalent of an acid is the quantity, in grams, which donates one mole of protons.1 equiv HCl = 1molHCl x 36.5g = 36.5g HCL 1mol H30+ mol mol H30+ 1equiv H2SO4 = 1mol H2SO4 x 98g = 49 g H2SO4

2mol H30+ mol mol H30+ A diprotic acid (H2SO4 ) has 2 atoms of ionizable hydrogen per molecule.

72

Equiv., cont.One equivalent of a base is the quantity, in grams, which accepts one mole of protons, or supplies one mole of OH- ions.1equiv KOH = 1 mol KOH x 56g = 56g KOH 1 mol OH- mol mol OH-

1equiv Ca(OH)2 =1 mol Ca(OH)2 x 74g = 37 g Ca(OH)2 2 mol OH- mol mol OH-

1 Mole of Ca(OH)2 supplies 2 equiv of OH- ions, therefore ½ mole of Ca(OH)2 is 1 equiv of Ca(OH)2

73

Titration

Titration is the controlled addition and measurement of the amount of a solution of known concentration required to react completely with a measured amount of a solution of unknown concentration.

More simply: it is using a known concentration of a solution to determine the concentration of a solution of unknown concentration.

74Picture retrieved from: cikguwong.blogspot.com

When a base is added to an acid the solution will become neutral and this will be shown by an indicator changing color.This will occur when equal numbers of H30+ and OH- are present.

75

Equivalence point The point at which two solutions used in a titration are present in chemically equivalent amounts.

End pointThe point in a titration at which an indicator changes color.

The figures below shows typical pH curves for various acid-base titrations. The equivalence points and end points are different for the various combinations of strong and weak acids and bases.

Retrieved from: chemguide.co.uk

76

In this section we are going to look at: indicators, pH meters, and titrations.

You saw 2 indicators demonstrated recently to determine how acidic or basic several solutions were.

Indicators used were litmus paper and pH paper (Hydrion).

Acid-base indicators are sensitive to pH of acids and bases. They will change color as a result of the ions present.

77

In acidic solutions, an indicator will be one color (litmus turns red) and in basic solutions, an indicator will be another color (litmus turns blue).

78

Indicator Samples

Methyl red, Bromthymol blue, Methyl orange, Phenolphthalein, Phenol red are indicator samples.

These will ionize in solution and, depending upon their acid or base strength will change color over a range of pH values until the end point is reached.

The range over which an indicator changes color is called its transition interval.

79

Reading Indicator Values

Litmus gives a very broad reading – a solution is either acidic or basic.

Indicators are more specific in reading the pH of an acid or base.

But the most accurate method of measuring pH is with a pH meter. A pH meter determines the pH of a solution by electrically measuring the voltage between the two electrodes placed in a solution.

80

Determination of the acidity/alkalinity of salt solutions produced by neutralization reactions.

• The relative pH of a neutralized solution can be determined utilizing the following scale showing A/B strengths.

pH 1-3 strong acid with a weak basepH 3-5 strong acid with a moderate basepH 5-7 moderate acid with a weak basepH 7(neutral)equal strength acid/base reactonpH 7-9 moderate base with a weak acidpH 9-12 strong base with a moderate acidpH 12-14 strong base with a weak acidSee your Ch 13 handout for more information.

81

Molarity and Titration

• Standard solution – the solution that contains the precisely known concentration of a solute.

• Primary standard – highly purified solid compound used to check the concentration of the known solution in a titration.

• Knowing the molarity and volume of a known solution used in a titration, the molarity of a given volume of a solution with unknown concentration can be found.

82

Titration Set up and Procedure

Retrieved from web.ysu.edu.

Your book has a more complete explanation. 1. Fill one buret with an acid. Record volume.2. Fill other buret with standard solution base.

Record volume.3. Indicator (Phenolphthalein) will be in a flask.4. Add a given amount of A to the flask.5. Begin adding B to the flask until the pink

color of the indicator begins to form. Swirl the contents constantly.

6. As the pink color begins to remain for longer periods of time, you are nearing the end point.

7. When the pink color remains after 30 seconds of swirling, the equivalence point is reached.

8. Record the exact volume of the base put in the flask.

83

Formulae to calculate molarityor normality in a titration.

Ma x Va = Mb x VbMa – molarity of acidVa – volume of acidMb – molarity of baseVb – volume of base

See Ch 13 handout for more info.

Na x Va = Nb x VbNa – normality of acidVa – volume of acidNb – normality of baseVb – volume of base

84

Titration ProblemWhat is the molarity of a 1.65 L solution of Al(OH)3 if it totally neutralizes 58.0 g of H3PO4 in a 6.00 L solution?Find M of the acid: H=3x1=3, P=1x31=31, O=4x16=64 for total of: 98g/mol58g x 1mol = 0.592 mol of acid 98 gMacid = 0.592 mol = 0.0986M 6.00 LMb = MaVa = 0.0986M(6.00L) = 3.59 x 10 -1 M Al(OH)3

Vb 1.65L

85

Titration using NormalityDetermine the volume of a 0.85 N sulfuric acid needed to neutralize 625 g of Sn(OH)4 in 8.5 L of water. Notice: the base provides 4 equiv of OH-.Find Nb of : Sn=119, O=4(16)=64, H=4(1)=4 for a total of 187g/4 equiv/mol=46.8g/equiv.Nb = 625g x 4 equiv x mol = 1.57 Nb 8.5L mol 187gVa = NbVb = 1.57N (8.5L) = 15.7 L acid Na 0.85N

86

Asgns for Acids and Bases:

Finish Ch13 (14&15) handout pages 125-30. This is due on Friday.Complete Acid-Base Reactions Worksheet