chapter 15 n.
Skip this Video
Loading SlideShow in 5 Seconds..
Chapter 15 PowerPoint Presentation
Download Presentation
Chapter 15

Loading in 2 Seconds...

play fullscreen
1 / 67

Chapter 15 - PowerPoint PPT Presentation

  • Uploaded on

Chapter 15. Aqueous Equilibria: Acids and Bases. Everyday Acids and Bases. Acids: vinegar, lemon juice, sulfuric acid Bases: antacids, ammonia. A. Acid-Base Concepts: The Bronsted-Lowry Theory. So far, we’ve discussed the Arrhenius theory of acids and bases:

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
Download Presentation

PowerPoint Slideshow about 'Chapter 15' - rory

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
chapter 15
Chapter 15

Aqueous Equilibria:

Acids and Bases

everyday acids and bases
Everyday Acids and Bases.
  • Acids: vinegar, lemon juice, sulfuric acid
  • Bases: antacids, ammonia
a acid base concepts the bronsted lowry theory
A. Acid-Base Concepts: The Bronsted-Lowry Theory
  • So far, we’ve discussed the Arrhenius theory of acids and bases:
    • Acids dissociate to produce H+ (examples: HCl, H2SO4)
    • Bases dissociate to produce OH- (examples: NaOH, Ba(OH)2 )

But -- as some bases don’t contain OH, this is limiting.

bronsted lowry theory of acids and bases
Bronsted-Lowry Theory of Acids and Bases

An acid is a substance that can transfer H+

A base is a substance that can accept H+

HA + B BH+ + A- Note conjugate

acid base acid base acid/base pairs.

conjugate pairs
Conjugate Pairs…

In the previous reaction, A- is the conjugate base of the acid HA

B is the conjugate base of the acid HB+

HA + B BH+ + A-

In an acid/base reaction, keep your eye on the proton!


Write balanced equations for the dissociation of the following Bronsted-Lowry acids in water:

a. H2SO4

b. H3O+

examples answer
Examples -- answer
  • H2SO4 + H2O HSO4- + H3O+
  • H3O+ + H2O H2O + H3O+

BOOKKEEPING -- be able to label acid, base, conjugate acid, conjugate base! (by convention, CA/CB are on the product side)


What is the conjugate acid of…

  • HCO3-
  • CO32-
b acid strength and base strength
B. Acid Strength and Base Strength

Think of the acid-base reaction as a “tug of war” between the two bases for a proton:

HA + H2O H3O+ + A-

Which is the stronger base: H2O or A-? Which wants the proton more? This will determine whether the equilibrium lies more to the right or to the left.

acid base strength cont
Acid/Base Strength cont.
  • If H2O is a stronger proton acceptor than A-, H2O will get the protons, and the solution will mostly contain H3O+ and A-.
  • Conversely, if A- is the stronger proton acceptor,, the solution will mostly contain HA and H2O.

**The proton is always transferred to the stronger base.

acid base strength equilibrium
Acid/Base Strength: Equilibrium

The strength of the acids and bases in an acid-base reaction will dictate the position of the equilibrium in an acid-base reaction.

HA + H2O A- + H3O+

if this is the stronger the equilibrium will lie to the right


Because a strong acid will dissociate more thoroughly/completely.

what does it mean to be a strong acid
What does it mean to be a strong acid?
  • Almost completely dissociated in water
  • Equilibrium almost entirely to the right
  • Solution consists almost entirely of H3O+ and A- ions -- almost no HA molecules

Strong acids have very weak conjugate bases. If HA has a strong tendency to lose its proton, A- will not be a good proton acceptor.

what does it mean to be a weak acid
What does it mean to be a weak acid?
  • Only partially dissociated in water.
  • Solution contains mostly undissociated HA.
  • Not much H3O+ and A- ion present in solution.
  • Equilibrium lies toward left.

Weak acids have strong conjugate bases. If HA does not have a strong tendency to lose its proton, A- will be a good proton acceptor.

c hydrated protons and hydronium ions
C. Hydrated Protons and Hydronium Ions

H+ does not exist by itself in aqueous solution, despite the fact that you frequently will see H+ (aq) in acid-base equilibria.

In aqueous solution, H+ binds to a water molecule to form H3O+ ==> hydronium ion.

d dissociation of water
D. Dissociation of Water

Water can act either as an acid or base, depending on the situation.

If acid is present, water is a base:

HA + H2O A- + H3O+

If base is present, water is an acid:

B + H2O HB + OH-

dissociation of water cont
Dissociation of Water cont.

Or, water can act as both an acid and a base!

H2O + H2O H3O+ + OH-

acid base acid base

This process is referred to as the dissociation of water, and has a dissociation constant, Kw.

dissociation of water cont1
Dissociation of Water cont.

For the dissociation of water,

Kw = [H3O+][OH-]

Remember that pure liquids, such as H2O, are not included in equilibrium expressions.

= equilibrium constant for the dissociation of water (also referred to as the ion product constant for water)

dissociation of water cont2
Dissociation of Water cont.

Very little of water is ionized; the equilibrium lies far to the left.

At 25oC, [H3O+] = [OH-] = 1.0 * 10-7 M


Kw = [H3O+][OH-] = 1.0 * 10-14 M

**This is true for any aqueous solution at 25oC.

h 3 o vs oh
[H3O+] vs. [OH-]

Defining acidic/basic/neutral solutions:

In an acidic solution, [H3O+] > [OH-]

In a basic solution, [H3O+] < [OH-]

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


The concentration of OH- in a sample of 25oC seawater is 5.0 * 10-6 M. Calculate the concentration of H3O+ ions, and classify the solution as acidic, neutral, or basic.

example answer
Example -- answer

Kw = [H3O+][OH-]

At 25oC, Kw = 1.0 * 10-14 M

Given [OH-] = 5.0 * 10-6 M

Solve for [H3O+]

[H3O+] = Kw/[OH-]

= (1.0 * 10-14)/(5.0 * 10-6)

= 2 * 10-9 M

Since [OH-] > [H3O+] -- basic

e the ph scale
E. The pH Scale

pH is a more convenient way to express the concentration of hydronium ion in a solution.

pH = -log[H3O+]

pH < 7 -- acidic

pH > 7 -- basic

pH = 7 -- neutral


Calculate the pH of a solution that has an [H3O+] of 6.0 * 10-5 M.

example answer1
Example -- answer

If pH = -log (6.0 * 10-5)

then pH = 4.22

This solution is acidic.

f measuring ph
F. Measuring pH

Color indicators are often used for approximation

Examples: phenolphthalein (changes from colorless --> pink when going from acidic --> basic), bromothymol blue

However, pH meters give you a more precise number, measuring the electrical potential of the solution (chapter 18)

g equilibria in solutions of weak acids
G. Equilibria in Solutions of Weak Acids

Beware: a weak acid is not the same thing as a dilute solution of a strong acid!

Even when dilute, the equilibrium for the strong acid will lie to the right (not for the weak acid)

equilibrium constant for a weak acid dissociation
Equilibrium Constant for a Weak Acid Dissociation

For the reaction

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

Ka = [H3O+][A-]/[HA]

In dilute aqueous solution, [H2O] is essentially constant.

pKa = -log Ka


The pH of 0.10 M HOCl is 4.23. Calculate Ka for hypochlorous acid. How close did you come to the true value?

example answer2
Example -- answer

Ka = [H3O+][-OCl]/[HOCl]

[H3O+] = 10-pH = 10-4.23 = 5.89 * 10-5 M

= [-OCl]

[HOCl] = 0.10 - 5.89 * 10-5 = 0.0999 M

Ka = (5.89 * 10-5)2/0.0999 = 3.47 * 10-8

h calculating equilibrium concentrations in solutions of weak acids
H. Calculating Equilibrium Concentrations in Solutions of Weak Acids

How are Ka values useful?

Allows us to calculate the pH of an acidic solution, as well as equilibrium concentrations of all species present

Example: Calculate concentrations of all species present, and the pH of, a 0.10 M HCN solution.

concentration of all species in 0 10 m hcn soln
Concentration of all species in 0.10 M HCN soln.

Step 1 List species present before any dissociation happens, and identify them as either acid or base.


acid acid or base

concentration of in 0 10 m hcn cont
Concentration of… in 0.10 M HCN cont.

Step 2 What proton transfer reactions can occur, given the aforementioned molecules?

HCN + H2O H3O+ + CN-

Ka = 4.9 * 10-10

H2O + H2O H3O+ + OH-

Kw = 1.0 * 10-14

**Ka values in Table 15.2

concentration of in 0 10 m hcn cont1
Concentration of… in 0.10 M HCN cont.

Step 3 Label the reaction that proceeds farther to the right (larger equilibrium constant) as the principal reaction; the other reaction(s) is the subsidiary reaction.

HCN reaction: principal

dissociation of water: subsidiary

concentration of in 0 10 m hcn cont2
Concentration of… in 0.10 M HCN cont.

Step 4 Create an ICE table, expressing changes in concentration in terms of x.

Principal rxn HCN + H2O H3O+ + CN-


Initial (M) 0.10 ~0 0

Change (M) -x +x +x

Equil. (M) 0.10 - x x x

H2O is not part of the equilibrium expression, and is present in excess.

concentration of in 0 10 m hcn cont3
Concentration of… in 0.10 M HCN cont.

Step 5 Place the equilibrium values into the Ka expression.

Ka = 4.9 * 10-10 = [H3O+][CN-]/[HCN]

= (x)(x)/(0.10 - x)

concentration of in 0 10 m hcn cont4
Concentration of… in 0.10 M HCN cont.

Step 5 continued In this particular case -- Ka is small. This means the reaction does not proceed very far to the right.

If this is the case, x is small, and to simplify our math, we can say that (0.10 - x) ~ 0.10

and 4.9 * 10-10 ~ x2/0.10

Thus x2 = 4.9 * 10-11 and x = 7.0 * 10-6

concentration of in 0 10 m hcn cont5
Concentration of… in 0.10 M HCN cont.

Step 6 Now, you can use x to find all equilibrium concentrations.

[H3O+] = [CN-] = x = 7.0 * 10-6 M

[HCN] = 0.10 - x = 0.10 M

**x was small/negligible here. This is not always the case!

concentration of in 0 10 m hcn cont6
Concentration of… in 0.10 M HCN cont.

Step 7 The only concentration left to calculate is [OH-] from the subsidiary reaction.

[OH-] = Kw/[H3O+] = 1.0*10-14/7.0*10-6

= 1.4 * 10-9 M

Since [H3O+] from the dissociation of water is also 1.4*10-9 M, our assumption in the ICE table that [H3O+] = 0 was valid.

concentration of in 0 10 m hcn cont7
Concentration of… in 0.10 M HCN cont.

Step 8 Finally… calculate pH!

pH = -log [H3O+]

= -log (7.0*10-6)


j percent dissociation in solutions of weak acids
J. Percent Dissociation in Solutions of Weak Acids
  • Another way to measure/express acid strength: percent dissociation.
  • The stronger the acid, the more dissociated it will be in aqueous solution.

percent dissociation =

([HA] dissoc./[HA] initial) * 100%

k polyprotic acids
K. Polyprotic Acids
  • Are acids that contain more than one dissociable proton
  • Examples: H2SO4 H3PO4
  • Dissociate in a stepwise manner, and each dissociation step has its own Ka
  • Note that each successive Ka value decreases. After the first proton is removed, the remaining conjugate base will have a negative charge, making the next proton harder to remove.
l equilibria in solutions of weak bases
L. Equilibria in Solutions of Weak Bases

Consider the reaction

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

There is a base-dissociation constant similar to that for an acid:

Kb = [BH+][OH-]/[B] = [NH4+][OH-]/[NH3]

weak bases cont
Weak Bases… cont.
  • Weak bases are frequently amines
  • Amines are derivatives of ammonia where one or more of the H has been replaced by a hydrocarbon group

Calculate the pH and the concentrations of all species present in 0.40 M NH3 (Kb = 1.8 * 10-5).

Step 1 What species are present prior to dissociation?


base acid or base

example cont
Example -- cont

Step 2/3 Seeing that Kb for NH3is 1.8 * 10-5, this is considered the principal reaction (as opposed to Kw)

Step 4

NH3 + H2O NH4 + OH-


Initial(M) 0.40 0 ~0

Change(M) -x +x +x

Equil.(M) 0.40-x x x

example cont1
Example -- cont.

Step 5

Kb = [NH4+][OH-]/[NH3] = x2/0.40-x ~x2/0.40

=1.8 * 10-5

x = 2.7 * 10-3 M = [NH4+] = [OH-]

[NH3] = 0.40 - x = 0.40 M

example cont2
Example -- cont

[H3O+] = Kw/[OH-]

= 1.0 * 10-14/2.7 * 10-3

= 3.7 *10-12 M

Use this information to determine pH

pH = -log[H3O+] = -log(3.7 * 10-12) = 11.43

makes sense… NH3 is basic!

m relationship between k a and k b
M. Relationship Between Ka and Kb
  • When dealing with a conjugate pair, you can calculate one from the other. Consider the following…

NH4+(aq) + H2O(l) H3O+(aq) + NH3(aq)

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


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

k a and k b cont
Ka and Kb cont.
  • Also consider equilibrium constants

Ka = [H3O+][NH3]/[NH4+] = 5.6 * 10-10

Kb = [NH4+][OH-]/[NH3] = 1.8 * 10-5

and Kw = [H3O+][OH-] = 1.0 * 10-14

Net equilibrium constant of two reactions when added: Ka * Kb

k a and k b cont1
Ka and Kb cont.

In general, when you add two chemical reaction together, the net equilibrium constant is the product of the two individual equilibrium constants.

Ka * Kb = (5.6 * 10-10)(1.8 * 10-5) = 1.0 * 10-14

= [H3O+][NH3]/[NH4+] * [NH4+][OH-]/[NH3]

In aqueous solution, Ka * Kb = Kw

n factors that affect acid strength
N. Factors That Affect Acid Strength

What makes one acid stronger than another?

Often determined by strength and polarity of the H-A bond.

How easily is the H-A bond broken? The more easily the bond is broken, the stronger the acid.

acid strength cont
Acid Strength cont.
  • How easily is the H-A bond broken?
    • The weaker the H-A bond, the more easily it is broken.
    • The more polar the H-A bond, the more easily the bond is broken. In a more polar bond, A is more electronegative. After the H-A bond is broken, A bears the negative charge. The more electronegative A is, the more stable A is in bearing the negative charge.

The more easily the H-A bond is broken, the stronger the acid.

acid strength cont1
Acid Strength cont.
  • Bond strength is the prevalent factor in groups(columns).
  • As you go down a group on the periodic table, acid strength increases (HI is a stronger acid than HF, etc).
  • This is due to atomic radius increasing and bond strength decreasing farther down the periodic table.
acid strength cont2
Acid Strength cont.
  • Polarity is the prevalent factor comparing within a row.
  • Within the same row of the periodic table, atomic radius does not appreciably change, but electronegativity increases going right.
  • When H is bonded to a more EN atom, the acid is stronger. An -OH containing molecule is more acidic than an -NH containing molecule.
strength of oxoacids
Strength of Oxoacids
  • Oxoacid: contains an -OH bond, which also contains the acidic H
  • Anything that might weaken the O-H bond increases the strength of the acid.

Case 1 Increase EN of Y, increase acid strength.

Shifts electron density toward Y.


--Y--O--H O will “feel” less negative charge

| when Y is more EN = greater


strength of oxoacids cont
Strength of Oxoacids cont.

Case 1 cont.

Examples… HOCl > HOBr >HOI

Case 2

If the identity of Y stays the same but more bonds are added to O, acid strength will also increase.

perchloric acid > hypochlorous acid

strength of oxoacids cont1
Strength of Oxoacids cont.

To continue the example…

H-O-Cl is weaker than H-O-Cl-O which is weaker than O

| The additional EN O’s draw

H-O-Cl-O electron density away from the site of deprotonation.

acid strength
Acid Strength
  • Any time you’re considering the strength of an acid, think about the stability of the corresponding anion.
  • If the corresponding anion (conjugate base) is particularly stable, the acid is more likely to dissociate.
  • If there are more EN atoms nearby, there are more places to distribute the negative charge resulting from deprotonation. This results in a more stable anion.
o lewis acids and bases
O. Lewis Acids and Bases

Another, more generalized, acid/base definition.

Lewis acid: electron pair acceptor

Lewis base: electron pair donor

Lewis acids are frequently metal cations.

lewis acids and bases
Lewis Acids and Bases


Cu2+ + 4 :NH3 --> Cu(NH3)42+

Lewis Lewis complex ion

acid base (deep blue)