Chapter 19Chapter 19

Acids, Bases, and SaltsAcids, Bases, and Salts

Anything in black letters = write it in your notes (‘knowts’)

19.1 – Acid-Base Theories19.1 – Acid-Base Theories

Taste sourDissolve active metals to produce hydrogen gas

Turns litmus paper RED

BasesTaste bitterFeels slippery on skin (dissolves oils on skin)

Turns litmus paper BLUE

Have you seen the litmus paper yet??

Acids

These are experimental definitions, they do not explain (theory) how an acid is different from a base.

Svante Arrhenius (1857 – 1927)

Arrhenius defined an acid and base theoretically.

Dissociate - to split or separate from another

First, a vocab word…

ACID – substance that dissociates in water to form hydrogen ions (H+).

BASE – substance that dissociates in water to form hydroxide ions (OH-).

HCl (aq) H+ (aq) + Cl- (aq)

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

Arrhenius Definition (~1887)

When an acid is placed in water, H+ ions are produced.

Hydrogen ions can also be thought of as H3O+ ions.

H3O+ = hydronium ion

HCl (aq) H+ (aq) + Cl- (aq)

HCl + H2O H3O+ + Cl-

or equivalently,

Brønsted-Lowry Definition (~1923)

ACID – donates H+

BASE – accepts H+

B-L definition covers more examples than the Arrhenius definition.

Johannes Bronstad (1879 – 1947)

Thomas Lowry (1874 – 1936)

ammonia ammonium ion

water donates a H+ and so is a B-L acid

ammonia accepts a H+ and so is a B-L base

Conjugate Acid – formed when a base accepts a H+

Conjugate Base – formed when an acid donates a H+

ACIDS donate H+, BASES accept H+

EXAMPLES…

HCO3- + H2O CO3

-2 + H3O+

HF + HCO3- F- + H2CO3

acid + base c. base + c. acid

acid + base c. base + c. acid

OH- + HCO3- H2O + CO3

-2

base + acid c. acid + c. base

label each as acid, base, c. acid, c. base

Amphoteric – substance that can be an acid or a base – depending on what it reacts with.

Water is amphoteric

ACIDS donate H+, BASES accept H+

label each reactant as an acid and base, label the products as conjugate acids or conjugate bases.

HNO3 + H2O H3O+ + NO3-

CH3COOH + H2O H3O+ + CH3COO-

NH3 + H2O NH4+ + OH-

H2O + CH3COO- CH3COOH + OH-

Lewis Acids and Bases (not covered)

Acid/Base Indicators

Litmus

Acid – red, Base – blue, Neutral - colorless

Phenolphthalein

Acid – colorless, Base – pink, Neutral - colorless

Cabbage

Acid – red/pink, Base – yellow/green, Neutral – blue/purple

Assignment:

Chapter 19 Worksheet #1 (FRONT)

19.2 – Hydrogen Ions and Acidity19.2 – Hydrogen Ions and Acidity

Molarity (M) – unit used to express the concentration of a solution

Molarity = mol solute (mol)

liters of soln (L)

[H+] = ‘the hydrogen ion concentration’

anything in [brackets] means the concentration in molarity

[OH-] = ‘the hydroxide ion concentration’

Self-Ionization of WaterSelf-Ionization of Water

Water ionizes to produce a small amount of H+ and OH- ions.

H2O H+ + OH-

[H+] = [OH-] = 1 x 10-

7 M

In pure water at 25 ̊C

Ion-product constant for water (Kw)

Kw = [H+][OH-] = 1.0 x 10-14

remember…anything in [brackets] represents the concentration in molarity

A solution is acidic if [H+] > 1.0 x 10-7 M …or if the pH of the solution is below 7

pH = ‘power of the hydrogen ion’pH = -log[H+]

Just as the mole was used to simplify large numbers of atoms, pH is used to simplify small concentration

numbers

[H+] = 1 x 10-7 M

Instead of writing out numbers like these…

[H+] = 2.4 x 10-4 M

[H+] = 7.3 x 10-10 M

pH = 7.00pH = 3.62pH = 9.14

we can write number like these

We can just say “This solution has a pH of 3.62”.

Instead of saying “This solution has a hydronium ion concentration of 2.4 x 10-4 M”.

Not only is pH an easier number to talk about, pH is understood by most people, whereas molarity is not.

The pH scale is used to describe how acidic or basic (alkaline) a substance is.

Pure water has [H+] = 1.00 x 10-

7 M

pH = - log [1.00 x 10-7]

pH = 7

The pH of water would be

pH = -log[H+]

Examples

Examples

[H+] = 2.3 x 10-5 M. Calculate the pH.

pH = - log [H+]

pH = - log [2.3 x 10-5]

pH = 4.64

Examples

[H+] = 1.0 x 10-5 M. Calculate the pH.

pH = - log [H+]

pH = - log [1.0 x 10-5]

pH = 5.0

Examples

-4.2 = log [H+]

10-4.2 = 6.31 x 10-5 M = [H+]

pH = - log [H+]

10-4.2 = 10log [H+]

pH = 4.2. Calculate [H+]

Summary of pHSummary of pH

[H+] = 10-pHpH = - log [H+]

0 7 14

[H+] 100 M 10-7 M 10-14 M

pH

[OH-

]10-14 M 10-7 M 100

M

Acidic Neutral Basic

Kw = [H+][OH-] = 1.0 x 10-

14

Strong acids & bases completely dissociate (split into ions) in water;

Weak acids & bases partly dissociate and only form a small amount of ions.

19.3 – Strengths of Acids and Bases19.3 – Strengths of Acids and Bases

HCl is a strong acid,

Concentration [HCl] [H3O+] [Cl-]

HCl (aq) H+ (aq) + Cl- (aq)

Initial 0.10 M 0 0

at Equilibrium 0 0.10 M 0.10 M

it completely ionizes

CH3COOH is a weak acid,

CH3COOH (aq) CH3COO- (aq) + H+ (aq)

Concentration [CH3COOH] [CH3COO-] [H3O+]

Initial 0.10 M 0 0

at Equilibrium 0.0987 M 1.34 x 10-3 M 0.00134 M

1.34 x 10-3 M

0.00134 M

it forms a small amount of ions

Chemical Equilibrium – occurs when forward rxn rate equals reverse rxn rate; dynamic

Le Chatelier’s Principle – at equilibrium, a rxn will shift forward or backward in response to any change in conditions (temp, pressure, concentration)

CH3COOH + H2O CH3COO- + H3O+

Increase [CH3COOH], rxn shifts the rxn to the right.

Increase [CH3COO-], rxn hifts rxn to the left.

CH3COOH + H2O CH3COO- + H3O+

Doing this, increases [CH3COO-] and causes a shift in the rxn to the left; increasing the pH.

Doing this, decreases [H3O+] causing a shift in the rxn to the left; increasing the pH.

1.Add methyl orange to acetic acid; divide into 3’s

red pH 4.0, orange pH 5.0, yellow pH 6.0yellow pH 6.0

2. Add Na+CH3COO- to acetic acid

3. Add NaOH to the acetic acid

Le Chatelier’s Principle Example 1

the rxn will shift to the left because the [Cl-] concentration increased.

NaCl (aq) Na+ + Cl-

What will happen when drops of HCl (aq) are added to a saturated solution of NaCl?

Le Chatelier’s Principle Example 2

HCl (aq) H+ + Cl-

Co(H2O)6+2 (aq) + 4Cl- (aq) + heat CoCl4

-2 (aq) + 6H2O

1. Add ~ 3 mL of conc. HCl to about 2 mL of 0.1 M CoCl2

Le Chatelier’s Principle Example 3

2. Add water to reverse the rxn

3. Add ~ 2 mL of 0.1M AgNO3(aq).

Ag+ (aq) + Cl- (aq) AgCl (s)

Buffer – solution that maintains a fairly stable pH when small amounts of acid or base are added.

Buffer solutions consist of a

weak acid and its conjugate base

a weak base and its conjugate acid.

or

Acetic Acid/Acetate Ion Buffer

CH3COOHCH3COO- + H+

CH3COOH + OH- CH3COO- + H2O

The acetate ion reacts with any added acid.

The acetic acid reacts with any added base.

CH3COO-CH3COOH

Universal pH Indicator Color Chart

Blood is buffered at pH b/w 7.35 - 7.45.

This is done mainly by the carbonic acid/bicarbonate buffer.

Carbonic acid (H2CO3) neutralizes added bases.

Bicarbonate ion (HCO3-) neutralizes added acids.

H2CO3 (aq) H2O + CO2

Hold your breath, CO2 level builds up in blood,

causing [H2CO3] ↑, pH ↓

Hyperventilate, CO2 level drops in blood,

causing [H2CO3] ↓, pH ↑

What determines if an acid or base is strong or weak?

Describe Le Chateliers Principle.

What occurs during chemical equilibrium?

Which direction will the rxn shift if [CH3COO-] increases?

CH3COOHCH3COO- + H+

What would happen to the pH of the solution?

What is a buffer?

What two things must a buffer contain?

How does a buffer work?

Questions

1. (a) H2SO4 caused the solution to turn from yellow to orange.

(b) Adding H2SO4 caused the reaction to shift to the right.

(c) Adding NaOH caused the reaction to shift to the left.

1. (a) Fe+3 (ferric or iron(III)) & NO3-1 (nitrate)

(b) K+1 (potassium) & SCN-1 (thiocyanate)

(c) The Fe(SCN)+2 caused the solution to turn dark red.

(d) Adding Fe3(SO4)2 increased the concentration of Fe(SCN)+2.

(f) Adding NaOH decreased [Fe(SCN)+2]

Le Chatelier’s Principle and Reversible Reactions ANALYSIS

Part 1

Part 2

(e) Adding Fe(NO3)3 increased the concentration of Fe(SCN)+2.

(g) Reducing [SCN-] caused the rxn to shift to the left (yellow)

19.4 – Neutralization Reactions19.4 – Neutralization Reactions

Neutralization Rxn – complete rxn of a strong base with a strong acid

A neutralization rxn will produce a salt and water.

Acid + Base Salt + H2O

HCl + NaOH NaCl + H2O

Titration – determining the concentration of an unknown solution using a solution whose concentration is known.

Equivalence Point – point where the amount of acid equals the amount of base

Standard – solution that has a known concentration.

End Point – point where the indicator changes color

ASSIGNMENT:

Read Section 19.4 (p. 672 – 675)

Answer Questions #35-43

EXAMPLE10.0 mL of 0.5 M HCl solution is added to 20.0 mL of NaOH of unknown concentration. What is the concentration of the NaOH?

HCl + NaOH NaCl + H2O0.5 M

10.0 mL 20.0 mLx M

Since the reaction of HCl and NaOH is 1:1 and twice the volume of NaOH was used, the NaOH must half as strong as HCl; [0.25 M].

EXAMPLEWhat volume of 0.10 M KOH is required to neutralize 20.0 mL of 0.20 M H2SO4 solution?

H2SO4 + 2KOH K2SO4 + 2H2O

0.20 M20.0 mL x mL

0.10 M

Since KOH requires twice as many moles as H2SO4, you should double your answer.

19.5 – Salts in Solution19.5 – Salts in Solution

not covered…

Sulfuric acid + magnesium hydroxide

Write the balanced chemical equation for each neutralization reaction

Phosphoric acid + calcium hydroxide

Nitric acid + ammonium hydroxide

Chapter 19 Quiz #2

1. What color will litmus paper be in an acidic solution?

3. What does [H+] mean?

4. What two products are always formed in an acid-base neutralization reaction?

2. What color will phenolphthalein indicator be in an basic solution?

5. Explain the difference between a strong acid and a weak acid.

7. What is a buffer?

8. How is the molarity of a solution calculated?

6. Explain why the pH of pure water is 7.00

9. A student titrated 10.0 mL of an HCl solution. The titration required 23.3 mL of 0.24M NaOH solution.

a. Which solution was the standard?

b. Which solution was more concentrated?

c. Convert both volumes to liters

d. Calculate the number of moles of NaOH that reacted.

e. Calculate the number of moles of HCl that reacted.

f. Calculate the molarity of the HCl solution.

10. Calculate the pH of solutions with the following hydrogen ion concentrations.

a. [H+] = 1.23 x 10-4M

b. [H+] = 3.42 x 10-7M

11. Calculate the hydrogen ion concentrations of solutions with the given pH.

a. pH = 3.14

b. pH = 9.2