the chemistry of acids and bases - texas tech university · 2013. 3. 12. · strong and weak...

© 2009 Brooks/Cole - Cengage

1The Chemistry of Acids and Bases


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Acid and Bases


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Acid and Bases


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Acid and Bases


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Strong and Weak Acids/Bases• Generally divide acids and bases into

STRONG or WEAK ones.STRONG ACID: HNO3(aq) + H2O(liq)

f H3O+(aq) + NO3-(aq)

HNO3 is about 100% dissociated in water.

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HNO3, HCl, H2SO4 and HClO4 are among the only known strong acids.

Strong and Weak Acids/Bases


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• Weak acids are much less than 100% ionized in water.

One of the best known is acetic acid = CH3CO2H



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• Strong Base: 100% dissociated in water.

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


Other common strong bases include KOH and Ca(OH)2. (limited solub.)

CaO (lime) + H2O f

Ca(OH)2 (slaked lime)CaO


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• Weak base: less than 100% ionized in water

One of the best known weak bases is ammonia

NH3(aq) + H2O(liq) e NH4+(aq) + OH-(aq)


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ACID-BASE THEORIES• The most general theory for common

aqueous acids and bases is the BRØNSTED - LOWRY theory

• ACIDS DONATE H+ IONS

• BASES ACCEPT H+ IONS


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The Brønsted definition means NH3 is a BASE in water — and water is itself an ACID

ACID-BASE THEORIES

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ACID-BASE THEORIESNH3 is a BASE in water — and water is itself

an ACID

NH3 / NH4+ is a conjugate pair — related

by the gain or loss of H+

Every acid has a conjugate base - and vice-versa.


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Conjugate Pairs


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More About WaterH2O can function as both an ACID and a BASE.

In pure water there can be AUTOIONIZATION

Equilibrium constant for autoion = Kw

Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC

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More About Water

Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC

In a neutral solution [H3O+] = [OH-]

so Kw = [H3O+]2 = [OH-]2

and so [H3O+] = [OH-] = 1.00 x 10-7 M


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Calculating [H3O+] & [OH-]You add 0.0010 mol of NaOH to 1.0 L of

pure water. Calculate [H3O+] and [OH-].Solution2 H2O(liq) e H3O+(aq) + OH-(aq)Le Chatelier predicts equilibrium shifts to

the ____________. [H3O+] < 10-7 at equilibrium.Set up a ICE table.


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pure water. Calculate [H3O+] and [OH-].Solution

2 H2O(liq) e H3O+(aq) + OH-(aq) initial 0 0.0010 equilib x 0.0010 + xKw = (x) (0.0010 + x)Because x << 0.0010 M, assume [OH-] = 0.0010 M[H3O+] = x = Kw / 0.0010 = 1.0 x 10-11 M


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pure water. Calculate [H3O+] and [OH-].Solution2 H2O(liq) e H3O+(aq) + OH-(aq)[H3O+] = Kw / 0.0010 = 1.0 x 10-11 M

This solution is _________

because [H3O+] < [OH-]


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[H3O+], [OH-] and pHA common way to express acidity and

basicity is with pH

pH = log (1/ [H3O+])

= - log [H3O+]In a neutral solution,

[H3O+] = [OH-] = 1.00 x 10-7 at 25 oC

pH = -log (1.00 x 10-7)


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[H3O+], [OH-] and pHWhat is the pH of the

0.0010 M NaOH solution? [H3O+] = 1.0 x 10-11 M

pH = - log (1.0 x 10-11) = 11.00

General conclusion —

Basic solution pH > 7 Neutral pH = 7 Acidic solution pH < 7


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The pH Scale

See Active Figure 17.1


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[H3O+], [OH-] and pHIf the pH of Coke is 3.12, it is ____________.Because pH = - log [H3O+] then

log [H3O+] = - pH

Take antilog and get

[H3O+] = 10-pH

[H3O+] = 10-3.12 = 7.6 x 10-4 M


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pH of Common Substances


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Other pX ScalesIn general pX = -log Xand so pOH = - log [OH-] Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC

Take the log of both sides -log (10-14) = - log [H3O+] + (-log [OH-])

pKw = 14 = pH + pOH


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Equilibria Involving Weak Acids and Bases

Aspirin is a good example of a weak acid, Ka = 3.2 x 10-4


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Acid Conjugate Baseacetic, CH3CO2H CH3CO2

-, acetateammonium, NH4

+ NH3, ammoniabicarbonate, HCO3

- CO32-, carbonate

A weak acid (or base) is one that ionizes to a VERY small extent (< 5%).


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Consider acetic acid, CH3CO2H (HOAc)HOAc + H2O e H3O+ + OAc-

Acid Conj. base

(K is designated Ka for ACID)

Because [H3O+] and [OAc-] are SMALL, Ka << 1.


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Equilibrium Constants for Weak Acids

Weak acid has Ka < 1 Leads to small [H3O+] and a pH of 2 - 7


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Equilibrium Constants for Weak Bases

Weak base has Kb < 1 Leads to small [OH-] and a pH of 12 - 7


30Ionization Constants for Acids/Bases

Acids ConjugateBases

Increase strength

Increase strength


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Relation

of Ka, Kb,

[H3O+]


32K and Acid-Base ReactionsA strong acid is 100% dissociated.

Therefore, a STRONG ACID—a good H+ donor—must have

a WEAK CONJUGATE BASE—a poor H+ acceptor.

HNO3(aq) + H2O(liq) e H3O+(aq) + NO3-(aq)

STRONG A base acid weak B •Every A-B reaction has two acids and two bases.•Equilibrium always lies toward the weaker pair.•Here K is very large.


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K and Acid-Base Reactions

We know from experiment that HNO3 is a strong acid.

1. It is a stronger acid than H3O+

2. H2O is a stronger base than NO3-

3. K for this reaction is large


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K and Acid-Base ReactionsAcetic acid is only 0.42% ionized when [HOAc] =

1.0 M. It is a WEAK ACIDHOAc + H2O e H3O+ + OAc-

WEAK A base acid STRONG BBecause [H3O+] is small, this must mean

1. H3O+ is a stronger acid than HOAc

2. OAc- is a stronger base than H2O

3. K for this reaction is small


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Types of Acid/Base Reactions

Strong acid (HCl) + Strong base (NaOH)H+ + Cl- + Na+ + OH- e H2O + Na+ + Cl-

Net ionic equationH+(aq) + OH-(aq) e H2O(liq)

K = 1/Kw = 1 x 1014

Mixing equal molar quantities of a strong acid and strong base produces

a neutral solution.


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Weak acid (acetic ac.) + Strong base (NaOH)CH3CO2H + OH- e H2O + CH3CO2

-

• This is the reverse of the reaction of CH3CO2

- (conjugate base) with H2O.

• OH- stronger base than CH3CO2-

• K = 1/Kb = 1/(5.6 x 10-10) = 1.8 x 109

Mixing equal molar quantities of a weak acid and strong base produces the acid’s conjugate base. The solution is basic.


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Strong acid (HCl) + Weak base (NH3)

H3O+ + NH3 e H2O + NH4+

This is the reverse of the reaction of NH4+

(conjugate acid of NH3) with H2O.H3O+ stronger acid than NH4

+

K = 1/Ka = 1.8 x 109

Mixing equal molar quantities of a strong acid and weak base produces the bases’s

conjugate acid. The solution is acid.


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Weak acid + Weak base

•Product cation = conjugate acid of weak base.•Product anion = conjugate base of weak acid.•pH of solution depends on relative strengths of cation and anion.

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Types of Acid/Base Reactions: Summary


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Calculations with Equilibrium Constants

pH of an acetic acid solution.

What are your observations?

a pH meter

2.0 M

0.0001 M

0.003 M

0.06 M


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Equilibria Involving A Weak AcidYou have 1.00 M HOAc. Calc. the equilibrium

concs. of HOAc, H3O+, OAc-, and the pH.Step 1. Define equilibrium concs. in ICE

table. [HOAc] [H3O+] [OAc-]

initial change equilib


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Equilibria Involving A Weak AcidYou have 1.00 M HOAc. Calc. the equilibrium

concs. of HOAc, H3O+, OAc-, and the pH.Step 1. Define equilibrium concs. in ICE

table. [HOAc] [H3O+] [OAc-]initial 1.00 0 0change -x +x +xequilib 1.00-x x xNote that we neglect [H3O+] from H2O.


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Equilibria Involving A Weak Acid

Step 2. Write Ka expression

You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.

This is a quadratic. Solve using quadratic formula or method of

approximations (see Appendix A).


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Step 3. Solve Ka expression


First assume x is very small because Ka is so small. Therefore,

Now we can more easily solve this approximate expression.


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Step 3. Solve Ka approximate expression


x = [H3O+] = [OAc-] = [Ka • 1.00]1/2

x = [H3O+] = [OAc-] = 4.2 x 10-3 M

pH = - log [H3O+] = -log (4.2 x 10-3) = 2.37


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Equilibria Involving A Weak AcidConsider the approximate expression

For many weak acids

[H3O+] = [conj. base] = [Ka • Co]1/2

where C0 = initial conc. of acid

Useful Rule of Thumb:

If 100•Ka < Co, then [H3O+] = [Ka•Co]1/2


47Equilibria Involving A Weak AcidCalculate the pH of a 0.0010 M solution of

formic acid, HCO2H.

HCO2H + H2O e HCO2- + H3O+

Ka = 1.8 x 10-4 Approximate solution [H3O+] = [Ka • Co]1/2 = 4.2 x 10-4 M, pH = 3.37Exact Solution [H3O+] = [HCO2

-] = 3.4 x 10-4 M [HCO2H] = 0.0010 - 3.4 x 10-4 = 0.0007 M


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Weak Bases


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Equilibria Involving A Weak BaseYou have 0.010 M NH3. Calc. the pH.

NH3 + H2O e NH4+ + OH-

Kb = 1.8 x 10-5

Step 1. Define equilibrium concs. in ICE table [NH3] [NH4

+] [OH-]

initial change equilib


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Kb = 1.8 x 10-5

Step 1. Define equilibrium concs. in ICE table [NH3] [NH4

+] [OH-]

initial 0.010 0 0change -x +x +xequilib 0.010 - x x x


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Kb = 1.8 x 10-5

Step 2. Solve the equilibrium expression

Assume x is small (100•Kb < Co), so x = [OH-] = [NH4

+] = 4.2 x 10-4 Mand [NH3] = 0.010 - 4.2 x 10-4 ≈ 0.010 MThe approximation is valid !


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Kb = 1.8 x 10-5

Step 3. Calculate pH[OH-] = 4.2 x 10-4 Mso pOH = - log [OH-] = 3.37Because pH + pOH = 14,pH = 10.63


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MX + H2O f acidic or basic solution?Consider NH4ClNH4Cl(aq) f NH4

+(aq) + Cl-(aq)(a) Cl- ion is a VERY weak base because

its conjugate acid is strong. Therefore, Cl- f neutral solution• Cl- does not “hydrolyze”, i.e. does not

react with water appreciably• So, conjugates of weak acids and bases

determine pH of salt solutions because

Acid-Base Properties of Salts


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NH4Cl(aq) f NH4+(aq) + Cl-(aq)

(b) Reaction of NH4+ with H2O

NH4+ + H2O f NH3 + H3O+

acid base base acidNH4

+ ion is a moderate acid because its conjugate base is weak.

Therefore, NH4+ f acidic solution

See TABLE 17.4 for a summary of acid-base properties of ions.



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Calculate the pH of a 0.10 M solution of Na2CO3. Na+ + H2O f neutralCO3

2- + H2O e HCO3- + OH-

base acid acid base Kb = 2.1 x 10-4

Step 1. Set up ICE table [CO3

2-] [HCO3-] [OH-]

initial change



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Step 1. Set up ICE table [CO3

2-] [HCO3-] [OH-]

initial 0.10 0 0 change -x +x +x equilib 0.10 - x x x

Acid-Base Properties of SaltsCalculate the pH of a 0.10 M solution of Na2CO3. Na+ + H2O f neutralCO3

2- + H2O e HCO3- + OH-



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2- + H2O e HCO3- + OH-



Assume 0.10 - x ≈ 0.10, because 100•Kb < Cox = [HCO3

-] = [OH-] = 0.0046 M

Step 2. Solve the equilibrium expression


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2- + H2O e HCO3- + OH-

base acid acid base


Step 3. Calculate the pH[OH-] = 0.0046 MpOH = - log [OH-] = 2.34pH + pOH = 14, so pH = 11.66, and the solution is ________.


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Polyprotic acids

• Polyprotic acids have more than one ionizable proton.

• The protons are removed in steps, not all at once.

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

• Therefore, Ka1 > Ka2 > Ka3 etc.


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Lewis acid a substance that accepts

an electron pair

Lewis Acids & Bases

Lewis base a substance that donates an electron pair

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Reaction of a Lewis Acid and Lewis Base

• New bond formed using electron pair from the Lewis base.

• Coordinate covalent bond

• Notice geometry change on reaction.

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Formation of hydronium ion is also an excellent example.

Lewis Acids & Bases

•Electron pair of the new O-H bond originates on the Lewis base.


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Lewis Acid/Base Reaction


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Other good examples involve metal ions.

Lewis Acids & Bases


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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX IONS

Lewis Acids & Bases


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[Ni(H2O)6]2+ + 6 NH3 f [Ni(NH3)6]2+

+ DMG

Lewis Acids & Bases

DMG = dimethylglyoxime, a standard reagent to detect nickel(II)


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Lewis Acid-Base Interactions in Biology

• The heme group in hemoglobin can interact with O2 and CO.

• The Fe ion in hemoglobin is a Lewis acid

• O2 and CO can act as Lewis bases

Heme group

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Many complex ions containing water undergo HYDROLYSIS to give acidic solutions.

[Cu(H2O)4]2+ + H2O f [Cu(H2O)3(OH)]+ + H3O+

Lewis Acids & Bases


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This explains why water solutions of Fe3+, Al3+, Cu2+, Pb2+, etc. are acidic.

Lewis Acids & Bases

This interaction weakens this bond

Another H2O pulls this H away as H+


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Amphoterism of Al(OH)3


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This explains AMPHOTERIC nature of some metal hydroxides.

Al(OH)3(s) + 3 H3O+ f Al3+ + 6 H2O

Here Al(OH)3 is a Brønsted base.

Al(OH)3(s) + OH- f Al(OH)4-

Here Al(OH)3 is a Lewis acid.

Lewis Acids & Bases


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Formation of complex ions explains why you can dissolve a ppt. by forming a complex ion.

AgCl(s) e Ag+ + Cl- Ksp = 1.8 x 10-10

Ag+ + 2 NH3 f Ag(NH3)2+ Kform = 1.6 x 107

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AgCl(s) + 2 NH3 e Ag(NH3)2+ + Cl-

Knet = __________________

Lewis Acids & Bases


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Why?• Why are some compounds

acids?• Why are some compounds

bases?• Why do acids and bases vary in

strength?• Can we predict variations in

acidity or basicity?


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Why is CH3CO2H an Acid?

1. The electronegativity of the O atoms causes the H attached to O to be highly positive.

2. The O—H bond is highly polar.3. The H atom of O—H is readily attracted to polar H2O.

–0.32

+0.24

+0.12

See Figure 17.12


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• Trichloroacetic acid is a much stronger acid owing to the high electronegativity of Cl.

• Cl withdraws electrons from the rest of the molecule.

• This makes the O—H bond highly polar. The H of O—H is very positive.

Acetic acid Trichloroacetic acid

Ka = 1.8 x 10-5 Ka = 0.3


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Hydrohalic Acids

• Acid strength of HX increases as you go down the group; HI>HBr>HCl>>HF

• This is generally attributable to decreasing HX bond strength as you go down a group

• Going across a period, increasing acidity is generally attributable to increasing electronegativity.

• HF>H2O>NH3>CH4


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Oxoacids

• Generally when comparing homologous acids, the greater the number of oxygen atoms, the greater the acidity

• So nitric acid is stronger than nitrous acid, perchloric is stronger than chloric, etc.

• Due to more oxygens withdrawing electron density from the O-H bond.


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These ions are BASES.They become more and more basic as the negative

charge increases.As the charge goes up, they interact more strongly

with polar water molecules.

NO3- CO3

2-

Basicity of Oxoanions

PO43-

the chemistry of acids and bases - texas tech university · 2013. 3. 12. · strong and weak...

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