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Acids and BasesAcids and BasesChapter 16Chapter 16

Johannes N. Bronsted Thomas M. Lowry1879-1947. 1874-1936.Both independently developed Bronsted-Lowry theory of acids and bases.

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Acids and Bases: A Brief ReviewAcids and Bases: A Brief ReviewClassical Acids:Taste sourDonate H+ (called “H-plus” or “proton”)Turn litmus redGenerally formed from H-Z, where Z = nonmetal

Classical Bases:Taste bitter and feel soapy.Donate OH- (called “O-H-minus” or “hydroxide”)Turn litmus blueGenerally formed from MOH, where M = metal

Neutralization:Acid + Base Salt + waterH-Z + MOH MZ + HOH

H+ in water is actuallyin the form of H3O+,“hydronium”

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Brønsted-Lowry Acids and BasesBrønsted-Lowry Acids and BasesProton Transfer ReactionsProton Transfer Reactions• Brønsted-Lowry acid/base definition:

acid donates H+

base accepts H+.

• Brønsted-Lowry base does not need to contain OH-.

• Consider HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq):

– HCl donates a proton to H2O. Therefore, HCl is an acid.

– H2O accepts a proton from HCl. Therefore, H2O is a base.

• Water can behave as either an acid or a base.• Amphoteric substances can behave as acids and bases.

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Brønsted-Lowry Acids and BasesBrønsted-Lowry Acids and BasesConjugate Acid-Base PairsConjugate Acid-Base Pairs• Whatever is left of the acid after the proton is donated

is called its conjugate base.• Similarly, whatever remains of the base after it

accepts a proton is called a conjugate acid.• Consider

–After H2O (base) gains a proton it is converted into H3O+ (acid). Therefore, H2O and H3O+ are conjugate acid-base pairs.

–After HCl (acid) loses its proton it is converted into Cl- (base). Therefore HCl and Cl- are conjugate acid-base pairs.

•Conjugate acid-base pairs differ by only one proton.

)(Cl)(OH)O(H)HCl( 32 aqaqlaq

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Brønsted-Lowry Acids and BasesBrønsted-Lowry Acids and BasesConjugate Acid-Base PairsConjugate Acid-Base Pairs

In each of following Bronsted-Lowry Acid/Base reactions,Which is acid? base? conjugate acid? conjugate base?

CH3NH2 + H2SO4 CH3NH3+ + HSO4

-

HCl + H2O H3O+ + Cl-

NH3 + H2O NH4+ + OH-

acid

acid

acid

base

base

base

conj. acid

conj. acid

conj. acid

conj. base

conj. base

conj. base

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Brønsted-Lowry Acids and BasesBrønsted-Lowry Acids and BasesRelative Strengths of Acids Relative Strengths of Acids

and Basesand Bases• The stronger the acid, the

weaker the conjugate base.• The stronger the base, the

weaker the conjugate acid. • H+ is the strongest acid that

can exist in equilibrium in aqueous solution.

• OH- is the strongest base that can exist in equilibrium in aqueous solution.

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Strong acids: completely ionized in water:HCl, HNO3, H2SO4 (also HBr, HI)

Strong bases: completely ionized in water:MOH, where M = alkaliM(OH)2, where M = alkaline earth

Strong and Weak Acids and BasesStrong and Weak Acids and Bases

Weak acids: incompletely ionized in water:any acid that is not strong - acetic acid, etc.Ka is finite.

Weak bases: incompletely ionized in water:any base that is not strong – NH3, etc.Kb is finite.

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The Autoionization of WaterThe Autoionization of WaterThe Ion Product of WaterThe Ion Product of Water

• In pure water the following equilibrium is established

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

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-3

O][H

]][OHO[H

eqK

(at 25oC)

but [H2O]2 = constant

14-3

-3

22

100.1]OH][OH[

]OH][OH[]OH[

w

weq

K

KK

This is called the autoionization of water

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The Autoionization of WaterThe Autoionization of WaterIn pure water at 25oC, [H3O+][OH-] = 1 x 10-14

and also, [H3O+] = [OH-] = 1 x 10-7 (From now on, for simplification, let’s use the abbreviation: [H+] = [H3O+] which means [H]+ = [OH-] = 1 x 10-7

We define pH = -log [H+] and pOH = -log [OH-]

In pure water at 25oC, pH = pOH = 7.00pH + pOH = 14

Acidic solutions have pH < 7.00Basic solutions have pH > 7.00

pKw = 14

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The pH ScaleThe pH Scale

Battery acid

Conc. Drano

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The pH ScaleThe pH Scale

(a) For [H+] = 3.4 x 10-5M, calculate pH, pOH and [OH-] pH = 4.47, pOH = 9.53, [OH-] = 2.95 x 10-10

(b) For [OH-] = 4.4 x 10-3M, calculate pH, pOH, and [H+] pOH = 2.36, pH = 11.64, [H+] = 2.27 x 10-12

(c) For pH= 8.9, calculate, pOH, [H+], [OH-] pOH = 5.1, [H+] = 1.26 x 10-9, [OH-] = 7.94 x 10-6

(d) For pOH= 3.2, calculate pH, [H+], [OH-] pH = 10.8, [H+] = 1.58 x 10-11, [OH-] = 6.3 x 10-4

The pH meter is the most accurate way to measure pH values of solutions.

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Use this “decision tree” to calculate pHUse this “decision tree” to calculate pHvalues of solutions of specific solutions.values of solutions of specific solutions.

1. Is it pure water? If yes, pH = 7.00.2. Is it a strong acid? If yes, pH = -log[HZ]3. Is it a strong base? If yes, pOH = -log[MOH] or pOH = -log (2 x [M(OH)2]) 4. Is it a weak acid? If yes, use the relationship Ka = x2/(HZ – x), where x = [H+]

5. Is it a weak base? If yes, use the relationship Kb = x2/(base – x), where x = [OH-]6. Is it a salt (MZ)? If yes, then decide if it is neutral, acid, or base; calculate its K value by the relationship KaKb = Kw, where Ka and Kb are for a conjugate system; then treat it as a weak acid or base.7. Is it a mixture of a weak acid and its weak conjugate base? It is a buffer; see next chapter.

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2. Strong Acids2. Strong Acids

Calculate the pH of 0.2 M HCl . pH = -log[H+] = -log[0.2] = 0.70

3. Strong Bases3. Strong Bases

Calculate the pH of 0.2 M NaOH . pOH = -log[OH-] = -log[0.2] = 0.70pH = 14 – 0.70 = 13.30

Calculate the pH of 0.2 M Ba(OH)2.pOH = -log[OH-] = -log[0.4] = 0.40pH = 13.60

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Weak acids are only partially ionized in solution, and are in equilibrium:

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

HA(aq) H+(aq) + A-(aq)

[HA]

]][AO[H -3

aK

[HA]]][A[H -

aK

•Ka is the acid dissociation constant.

OR:

(shorthand):

(shorthand)

4. Weak Acids4. Weak Acids

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4. Weak Acids4. Weak Acids

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Calculate pH of 0.2 M solution of acetic acid, HC2H3O2

From previous slide, Ka= 1.8 x 10-5

From now on, we’ll use the shorthand notation for HC2H3O2 = HAc

Equilibrium is then: HAc H+ + Ac-

Initial conc (M): 0.2 0 0

at equilibrium: 0.2-x x x

Ka=

[HAc]

]][Ac[H

x0.2

x2

But, this is a quadratic equation!!

We can assume x is small if Ka < 10-4.

Then, x2/0.2= 1.8 x 10-5 x = 1.90 x 10-3 = [H+] = [Ac-]

pH = -log (1.90 x 10-3) = 2.72

change: -x +x +x

Denote the acetate ion, C2H3O2- as “Ac-”

4. Weak Acids4. Weak Acids

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Polyprotic AcidsPolyprotic Acids• Polyprotic acids have more than one ionizable proton.• The protons are removed in steps not all at once:

H2SO3(aq) H+(aq) + HSO3-(aq)

HSO3-(aq) H+(aq) + SO3

2-(aq)

Ka1 = 1.7 x 10-2

Ka2 = 6.4 x 10-8

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

•Therefore, Ka1 > Ka2 > Ka3 etc.

•Most H+(aq) at equilibrium usually comes from the first ionization (i.e. the Ka1 equilibrium).

4. Weak Acids4. Weak Acids

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Polyprotic AcidsPolyprotic Acids

4. Weak Acids4. Weak Acids

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Carbonic acid, H2CO3, Ka1=4.3 x 10-7

H2CO3 H+ + HCO3- Ka1 = 4.3 x 10-7

HCO3- H+ + CO3

2- Ka2 = 5.6 x 10-11

Since Ka1>>Ka2, nearly all H+ ions come from 1st equilibrium.Therefore, the 1st equilibrium determines the pH.Calculate pH of 0.4 M H2CO3 solution.Ka = x2/HZ-x4.3 x 10-7 = x2/0.4x = [H+] = 4.15 x 10-4

pH = -log (4.1 x 10-4) = 3.38

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

resulting ions:Weak base + H2O conjugate acid + OH-

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

]NH[

]OH][NH[

3

-4

bK

Example:

The base dissociation constant, KThe base dissociation constant, Kbb is defined as: is defined as:

5. Weak Bases5. Weak Bases

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• The larger Kb the stronger the base.

5. Weak Bases5. Weak Bases

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What is pH of 0.15 M solution of methylamine, NH2CH3, a weak base?

For convenience, denote it as “B”

Then: B + H2O BH+ + OH- Kb =4.4 x 10-4 Initially: 0.15 0 0change: -x +x +xAt equil: 0.15-x x x

Then: 422

4.4x10.15

x

x.15

x

][OH8.12 x 10x 3

2.09)0log(8.12x1]log[OHpOH 3

11.92.0914pOH14pH

5. Weak Bases5. Weak Bases

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Relationship Between KRelationship Between Kaa and K and Kbb

• For a conjugate acid-base pair

Ka Kb = Kw (constant)

• Therefore, the larger the Ka, the smaller the Kb. That is, the stronger the acid, the weaker the conjugate base.

• Taking negative logarithms:

-log Ka- log Kb= -log Kw

pKa + pKb = pKw

For HAc (acetic acid), the pKa = -log (1.8 x 10-5) = 4.74.Thus, the pKb for Ac- (acetate 0 is 14 – 4.74 = 9.26

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Salts may be acidic, basic or neutral.Salts made by a strong acid and a strong base are neutral, e.g., NaCl, KNO3.Salts made by a weak acid and a strong base are weakly basic,e.g., sodium acetate, NaAc, NaHCO3.Salts made by a strong acid and a weak base are weakly acidic,e.g., NH4Cl.

Calculate the pH of a 0.35 M solution of sodium acetate.Since Ka Kb = Kw, where Ka = 1.8 x 10-5 for acetic acid,Then Kb for NaAc (sodium acetate) is Kb = Kw/Ka = 1.00 x 10-14/1.8 x 10-5 = 5.56 x 10-10.Now treat this as a weak base problem, Kb = x2/base = 5.56 x 10-10 = x2/0.35x = [OH-] = 1.39 x 10-5

pOH = 4.85 and pH = 9.14

6. Salts6. Salts