Chapter 16 Acids and Bases. The Arrhenius Model acidhydrogen ions, H +, An acid is any substance...

Chapter 16

Acids and Bases

The Arrhenius Model An acidacid is any substance that produces hydrogen hydrogen

ions, Hions, H++,, in an aqueous solution. Example: when hydrogen chloride gas is dissolved

in water, the following ions are produced.HCl(g) H+

(aq) + Cl-(aq)

A basebase is an substance that produces hydroxide hydroxide ions, OHions, OH--,, in an solution.

Example: When solid sodium hydroxide is dissolved in water the following ions are produced.

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

(aq)Copyright © Houghton Mifflin Company

16-2

Another Theory: Brønsted-Lowry Model

The Arrhenius model is limiting in its classification of acids and bases, suggesting there is only one kind of acid or base. A more general definition was suggested by a Danish chemist named Johannes Brønsted and an English chemist named Thomas Lowry.

The BrThe Brøønsted-Lowry Model states:nsted-Lowry Model states: An acid is a proton (HAn acid is a proton (H++) donor) donor A base is a proton (HA base is a proton (H++) acceptor.) acceptor.

Copyright © Houghton Mifflin Company

16-3

Brønsted-Lowry Model Let’s look at a general reaction for an acid (HA) in water.

HAHA(aq)(aq) + H + H22OO(l)(l) H H33OO++(aq)(aq) + A + A--

(aq)(aq)

In the reaction which reactant was the proton donor (acid)?

Which reactant was the proton acceptor (base)? If we look at the products, we see that now we have role

reversal. Which product is a proton donor? Which product is a proton acceptor? In this case, the In this case, the proton donor proton donor in the products side is in the products side is

known as the known as the conjugate acidconjugate acid, and the , and the proton acceptor proton acceptor is is known as the known as the conjugate baseconjugate base..


16-4

PracticeIn the following reactions identify the acid, base, conjugate acid,

and conjugate base.

1. HF(aq) + H2O(l) H3O+(aq) + F-

(aq)

2. H2S(aq) + H2O(l) H3O+(aq) + HS-

(aq)

3. NH3(aq) + H2O(aq) NH4+

(aq) + OH-(aq)

4. HSO4- + C2O4

2- SO42- + HC2O4

-

5. Write the conjugate bases for the following acids.a. HClO4

b. HNO3

c. HC2H3O2


16-5

16.2 Acid Strength When we put an acid or base in water, the

compound breaks apart into it’s respective ions is called dissociationdissociation..

The degree which a compound dissociatesdegree which a compound dissociates in water determines the strength of the acid or determines the strength of the acid or base.base.

Strong acids and basesStrong acids and bases are substances that are completely ionizedcompletely ionized, or completely dissociatedcompletely dissociated in solution.

Weak acids and basesWeak acids and bases are substances that only ionize or dissociate partiallyionize or dissociate partially in solution.


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Figure 16.1: Representation of the behavior of acids of different strengths in aqueous solution.


16-7

Acids vs Conjugate Bases


16-8

AcidAcid Conjugate BaseConjugate Base

HClO4 ClO4-

HI I-

HBr Br-

HCl Cl-

H2SO4 HSO4-

HNO3 NO3-

H3O+ H2O

HSO4- SO4

2-

HF F-

HNO2 NO2-

HCOOH HCOO-

CH3COOH CH3COO-

NH4+ NH3

HCN CN-

H2O OH-

NH3 NH2-

Acid Strength Acid Strength IncreasesIncreases

Base Strength Base Strength IncreasesIncreases

Strong AcidsStrong Acids

Strong BasesStrong Bases

Weak AcidsWeak Acids

Weak BasesWeak Bases

16.3 Water as an Acid and a Base A substance is said to be amphoteric if it can

behave either as an acid or as a base. Water is the most common amphoteric

substance. Let’s look at the ionization of 2 water molecules.

HH22OO(l)(l) + H + H22OO(l)(l) ↔ ↔ HH33OO++(aq) (aq) + OH+ OH--

(aq)(aq)

In this reaction one water molecule acts as an acid, and one acts as a base.

At 25°C the concentrations of these ions have been calculated repeatedly to be

[H[H33OO++]=[OH]=[OH--]= 1.0 x 10]= 1.0 x 10-7-7 M MCopyright © Houghton Mifflin Company

16-9

Ion-Product Constant, Kw• [H3O+][OH-] = 1.0 x 10-14 = Kw

• We call this constant, 1.0 x 10-14, Kw or the ion-product constant for water.

• Kw = [H+][OH-] = 1.0 x 10-14

• In any aqueous solution at 25°C, no matter what it contains, the product of [H+] and [OH-] must always equal 1.0 x 10-14.

• This means if the [H+] goes up, the [OH-] must go down so the product does not change.

• [H+] = the concentration of H ions


16-10

3 Types of Situations In an aqueous solution, there are 3 possible situations.

Each one always maintains a Kw=1.0 x 10-14.

1. A neutral solution, where [H+]=[OH-]2. An acidic solution, where [H+]>[OH-]3. A basic solution, where [H+]<[OH-]

Example: Calculate the [H+] and determine the type of solution if you have 1.0 x 10-5 M OH-

What information do you know from this equation? Kw = [H+][OH-]

Rearrange to find [H+] = Kw = 1.0 x 10-14 = ____ [OH-] 1.0 x 10-5

Is this solution acidic, basic, or neutral?Copyright © Houghton Mifflin Company

16-11

Practice ProblemsDetermine the [H+] and [OH-] and whether the solution is acidic,

basic or neutral.

1. 10.0 M H+

2. 1.0 x 10-7 M OH-

3. 3.4 x 10-4 M H+

4. 2.6 x 10-8 M H+


16-12

16.4 The pH Scale Calculating the [H+] and [OH-] brings a lot of

very small numbers which is rather inconvenient. In 1909 a French chemist, named Soren Sorensen, came up with the pH scale.

pH, literally means “power of hydrogen” When calculating the pH we use the following:

pH = - log [H+] Calculate the pH of a solution with [H+] = 1.0 x 10-7 M Acidic solution pH < 7.00 Basic solution pH > 7.00 Neutral solution pH= 7.00


16-13

The pH Scale• On the pH scale, each increase of 1 unit equals a

power of ten change in the [H+].• A solution with a pH of 3, has a [H+] = 1 x 10-3 M,

which is 10 greater than a solution with pH of 4, or [H+] of 1 x 10-4 M and 100 times greater than a pH of 5 or 1 x 10-5=[H+].

• As the [H+] increases, the pH decreases.• The pH scale runs values from 0 to 14, zero being

very acidic, 14 being very basic, and 7 being neutral.


16-14

Figure 16.3: The pH scale.

Use the scale to determine if the following are acidic, neutral or basic:

a. grapefruit pH=3.2b. orange juice pH=3.5

c. urine pH=4.8-7.5 (depends on H2O)d. saliva pH= 6.4-6.9

e. milk pH=6.5f. gastric juice in stomach pH=2.0

g. blood pH=7.35-7.45h. tears pH=7.4

i. milk of magnesia pH=10.6j. ammonia pH=11.4k. Draino pH=12.0


16-15

Calculating pOH• Sometimes we find it easier to calculate the pOH, since we

are given the [OH-].pOH = -log [OH-]

• Calculate the pOH of [OH-] = 1.0 x 10-2 M

• The pOH is not commonly used to describe a solution, so we must convert the pOH to the pH

• pH + pOH = 14

• What is the pH of our pOH solved at the top of this slide?


16-16

Calculating pH and pOH Practice Calculate the pH and pOH for the following:1. [H+] = 1.0 x 10-4 M2. [OH-] = 1.0 x 10-3 M3. [H+] = 5.60 x 10-12 M4. Determine whether each of the solutions above

is acidic, basic or neutral. NOTE: When determining sig figs for logs, the the

number of decimal places for a log must equal number of decimal places for a log must equal the number of significant figures in the original the number of significant figures in the original number.number.


16-17

Calculating the [H+] from the pH

• Sometimes the pH is available, but we need to know the [H+] of the solution. In this case we need to find the inverse log of the –pH.

[H+] = inverse log (-pH)• On your calculator you may enter the –pH, then push

the inverse key and then the log.• Your calculator may also have a 10x key, typically above

the log key. With this key you would enter your –pH, then push the 10x key.

• Calculate the [H+] if the pH is 5.00.• The same sequence is used for [OH-] and pOH.


16-18

Practice Problems• The pH of a human blood sample was measured

to be 7.41. What is the [H+] in this blood?• The pOH of a liquid drain cleaner was found to be

10.50. What is the [OH-] for this cleaner?


16-19

• The pH curve for the titration of 25.0 mL of 0.200 M HNO3 with 0.100 M NaOH. If you graphed the results from the computer simulation, it would look like this graph.

Copyright © Houghton Mifflin Company 16-20

Neutralization Reaction Net Ionic Equation

Since the acid and base in the reaction are strong, they will both dissociate completely in water. Write the following as an ionic equation:

HClHCl(aq)(aq) + NaOH + NaOH(aq)(aq) NaCl NaCl(aq)(aq) + H + H22OO(l)(l)

HH++(aq) (aq) +Cl+Cl--

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

(aq)(aq) Na Na+ + +Cl+Cl- - + H+ H22OO(l)(l)

We can then rewrite this as a NIE:We can then rewrite this as a NIE:HH++

(aq)(aq) + OH + OH--(aq)(aq) H H22OO(l)(l)

The above is a neutralization reaction. neutralization reaction. It is called so because if equal amounts of H+ and OH- are available for reaction, a neutral solution will result.


16-21

Calculating Volume in Neutralization Reactions

• What volume of 0.100 M HCl is needed to neutralize 25.0 mL of a 0.350 M NaOH solution?

• MacidVacid(#H+) = Mbase Vbase (#OH- )

• (0.100M) ?V (1) = 0.35M(.025L)(1)• V= (0.35M)(.025L)

0.100 MV= 0.0875L or 87.5mLOR YOU CAN CALCULATE USING STOICHIOMETRY!


16-22

Step 1:Step 1: Write a balanced net ionic equation for the Write a balanced net ionic equation for the reaction.reaction.

HH++(aq)(aq) + OH + OH--

(aq)(aq) H H22OO(l)(l)

Step 2:Step 2: Calculate the moles of reactants.Calculate the moles of reactants.25.0 mL NaOH x 1L x 0.350 mol NaOH 1 1000 mL 1 L NaOH= 8.75 x 108.75 x 10-3-3 mol OH mol OH--

Step 3:Step 3: Determine which reactant is limiting.Determine which reactant is limiting. This problem requires the addition of just enough H+ ions

to react exactly with the OH- ions present. The number of moles of OH- ions present determines the

number of moles of H+ ions. Therefore the OHOH-- are limiting. are limiting.


16-23

Step 4:Step 4: Calculate the moles of HCalculate the moles of H++ required. required. 8.75 x 10-3 mol OH- x 1 mol H+ = 8.75 x 108.75 x 10-3-3 mol H mol H++

1 1 mol OH-

Step 5:Step 5: Calculate the volume of 0.100 M HCl Calculate the volume of 0.100 M HCl required.required.

We know that HCl:H+ is a 1:1 ratio, so the [H+] is 0.100 M.

Volume x 0.100 mol H+ = 8.75 x 10-3 mol H+

1L Now we must solve for volume. Volume = 8.75 x 10-2 L or 87.5 mL


16-24

Practice Problem

1. Calculate the volume of 0.10 M HNO3 needed to neutralize 125 mL of 0.050 M KOH.


16-25

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Chapter 16 Acids and Bases. The Arrhenius Model acidhydrogen ions, H +, An acid is any substance...

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