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

1The Chemistry of Acids and Bases

Acid and Bases

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

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

One of the best known is acetic acid = CH3CO2H

• 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

• 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

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.

Conjugate Pairs

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

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.

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

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-]

[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)

[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

The pH Scale

See Active Figure 17.1

[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

pH of Common Substances

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

Equilibria Involving Weak Acids and Bases

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

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%).

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.

Equilibrium Constants for Weak Acids

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

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

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.

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

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

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.

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.

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.

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

Calculations with Equilibrium Constants

pH of an acetic acid solution.

What are your observations?

a pH meter

0.0001 M

0.003 M

0.06 M

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

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.

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).

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.

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

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

Weak Bases

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

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

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 !

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

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

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.

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

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-

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

-] = [OH-] = 0.0046 M

Step 2. Solve the equilibrium expression

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 ________.

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.

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.

Lewis Acid/Base Reaction

Other good examples involve metal ions.

Lewis Acids & Bases

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

Lewis Acids & Bases

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

Lewis Acids & Bases

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

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

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+

Amphoterism of Al(OH)3

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

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

-------------------------------------

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

Knet = __________________

Lewis Acids & Bases

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?

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

See Figure 17.12

• 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

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

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.

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

Basicity of Oxoanions

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