Acid Base Equilibriamaybe do packet 2 before

packet 3? Combine concepts from 1 and 3?

Arrhenius Acids & Bases(1st year chem definition!)

An Arrhenius acid is a substance that, when dissolved in water, increases the concentration of H+ (needs H+ in it)

Example: HCl (monoprotic)H2SO4 (diprotic)

An Arrhenius base is a substance that, when dissolved in water, increases the concentration of OH–

Example: NaOH (monobasic)Ca(OH)2 (dibasic)

Brønsted-Lowry Acids and Bases

Brønsted-Lowry acid is a species that donates H+ (proton)

Brønsted-Lowry base is a species that accepts H+ (proton)

Brønsted-Lowry definition of a base does not mention OH– and the reaction does not need to be aqueous.

The H+ Ion in Water• The H+(aq) ion is

simply a proton with no surrounding valence electrons.

• In water, clusters of hydrated H+(aq) ions form.

• The simplest cluster is H3O+(aq) We call this a hydronium ion.

• Larger clusters are also possible (such as H5O2

+ and H9O4+).

• Generally we use H+(aq) and H3O+(aq) interchangeably.

Proton-Transfer Reactions• Consider

NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)

• H2O donates a proton to ammonia. – Therefore, water is acting as an acid.

• NH3 accepts a proton from water. – Therefore, ammonia is acting as a base.

Amphoteric substances • Can behave as acids and bases.• Water is an example of an amphoteric

species.

Water as an acid (proton donor)H2O + NH3 NH4

+ + OH-

Water as a base (proton acceptor)H2O + HNO2 H30+ + NO2

-

Conjugate Acid-Base Pairs

• A conjugate acid is the substance formed by adding a proton to the base.

• A conjugate base is the substance left over after the acid donates a proton.

Within a pair the acid has more hydrogen!

Strong Acids & BasesStrong Acids Strong Basessulfuric acid H2SO4 Lithium hydroxide LiOH

Nitric acid HNO3 Sodium hydroxide NaOH

Perchloric acid HClO4 Potassium hydroxide KOH

Chloric acid HClO3 Rubidium hydroxide RbOH

Hydrochloric acid HCl Cesium hydroxide CsOH

Hydrobromic acid HBr Calcium hydroxide Ca(OH)2

Hydroiodic acid HI Strontium hydroxide Sr(OH)2

Barium hydroxide Ba(OH)2

Less common: O2- H- CH3

-

All other acids and bases are weak!

Ions

Neutral anions Neutral cationsHydrogen sulfate HSO4

- Lithium Li+

Nitrate NO3- Sodium Na+

Perchlorate ClO4- Potassium K+

Chlorate ClO3- Rubidium Rb+

Chloride Cl- Cesium Cs+

Bromide Br- Calcium Ca2+

Iodide I- Strontium Sr2+

Barium Ba2+

Most anions are weak basesMost cations are weak acids

Anions of strong acids and cations of strong bases are neutral

Strengths of Acids and BasesStrong acids completely ionize in water.

HCl + H2O H3O+ + Cl-

HCl H+ + Cl-

• Essentially no un-ionized molecules remain in solution so the equation usually does not contain equilibrium arrows. Keq >>1

• Their conjugate bases have negligible tendencies to become protonated

Cl- + H+ HCl.

The conjugate base of a strong acid is a neutral anion.

Strengths of Acids and BasesStrong bases completely dissociate in water

NaOH(aq) Na+(aq) + OH-(aq)

• Essentially no undissociated compound remains in solution so the equation usually does not contain equilibrium arrows. Keq >>1

• The ions have negligible tendencies to attract OH- in solution.

Na+(aq) + OH-(aq) NaOH(aq)

The cation of a strong base is a neutral cation.

All other acids are Weak acids. They only partially dissociate in aqueous solution.

HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2-(aq)

• They exist in solution as a mixture of molecules and component ions. (usually mostly molecules in equilibrium)

• Their conjugate bases are weak bases.– Besides non-neutral anions, weak bases tend to be

nitrogen containing organic compounds • Example: Acetic acid is a weak acid; acetate

ion (conjugate base) is a weak base.

acid conjugate basebase conjugate

acid

Non-acid compounds with hydrogen

Not all compounds containing hydrogen are acidic. These are extremely weak acids…so much that we

don’t consider them acids at all. Their conjugate bases are strong bases!

Negligible acidity: OH- H2 CH4

Strong bases: O2- H- CH3-

Page 657 in textbook

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

In acid-base reactions, the reaction favors the transfer of a proton from the stronger acid to the stronger base to form a weaker acid and weaker base.

We need a more specific way to determine acid strength!

Where does that acid/base ranking come from? Keq!!!

Since weak acids and bases are in equilibrium…we can write equilibrium constant expressions!

When looking at the reaction of a weak acid with water we label the equilibrium constant Ka:HF(aq) + H2O(l) H3O+(aq) + F-(aq)

When looking at the reaction of a weak base with water we label the equilibrium constant Kb:NH3(aq) + H2O(l) NH4

+(aq) + OH-(aq)

3[ ][ ][ ]a

H O FKHF

4

3

[ ][ ][ ]b

NH OHKNH

Ka and Kb

• The value of Ka or Kb indication the extent to which the weak acid or base ionizes/dissociates.– Larger Ka or Kb means more products!

• Weak acids with larger Ka values are stronger.

• Weak bases with larger Kb values are stronger. (Values in Appendix D)

19

Percent Ionization• Percent ionization is another method to assess

acid strength.• For the reaction: HA(aq) H+(aq) + A–

(aq)

• The higher the percent ionization, the stronger the

acid.

+equilibrium

initial

H% ionization = × 100

HA

Polyprotic Acids Polyprotic acids have more than one ionizable proton.• The protons are removed in successive steps.

Consider the weak acid, H2SO3 (sulfurous acid):H2SO3(aq) H+(aq) + HSO3

–(aq) Ka1 = 1.7 x 10–2

HSO3–(aq) H+(aq) + SO3

2–(aq) Ka2 = 6.4 x 10–8

• It is always easier to remove the first proton in a polyprotic acid than the second.

Ka1 > Ka2 > Ka3, etc

The Autoionization of WaterIn pure water the following equilibrium is established:

H2O(l) + H2O(l) H3O+(aq) + OH–(aq) acid base acid base• This process is called the autoionization of water.

We can write an equilibrium constant expression for the autoionization of water:

Kw = [H3O+] [OH–] = 1.0*10-14 @25°C Kw is called the “ion-product constant”Kw = Ka*Kb for conjugate pair

The Ion Product Constant

• This applies to pure water as well as to aqueous solutions. (for our purposes…)

• A solution is neutral if [OH–] = [H3O+].• If the [H3O+] > [OH–], the solution is acidic. • If the [H3O+] < [OH–], the solution is basic.

In a neutral solution at 25oC, [H+] = [OH-] = 1.0 x 10-7 M

Give the conjugate base of the following Bronsted-Lowry acids:

(a) HIO3 (b) NH4+1 (c) H2PO4

-1 (d) HC7H5O2

remove H+

(a) IO3-1

(b) NH3 (c) HPO4

-2 (d) C7H5O2

-

24

Designate the Bronsted-Lowry acid and the Bronsted-Lowry base on the left side of each of the following equations, and also designate the conjucate acid and base on the right side:

(a) NH4+1(aq) + CN-1(aq) HCN(aq) + NH3(aq)

acid base acid base

(b) (CH3)3N(aq) + H2O OH-1(aq) + (CH3)3NH+1(aq) base acid base acid

(c) HCHO2(aq) + PO4-3(aq) HPO4

-2(aq) + CHO2-1(aq)

acid base acid base

(a) The hydrogen oxalate ion (HC2O4-1) is amphoteric. Write a balanced

chemical equation showing how it acts as an acid toward water and another equation showing how it acts as a base towards water. (b) What is the conjugate acid and base of HC2O4

-1?

(a) Acid: HC2O4-1(aq) + H2O(l) C2O4

-2(aq) + H3O+1(aq) acid base conj base conj acid

Base: HC2O4-1(aq) + H2O(l) H2C2O4(aq) + OH-1(aq)

base acid conj acid conj base

(b)

26

Label each of the following as being a strong acid, a weak acid, or a species with negligible acidity. In each case write the formula of its conjugate base, and indicate whether the conjugate base is a strong base, a weak base or a species with negligible basicity: (a) HNO2 (b) H2SO4 (c) HPO4

-2 (d) CH4 (e) CH3NH3

+1

(a) HNO2 , (b) H2SO4 , (c) HPO4

-2 ,(d) CH4 , (e) CH3NH3

+1 ,

NO2-1 ,

HSO4-1 ,

PO4-3 ,

CH3-1 ,

CH3NH2 ,

ACID BASEstrong negligibleweak weaknegligible strong

IMPORTANT

weak acid weak basestrong acid negligible baseweak acid weak base

negligible acid strong baseweak acid weak base

(a) Which of the following is the stronger acid, HBrO or HBr?

(b) Which is the stronger base, F-1 or Cl-1?

Briefly explain your choices.

(a) HBr - It is one of the seven strong acids(b) F-1

HF is a weak acid so F- is a weak baseHCl is a strong acid so Cl- is neutral

Predict the products of the following acid-base reactions & determine whether equilibrium lies to the right or left:

(a) O-2(aq) + H2O(l) (b) CH3COOH(aq) + HS-1(aq) (c) NO3

-1(aq) + H2O(l) * You will need a chart like this or Ka/Kb values to determine equilibrium!

(a) O-2(aq) + H2O(l) OH-1(aq) + OH-1(aq)

base acid acid base

*OH- is the weaker acid so equilibrium lies to the right

(b) CH3COOH(aq) = HC2H3O2 (aq) HC2H3O2 (aq) + HS-1(aq) C2H3O2

-1(aq) + H2S(aq)

acid base base acid

*H2S is the weaker acid so equilibrium lies to the right

(c) NO3-1(aq) + H2O(l) HNO3(aq) + OH-1(aq)

base acid acid base

*H2O is the weaker acid so equilibrium lies (far!) to the left

The “p” Function• p is short for “– log10”

pH = -log10[H+] = -log[H+]pOH = -log10[OH-] = -log [OH-]

• Note that this is a logarithmic scale. • Thus a change in [H+] by a factor of 10

causes the pH to change by 1 unit. • Most pH values fall between 0 and 14.

• In neutral solutions at 25oC, [H+] = 1.0 x 10-7

pH = -log[1.0 x 10-7] = 7.00

Lower pH = more acidicHigher pH = more basic

[H+] pH

Acidic > 1.0 x 10–7 < 7.00

Neutral 1.0 x 10-7 7.00

Basic < 1.0 x 10–7 > 7.00

Another one: pK• We can use a similar system to describe the

equilibrium constantpK = - log[K]

• The value of Kw at 25oC is 1.0 x 10–14

pKw = - log (1.0 x 10–14)= 14.0

Kw=[H+][OH-]pKw= pH + pOH = 14.0

The pH Loop

pH

pOH

[H+]

[OH-]

[H+]=10-pH

pH = -log[H+]

[OH-]=10-pOH

pOH = -log[OH-]

pH + pOH=14.0[H+] [OH-]=10-14

Measuring pHThe most accurate method to measure pH is to use a pH meter.

Acid Base Indicators:• certain dyes change color as pH changes. • Indicators are less precise than pH meters.

– Many indicators do not have a sharp color change as a function of pH.• Most acid-base indicators can exist as either an acid or a

base.– These two forms have different colors.– The relative concentration of the two different forms is sensitive to the

pH of the solution.• Thus, if we know the pH at which the indicator turns color, we

can use this color change to determine whether a solution has a higher or lower pH than this value.

Example 1: Calculate [H+1] for each of the following solutions and indicate whether the solution is acidic, basic or neutral. (a) [OH-1] = 0.0007 M (b) a solution where [OH-1] is 100 times greater than [H+1]

]][[ OHHkw

14 4

11

1.0 10 [ ](7 10 )

[ ] 1 10 (basic)

x H x

H x M

][100][ HOH

2][100][100*][ HHHkw

(basic) 100.1100

100.1100

][ 814

MxxKH w

Example 2: (a) If NaOH is added to water, how does the [H+1] change? How does pH change?(b) If [H+1] = 0.005 M, what is the pH of the solution? Is the solution acidic or basic?(c) If pH = 6.3, what are the molar concentrations of H+1(aq) and OH-1(aq) in the

solution?

(a) Kw = [H+] [OH-]. the [OH-] will increase and the [H+] will decrease.

[H+] decreases, pH increases.

(b) pH = - log [H+] = - log (0.005) = 2.3 acidic

(c) pH = 6.3 [H+] = 10-pH = 10-6.3 = 5 x 10-7 M pOH = 14 – pH = 14 – 6.3 = 7.7

[OH-] = 10-pOH = 10-7.7 = 2 x 10-8 M

Strong Acid Calculations• In solution the strong acid is usually the only

source of H+

• The pH of a solution of a monoprotic acid may usually be calculated directly from the initial molarity of the acid.

Caution: If the molarity of the acid is less than 10–6 M then the autoionization of water needs to be taken into account.

Example 3: Calculate the pH of each of the following strong acid solutions: (a) 1.8 10-4 M HBr (b) 1.02 g HNO3 in 250 mL of solution (c) 2.00 mL of 0.500 M HClO4 diluted to 50.0 mL (d) a solution formed by mixing 10.0 mL of 0.0100 M HBr with 20.0 mL of 2.5 10-3 M HCl

(a) 1.8 x 10-4M HBr = 1.8 x 10-4M H+pH = -log(1.8 x 10-4) = 3.74

(b)

0.0647 M HNO3 = 0.0647 M H+ pH = -log (.0647) = 1.19

(c) M1V1 = M2V2

(0.500 M)(.00200 L) = (x M)(0.0500 L)x = .0200 M HCl

(d)

pH = -log(.0050) = 2.30

3 33

3

1.02 g HNO 1 mol HNO* 0.0647 M HNO0.250 L 63.0 g HNO

(.0100 *.0100 ) (.00250 *.0200 )[ ] .00500.0300total

M L M LH ML

[H+] = .0200 M pH = -log(.0200) = 1.70

Strong Base Calculations

• Strong bases are strong electrolytes and dissociate completely in solution.

• For example:

NaOH(aq) Na+(aq) + OH–(aq)

The pOH (and thus the pH) of a strong base may be calculated using the initial molarity of the base.

Example 4: Calculate [OH-1] and pH for (a) 3.5 10-4 M Sr(OH)2 (b) 1.50 g LiOH in 250 mL of solution (c) 1.00 mL of 0.095 M NaOH diluted to 2.00 L (d) a solution formed by adding 5.00 mL of 0.0105 M KOH to 15.0 mL of 3.5 10-3 M Ca(OH)2

(a) [OH-] = 2[Sr(OH)2] = 2(0.00035 M) = .00070 M OH-

pOH = -log(.00070) = 3.15pH = 14 – 3.15 = 10.85

(b)

pOH = -log (.251) = 0.601pH = 14 - .601 = 13.399

(c) M1V1 = M2V2 (0.095 M)(.00100 L) = (x M)(2.00 L)x = .000048 M NaOH = [OH-]

pOH = -log(.000048) = 4.32pH = 14 – 4.32 = 9.68

(d)

pOH = -log(.0079) = 2.1pH = 14 – 2.1 = 11.9

][OHLiOH M 251.0LiOH g 24.0LiOH mol 1*

L 0.250LiOH g 50.1 -

ML

LMLMOH total 0079.0200.

)0150.*0035(.2)00500.*0105(.][

Weak Acid Calculations • Weak acids are only partially ionized in

aqueous solution.• Therefore, weak acids are in equilibrium:

HA(aq) + H2O(l) H3O+(aq) + A–(aq)

OR HA(aq) H+(aq) + A–(aq)• We can write Ka for this dissociation:

HAAHK

HAAOHK aa

or 3

Calculating Ka from pH for weak acids

• In order to find the value of Ka, we need to know all of the equilibrium concentrations. (ICE Chart)

• The pH gives the equilibrium concentration of H+.

• We then substitute these equilibrium concentrations into the equilibrium constant expression and solve for Ka.

Example 4: A 0.20 M solution of niacin (a monoprotic weak acid) has a pH of 3.26. What is the Ka for niacin?

HA H+ A-

I 0.20 0 0

C -.000550 +.000550 +.000550

E .019945 10-3.26=.000550 .000550

5(.000550)(.000550) 1.5*10(.019945)aK

Using Ka to Calculate pH for weak acids

• Write the balanced chemical equation clearly showing the equilibrium.

• Write the equilibrium expression. Look up the value for Ka (in a table).

• Write down the initial and equilibrium concentrations for everything except pure water. (ICE table)

• We usually assume that the equilibrium concentration of H+ is x.

• Substitute into the equilibrium constant expression and solve.

• Remember to convert x to pH if necessary.

Polyprotic Acids • Polyprotic acids have more than one ionizable

proton.H2SO3(aq) H+(aq) + HSO3

–(aq) Ka1 = 1.7 x 10–2

HSO3–(aq) H+(aq) + SO3

2–(aq) Ka2 = 6.4 x 10–8

• The majority of the H+(aq) at equilibrium usually comes from the first ionization

• If the successive Ka values differ by a factor of 103, we can usually get a good approximation of the pH of a solution of a polyprotic acid by considering the first ionization only.

• If not, then we have to account for the successive ionizations

Example 5: The acid dissociation constant for benzoic acid, HC7H5O2 is 6.5 x 10-5. Calculate the equilibrium concentrations of H3O+, C7H5O2

-, and HC7H5O2 . The initial concentration of HC7H5O2 is 0.050M.

HC7H5O2 (aq) H+(aq) + C7H5O2- (aq)

I 0.050 0 0C -x x xE (.050-x) x x

x2 + (6.5*10-5)x – (3.25*10-6) = 0

x= .0018M = [H+]=[C7H5O2- ]

[HC7H5O2] = .050 - .0018 = .048 M

257 5 2

7 5 2

[ ][ ] 6.5*10[ ] (.050 )aH C H O xKHC H O x

What if I don’t have a quadratic equation program?Algebra Shortcut:

Assume x is much smaller (less than 5% of .050)

To simplify: .050-x = .050Now you don’t need the quadratic equation!

x= .0018M = [H+]=[C7H5O2- ]

[HC7H5O2] = .050 - .0018 = .048 M

If you made this assumption you need to check and make sure it’s valid – if this answer isn’t less than 5% use the quadratic equation!

2 257 5 2

7 5 2

[ ][ ] 6.5*10[ ] (.050 ) .050aH C H O x xKHC H O x

.0018 *100% 3.6%.050

Example 6: Calculate the pH of the following solution (Ka and Kb values are in Appendix D). 0.175 M hydrazoic acid, HN3

(a) HN3(aq) H+(aq) + N3-(aq)

I 0.175M 0 0 C -x x x

E 0.175-x x x

x2 + (1.9*10-5)x – (3.325*10-6) = 0x=.0018M H+

pH = -log(0.0018) = 2.74

253

3

[ ][ ]1.9*10

(0.175 )aH N xKHN x

Example 7: Calculate the percent ionization of 0.400 M hydrazoic acid, HN3, solution.

HN3(aq) H+(aq) + N3-(aq)

I 0.400M 0 0 C -x x x

E 0.400-x x x

x2 + (1.9*10-5)x – (7.6*10-6) = 0x=.0028M H+

[ ] 0.0028% *100% 0.70%[ ] 0.400

eq

initial

Hionization

HA

253

3

[ ][ ]1.9*10

(0.400 )aH N xKHN x

Example 8: Citric acid, which is present in citrus fruits, is a triprotic acid. Calculate the pH of a 0.050 M solution of citric acid.

H3C6H5O7(aq) H+(aq) + H2C6H5O7-1(aq) Ka = 7.4*10-4

H2C6H5O7-1(aq) H+(aq) + HC6H5O7

-2(aq) Ka = 1.7*10-5

HC6H5O7-2(aq) H+(aq) + C6H5O7

-3(aq) Ka = 4.0*10-7

To calculate the pH of a .050M solution, assume initially that only the first ionization is important

H3C6H5O7(aq) H+(aq) + H2C6H5O7-1(aq)

I 0.050 0 0C -x x xE 0.050-x x x

x = 0.0057 0.0057=[H+] = [H2C6H5O7

-1]

242 6 5 7

3

[ ][ ]7.4*10

(0.050 )aH H C H O xK

HN x

Does the second ionization have any effect?H2C6H5O7

-1(aq) H+(aq) + HC6H5O7-2(aq) Ka = 1.7*10-5

I 0.0057 0.0057 0C -y y yE 0.0057-y 0.0057+y y

51

7562

2756

2 10*7.10057.0

)0057(.][

]][[

yyy

OHCHOHHCHKa

y = 0.000017

This value is small compared to 0.0057 (think SDs)

Total [H+] = 0.0057 + .000017 = 0.0057

which indicates the 2nd (and any subsequent) ionizations can be ignored

pH = -log(0.0057) = 2.24

Weak Base Calculations• Weak bases remove protons from substances.• There is an equilibrium between the base and the

resulting ions:Example: NH3(aq) + H2O(l) NH4

+(aq) + OH–(aq).The base-dissociation constant, Kb, is

The larger Kb, the stronger the base.

4

3b

NH OHK

NH

16.8 Relationship Between Ka and Kb

• Generally only either Ka or Kb for a conjugate pair is reported in tables. If you know one you can find the other!

• Consider the following equilibria: NH4

+(aq) NH3(aq) + H+(aq)NH3(aq) + H2O(l) NH4

+(aq) + OH–(aq)

• We can write equilibrium expressions for these reactions: +

3a +

4

[NH ][H ]K =[NH ]

+ -4

b3

[NH ][OH ]K =[NH ]

• If we add these equations together:NH4

+(aq) NH3(aq) + H+(aq)Ka

NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)

Kb• The net reaction is the autoionization of water.

H2O(l) H+(aq) + OH–(aq)Kw

• When we add equations we multiply K.Kw = Ka x Kb

Alternatively, we can express this as:pKa + pKb = pKw = 14.00 (at 25oC)

Example 9: Calculate the molar concentration of OH-1 ions in a 0.050 M solution of hydrazine, H2NNH2, Kb = 1.3 10-6. What is the pH of this solution?

+1 12 32 2 2 (aq) H NNH (aq) + H O (l) H NNH (aq) + OH

initial 0.050 0 0change

- x x xequil. 0.050 - x x x

-1

2

+1 2-62 3

b2

[H NNH ][OH ] xK = = = 1.3 x 10[H NNH ] 0.050 - x

x = 0.00025 M

-1pOH = - log [OH ] = - log (.00025) = 3.60pH = 14 - pOH = 14 - 3.60 = 10.40

Example 10: Although the acid dissociation constant for phenol, C6H5OH, is listed in Appendix D, the base dissociation constant for the phenolate ion, C6H5O-1 is not. (a) Explain why it is not necessary to list both. (b) Calculate the Kb for the phenolate ion. (c) Is the phenolate ion a weaker or stronger base than ammonia, NH3?

a b w(a) For a conjugate acid/conjugate base pair K x K = K

-14-5w

b -10a

K 1.0 x 10(b) K = = = 7.7 x 10

K 1.3 x 10

-1 -5 -5b 6 5 b 3

-16 5

(c) K for C H O (7.7 x 10 ) > K for NH (1.8 x 10 )

C H O is a stronger base

Appendix D

59

16.9 Acid-Base Properties of Salt Solutions

• All soluble salts are strong electrolytes.- In solution, they exist nearly entirely of

ions.- Acid-base properties of salts are due to the

reactions of their ions in solution.

• Many ions can react with water to form OH– or H+.• This process is called hydrolysis.

Remember this?

Neutral anions Neutral cationsHydrogen sulfate HSO4

- Lithium Li+

Nitrate NO3- Sodium Na+

Perchlorate ClO4- Potassium K+

Chlorate ClO3- Rubidium Rb+

Chloride Cl- Cesium Cs+

Bromide Br- Calcium Ca2+

Iodide I- Strontium Sr2+

Barium Ba2+

Most anions are weak basesMost cations are weak acids

Anions of strong acids and cations of strong bases are neutral

SummaryAnions

– Of strong acids are neutral.•Example: Cl- of HCl

– Cl- + H2O X– Of weak acids are basic.

•Example: F- of HF– F- + H2O HF + OH-

– With ionizable protons are amphoteric.• Example: HSO4

–

Summary

Cations – Of strong bases are neutral.

•Example: Na+ of NaOH– Na+ + H2O X

–All other cations are weak acids•Example: Fe3+

• Cation hydrolysis reaction:

Smaller and more highly charged ions = stronger (weak) acids

(All are 1.00 M solutions)

The pH of a solution may be qualitatively predicted:

Salts from a weak acid and weak base can be either acidic or basic.

Compare Ka of the cation and Kb of the anion- If the Ka is larger the solution will be acidic- If the Kb is larger the solution will be basic

For example, consider NH4CN. The Kb of CN-1 is larger than the Ka of NH4

+1

so the solution will be basic.

Cation Anion Solution is: Example

Neutral(from strong base)

Neutral(from strong acid)

Neutral NaCl

Neutral(from strong base)

Basic(from weak acid)

Basic NaF

Acidic(from a weak base)

Neutral (from a strong acid)

Acidic FeCl3NH4Cl

Acidic (from a weak base)

Basic(from a weak acid)

Depends! NH4CN

Example 1: Predict whether aqueous solutions of the following compounds are acidic, basic or neutral.

(a) NH4Br (b) FeCl3 (c) Na2CO3

(d) KClO4 (e) NaHC2O4

Ka for acid HC2O4

-1 = 6.410-5

Kb for base HC2O4-1 = 1.710-13

acidic acidic basic

neutral acidic

Example 2: Using data from Appendix D, calculate [OH-1] and pH for the following solution 0.10 M NaCN

+1 -1

+1 -1NaCN is a soluble salt and will dissociate into Na and CN ions. Na ion does not interact with water, but CN acts as a weak base:

--5

b -1

21[HCN][OH ]K = = 2.0 x 10 =[C

x 0.] 10-xN

-14-1 w

b -105

a

-K 1 x 10K for CN = = =K for HCN 4.9

2.0 x 10

10 x

-1 -1

2 CN + H O HCN + OH I 0.10 0 0 C - x x x E 0.10 - x x x

14 - 2.85 = 1pH = 1.15

0.0pOH 014 = - l ) =og( 2.85

-1 0.0014 M = [x OH = ]

Ka for HCN (given in Appendix D) = 4.9x10-10

16.10 Acid-Base Behavior and Chemical StructureAcidity is directly related to the strength of attraction for

a pair of electrons to a central atom.

4 situations to consider:

1. IonsIonic Charge and Size

When comparing ions of similar structure:More positive ions are stronger acids.tie breaker: Smaller ion is stronger acid

Acid strength: Na+ < Ca2+ < Cu2+ < Al3+

PO43- < HPO4

2- < H2PO4- <

H3PO4

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Example 3: Predict which member of each pair produces the more acidic aqueous solution:

(a) K+1 or Cu+2 (b) Fe+2 or Fe+3 (c) Al+3 or Ga+3

(a) Cu+2 has higher charge (and K+1 is neutral)

(b) Fe+3 has higher charge

(c) Al+3 has a smaller size

2. Binary Acids: Bond Polarity (Electronegativity) & Strength

• The H–X bond strength is important in determining relative acid strength in any group in the periodic table. – The weaker the bond the easier it will break– The H–X bond strength tends to decrease down

a group - acid strength increases down a group

• H–X bond polarity is important in determining relative acid strength in any period of the periodic table. – The H-X bond polarity tends to increase across a period

- acid strength increases (from left to right) across a period

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more polar bond – stronger acid larger atomweaker bondstronger acid

3. Oxyacids (Acids with oxygen) with different central atoms

Generally, the larger the electronegativity of the central atom the stronger the acid.–The stronger the pull on electrons the less

tightly the H is held

Acid Strength: H3BO3 < H2CO3 < HNO3

The higher EN of central atom means more electron density shift away from H - H is easier to remove – stronger acid

4. Oxyacids with the same central atomGenerally, the more oxygens attached to the

central atom the stronger the acid.–The more atoms pulling on electrons the

less tightly the H is held

Acid Strength: HClO < HClO2 < HClO3 < HClO4

More O atoms means more electron density shift away from H and H is easier to remove – stronger acid

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Example 4: Explain the following observations: (a) HNO3 is a stronger acid than HNO2

(b) H2S is a stronger acid than H2O

(c) H2SO4 is a stronger acid than HSO4-1

(d) H2SO4 is a stronger acid than H2SeO4

(e) CCl3COOH is a stronger acid than CH3COOH(a) more oxygen atoms - electron density shifts away from H

bond - easier to remove H(b) bond strength decreases down a group - H easier to

remove from S(c) More positive ion is stronger (d) S has higher electronegativity than Se – pulls electrons from H –

easier to remove H(e) the more electronegative 3 Cl atoms (as opposed to the 3

H atoms) pull electron density more weakens the O-H bond and makes H easier to remove

16.11 Lewis Acids and Bases• A Brønsted-Lowry acid is a proton donor.• Lewis proposed a new definition of acids and

bases that emphasizes the shared electron pair. • A Lewis acid is an electron pair acceptor.• A Lewis base is an electron pair donor.

• Note: Lewis acids and bases do not need to contain protons.

• Therefore, the Lewis definition is the most general definition of acids and bases.

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Typical Lewis acid-base reaction

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• What types of compounds can act as Lewis acids?• Lewis acids must have a vacant orbital (into which the electron pair can be donated).• Lewis acids sometimes have an incomplete octet (e.g., BF3).• Transition-metal ions can be Lewis acids (empty d orbitals)• Compounds with multiple bonds can act as Lewis acids.

Example 5: Identify the Lewis acid and Lewis base in each of the following reactions:(a) Fe(ClO4)3 + 6 H2O Fe(H2O)6

+3 + 3 ClO4-1

(b) CN-1 + H2O HCN + OH-1

(c) (CH3)3N + BF3 (CH3)NBF3

(d) HIO + NH2-1 NH3 + IO-1

Acid Base

H donor H acceptore acceptor e donator

empty orbitals/mult. bonds has lone pairs

incomplete octet/cation often contains N

(a) Fe(ClO4)3 or Fe+3 H2O (b) H2O CN-1

(c) BF3 (CH3)3N (d) HIO NH2

-1