Acids and Bases - Folsom Cordova Unified School …...Acids and Bases The Arrhenius Model An acid is...

Chapter 16

Acids and Bases

The Arrhenius Model

An acid is any substance that produces hydrogen ions, 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 base is an substance that produces hydroxide ions, 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

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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 Brønsted-Lowry Model states:

An acid is a proton (H+) donor

A base is a proton (H+) acceptor.

Copyright © Houghton Mifflin

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

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

(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 proton donor in the products side is known as the conjugate acid, and the proton acceptor is known as the conjugate base.


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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- SO4

2- + HC2O4-

5. Write the conjugate bases for the following acids.

a. HClO4

b. HNO3

c. HC2H3O2


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

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

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

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

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


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


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Acids vs Conjugate Bases


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Acid Conjugate 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 Increases

Base Strength Increases

Strong Acids

Strong Bases

Weak Acids

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

H2O(l) + H2O(l) ↔ H3O+(aq) + OH-

(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

[H3O+]=[OH-]= 1.0 x 10-7 M


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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


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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?


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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+


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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


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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+] = 10-3 M, which is 10 greater than a solution with pH of 4, or [H+] of 10-4 M and 100 times greater than a pH of 5 or 10-5=[H+].

• As the [H+] increases, the pH decreases.

• The pH scale runs values from 0-14, zero being very acidic, 14 being very basic, and 7 being neutral.


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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=11.4

k. Draino pH=12.0


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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?


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Calculating pH and pOH Practice

Calculate the pH and pOH for the following:

1. [H+] = 1.0 x 10-4 M

2. [OH-] = 1.0 x 10-3 M

3. [H+] = 1.0 x 10-12 M

4. Determine whether each of the solutions above is acidic, basic or neutral.

NOTE: When determining sig figs for logs, the number of decimal places for a log must equal the number of significant figures in the original number.


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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.


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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?


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• 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:

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

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

(aq) +OH-(aq) Na+ +Cl- + H2O(l)

We can then rewrite this as a NIE:

H+(aq) + OH-

(aq) H2O(l)

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


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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 M

V= 0.0875L or 87.5mL

OR YOU CAN CALCULATE USING STOICHIOMETRY!


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

H+(aq) + OH-

(aq) H2O(l)

Step 2: 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 10-3 mol OH-

Step 3: 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 OH- are limiting.


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

H+

1 1 mol OH-

Step 5: Calculate the volume of 0.100 M HClrequired.

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


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Practice Problem

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


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