Electrolytes and Nonelectrolytes (1 _{of 2)}

Electrolytes: compounds that conduct electricity when dissolved in water.

Nonelectrolytes: substances that do not conduct electricity when dissolved in water.

Comparing Solution Conductivity

(Distilled water) (Sugar solution) (Salt solution)

Electrolytes and Nonelectrolytes (2 _{of 2)}

Ion movement causes conduction of electricity in water.

3 classes of compounds, acids, bases, and salts are electrolytes because they produce ions in water when they dissolve.

Table 15.2 Representative Electrolytes and Nonelectrolytes

Electrolytes and Nonelectrolytes Practice (1 of 2)

Which compound will not dissociate in water?

1. HCl
2. KBr
3. NaOH
4. CH₃OH

Electrolytes and Nonelectrolytes

Practice (2 of 2)

Which compound will not dissociate in water?

Dissociation of Electrolytes

Salts dissociate into their respective cations and anions when dissolved in water.

NaCl(s)→ Na⁺(aq)+Cl⁻(aq)

Hydrated sodium (purple) and chloride (green) ions

The negative end of The positive end of the water dipole is the water dipole is attracted to the attracted to the positive Na ion.⁺positive Cl ion.⁻

When NaCl dissolves in water, each ion is surrounded by several water molecules.

The permanent dipoles in the water molecules cause specific alignment around the ions.

Electrolyte Ionization

Ionization: process of ion formation in solution. Ionization results from the chemical reaction between a compound and water.

Acids ionize in water, producing the hydronium ion (HO3 +) and a counter anion.

HCl(g) + H O₂ (l) → H O₃ ⁺ (aq)+Cl⁻(aq)

H PO_{3 4}(aq) + H O ₂ (l)€ H PO₂ ^–₄ (aq) + H O₃ ⁺(aq)

Bases ionize in water, producing the hydroxide ion (OH⁻) and a counter cation.

NH₃(aq) + H O ₂ (l)€ OH^– (aq) + NH⁺₄ (aq)

Strong and Weak Electrolytes (1 _{of 2)}

Strong electrolytes: undergo complete ionization in water.

Example: HCl (strong acid)

Weak electrolytes: undergo incomplete ionization in water.

Example: CH₃COOH (weak acid)

HCl (left) is 100% ionized.

CH₃COOH exists primarily in the unionized form.

Strong and Weak Electrolytes (2 _{of 2)}

Table 15.3 Strong and Weak Electrolytes

Double arrows indicate incomplete ionization (typically weak electrolytes).

HF (aq) + H O ₂ (l)€ F^– (aq) + H O₃ ⁺(aq)

Salts (2 of 2)

Salts can dissociate into more than 2 ions, depending upon the compound.

A 1 M solution of NaCl produces a total of 2 M of ions.

NaCl(s)→ Na ⁺(aq) +Cl⁻ (aq)

1 M 1 M 1 M

A 1 M solution of CaCl₂ produces a total of 3 M of ions.

CaCl₂(s) → Ca²⁺(aq) + 2Cl⁻(aq)

1 M 1 M 2 M

Salts Practice (3 _{of 4)}

What is the concentration of bromide ion in a 1.5 M solution of magnesium bromide?

1. 0.75M
2. 1.5M
3. 3.0M
4. 4.5M

Salts Practice (4 of 4)

What is the concentration of bromide ion in a 1.5 M solution of magnesium bromide?

1. 0.75M
2. 1.5M
3. Answer: 3.0M
4. 4.5M

For every one mole of MgBr₂, 2 moles of Br⁻ ionize.

MgBr₂ (s)→ Mg²⁺(aq) + 2 Br⁻(aq)

1.5 M 1.5 M 3.0M

Colligative Properties of Electrolyte Solutions

Colligative properties: depend only on the number of moles of dissolved particles present.

This must be taken into consideration when calculating freezing point depression or boiling point elevation due to the presence of solute particles.

Example:

What is the boiling point elevation of a 1.5 m aqueous solution of CaCl₂? (K_b for water is 0.512 degrees Celsius/m).

Because CaCl₂ is a strong electrolyte, 3 mol of ions

(1 mol Ca and 2 mol Cl ions^{2+ −} ) will be present in the solution.

3 mol ions 0.512C

 =T_b 1.5 m CaCl₂   = 2.3C

1 mol CaCl₂ 1 m

Colligative Properties of Electrolytes Practice (1 of 2)

What is the boiling point of a 2.0 m aqueous solution of NaCl? (K_b for water = 0.512 degrees Celsius/m) a. 101.02 degrees Celsius

1. 1.02 degrees Celsius
2. 2.05 degrees Celsius
3. 102.05 degrees Celsius

Colligative Properties of Electrolytes

Practice (2 of 2)

What is the boiling point of a 2.0 m aqueous solution of NaCl? (K_b for water = 0.512 degrees Celsius/m) a. 101.02 degrees Celsius

1. 1.02 degrees Celsius
2. 2.05 degrees Celsius
3. Answer: 102.05 degrees Celsius

Because NaCl is a strong electrolyte, 2 mol of ions (1 mol Na and 1 mol Cl ions^{+ −} ) will be present in the solution.

2 mol ions 0.512°C

 =T_b 2.0 m NaCl   = 2.05°C

1 mol NaCl 1 m

T = 100.0 ºC + 2.05 ºC = 102.05 ºC_b

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Autoionization of Water

Pure water auto(self) ionizes according to the reaction:

H O ₂ (l)+ H O ₂ (l)€ H O₃ ⁺(aq)+ OH^–(aq) Based on the reaction stoichiometry:

Concentration H O = Concentration OH = 1 10 M₃ ^{+ –}  ^–7

H O3 +   OH−=(1 10 –7)2 = 1 10 –14 When acid or base is present in water, [H O₃ ⁺] and OH[ ⁻] change.

In acidic solutions, _H O₃ ⁺ _{ }> OH^–_.

In basic solutions, _H O₃ ⁺ _{ }< OH^–_.

Copyright ©2016 John Wiley & Sons, Inc.

Neutralization

General Reaction acid + base → salt + water

Example

HCl(aq)+ NaOH(aq)→ NaCl(aq)+H O₂ (l)

Overall Ionic Equation:

H⁺(aq)+Cl⁻(aq)+ Na⁺(aq)+OH⁻(aq)→ Na⁺(aq)+Cl⁻(aq)+H O₂ (l)

All species are included; soluble compounds shown as ions.

Net Ionic Equation:

H⁺(aq)+ OH⁻(aq)→ H O ₂ (l)

Spectator ions (green) are removed from both sides.

Titration

Titration: experiment where the volume of one reagent (titrant) required to react with a measured amount of another reagent is measured.

Titrations allow the amount of an acid or base present in a sample to be determined.

Indicators are used to signal the endpoint of a titration, the point when enough titrant is added to react with the acid/base present.

Burets deliver measured amounts of the titrant into a solution of the unknown reagent.

Endpoint

Titration Calculations (1 _{of 2)}

If 22.59 mL of 0.1096 M HCl is used to titrate 25.00 mL of NaOH, what is the molarity of the base?

Reaction HCl (aq)+ NaOH(aq)→ NaCl(aq)+ H O ₂ (l)

Knowns 22.59 mL of 0.1096 M HCl

25.00 mL NaOH

Solving for Molarity of base

Calculate

0.1096 mol HCl 1 mol NaOH

22.59 mL   = 0.002476 mol NaOH

1000 mL soln 1 mol HCl

0.002476 mol NaOH

Molarity NaOH = = 0.09903 M NaOH

0.02500 L soln

Titration Calculations (2 _{of 2)}

What is the molarity of a NaOH solution if 21.93 mL of base is required to titrate 0.243 g of oxalic acid (H₂C₂O₄)?

Reaction H C O _{2 2 4} (aq)+ 2 NaOH(aq)→ NaC O_{2 4}(aq)+ 2 H O ₂ (l)

Knowns 0.243 g H₂C₂O₄

21.93 mL NaOH

Solving for Molarity of base

Calculate

1 mol H C O^{2 2 4} 2 mol NaOH

0.243 g H C O_{2 2 4}   = 0.005398 mol NaOH

90.04 g H C O_{2 2 4} 1 mol H C O_{2 2 4}

0.005398 mol NaOH

Molarity NaOH = = 0.246 M NaOH

0.02193 L soln Always check reaction stoichiometry!

Titration Calculations Practice (1 _{of 2)}

What is the concentration of a nitric acid solution if 10.0 mL of the solution is neutralized by 3.6 mL of 0.20 M NaOH?

1. 0.072 M
2. 53.6 M
3. 0.56 M
4. 5.6 M

Titration Calculations Practice _{(2 of 2)}

What is the concentration of a nitric acid solution if 10.0 mL of the solution is neutralized by 3.6 mL of 0.20 M NaOH?

1. Answer: 0.072 M Reaction
2. 53.6 M HNO₃(aq)+ NaOH(aq)→ NaNO₃(aq)+ 2 H O ₂ (l)
3. 0.56 M Knowns 3.60 mL, 0.20 M NaOH
4. 5.6 M 10.00 mL HNO₃

Solving for Molarity of acid

Calculate

0.20 mol NaOH 1 mol HNO³

3.60 mL NaOH   = 0.00072 mol HNO₃

1000 mL NaOH 1 mol NaOH

0.00072 mol HNO³

Molarity HNO₃ = = 0.072 M HNO

0.01000 L soln ³

Copyright ©2016 John Wiley & Sons, Inc.

Net Ionic Equations

Rules for Writing Net Ionic Equations

1. Strong electrolytes are written as the corresponding ions. Example: NaOH (aq) is written as Na⁺(aq) + OH⁻(aq)
2. Weak electrolytes and nonelectrolytes are written as molecules.

Example: CH₃OH(aq), CH₃COOH(aq), etc.

1. Solids and gases are written as their molecular forms.
2. The net ionic equation does not include spectator ions.
3. The net ionic equation must balance atoms and charge.
   1. of 4)

Write the net ionic equation (NIE) for the following reaction:

BaCl₂ (aq) + Na SO_{2 4} (aq)→ BaSO₄ (s)+ 2 NaCl (aq) Total Ionic Equation

Ba²⁺ (aq) + 2 Cl⁻(aq) + 2 Na⁺ (aq) + SO₄²⁻(aq) → BaSO₄ (s + ) 2 Na⁺ (aq) + 2 Cl⁻(aq)

Spectator Ions

Net Ionic Equation

Ba²⁺(aq) + SO₄²⁻(aq)→ BaSO₄(s)

• 1. of 4)

Write the net ionic equation (NIE) for the following reaction:

Na CO_{2 3}(aq) + 2 HCl aq( )→ CO₂ (g)+ 2 NaCl (aq) + H O ₂ (l) Total Ionic Equation

2 Na⁺(aq) +Co₃²⁻(aq)+ 2 H⁺(aq)+ 2 Cl⁻(aq)→ 2 Na⁺(aq) + 2 Cl⁻(aq) +CO₂(g)+ H O₂ (l)

Spectator Ions

Net Ionic Equation

2 H⁺(aq) + CO₃²⁻(aq)→ CO₂(g)+ H O₂ (l)

(3 of 4)

What is the net ionic equation when hydrobromic acid reacts with potassium hydroxide?

1. H⁺ (aq) + Br⁻(aq) → HBr(l)
2. K⁺ (aq) + OH⁻(aq)→ KOH (s)
3. K⁺(aq) + H⁻(aq)→ KH(s)
4. H⁺(aq) + OH⁻(aq)→ H O₂ (l)

Net Ionic Equations Practice _{(4 of 4)}

What is the net ionic equation when hydrobromic acid reacts with potassium hydroxide?

1. H⁺(aq) + Br⁻(aq) → HBr(l)
2. K⁺ (aq) + OH⁻(aq)→ KOH (s)
3. K⁺(aq) + H⁻(aq)→ KH(s)
4. Answer: H⁺(aq) + OH⁻(aq)→ H O₂ (l)

Equation

HBr(aq)+ KOH (aq)→ KBr(aq) + H O₂ (l)

Total Ionic Equation

H⁺(aq)+Br⁻(aq)+K⁺(aq)+OH⁻(aq)→K⁺(aq)+Br⁻(aq)+H O₂ (l)

Acid Rain

Acid rain: atmospheric precipitation more acidic than typical.

General Process for Acid Rain Formation:

1. Emission of nitrogen or sulfur oxides.
2. Transportation of these chemicals throughout the atmosphere.
3. Chemical reaction of the oxides with water. This forms sulfuric and nitric acids.
4. Precipitation carries the acids to the ground.

Foundations of College Chemistry

Fifteenth Edition

Morris Hein, Susan Arena, and Cary Willard

Chapter 15

Acids, Bases, and Salts

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Acids, Bases, and Salts

Lemons and limes are examples of foods that contain acidic solutions.

Arrhenius Acids

Arrhenius Acid: An acid solution contains an excess of H+ ions.

Common Properties of Acids

1. Sour taste
2. Turns litmus paper pink
3. Reacts with:

Metals to produce H₂ gas Bases to yield water and a salt Carbonates to give carbon dioxide

Arrhenius Bases

Arrhenius Bases: A basic solution contains an excess of

OH− ions.

Common Properties of Acids

1. Bitter/caustic taste
2. Turns litmus paper blue
3. Slippery, soapy texture
4. Neutralizes acids

Brønsted-Lowry Acids and Bases (1 _{of 5)}

A Brønsted-Lowry acid is a proton (H ) donor.⁺

A Brønsted-Lowry base is a proton (H ) acceptor.⁺

HCl(g) + H O₂ (l) → 2 H O₃ ⁺ (aq) +Cl⁻ (aq) ^Acid Base Hydronium ion^_hydrated H+^_

 

Conjugate Acid/Base Pairs: Two species that differ from each other by the presence of one hydrogen ion.

Brønsted-Lowry Acids and Bases (2 _{of 5)}

Identify the base in the following reaction and the compound’s conjugate acid.

H PO₂ ^–₄ (aq) + H O ₂ (l)€ HPO₄^2- (aq) + H O₃ ⁺(aq)

Base

1. HPO_{2 4}⁻(aq)
2. H₂O (l)
3. HPO²₄⁻(aq)
4. HO₃ ⁺ (aq)

Brønsted-Lowry Acids and Bases _{(3 of 5)}

Identify the base in the following reaction and the compound’s conjugate acid.

H PO₂ ^–₄ (aq) + H O ₂ (l)€ HPO₄^2- (aq) + H O₃ ⁺(aq)

Base

1. HPO_{2 4}⁻(aq)
2. Answer: H₂O (l)
3. HO3 + (aq)
4. HPO²₄⁻(aq)

Brønsted-Lowry Acids and Bases (4 _{of 5)}

Identify the base in the following reaction and the compound’s conjugate acid.

H PO₂ ^–₄ (aq) + H O ₂ (l)€ HPO₄^2- (aq) + H O₃ ⁺(aq) Conjugate Acid a. HPO2 4−(aq)

1. H₂O (l)
2. HPO24−(aq)
3. HO₃ ⁺ (aq)

Brønsted-Lowry Acids and Bases _{(5 of 5)}

Identify the base in the following reaction and the compound’s conjugate acid.

H PO₂ ^–₄ (aq) + H O ₂ (l)€ HPO₄^2- (aq) + H O₃ ⁺(aq) Conjugate Acid a. HPO_{2 4}⁻(aq)

1. H₂O (l)
2. HPO24−(aq)
3. Answer: HO₃ ⁺ (aq)

Lewis Acid-Bases

Lewis Acid: electron pair acceptor. Lewis Base: electron pair donor.

The _H+ is a Lewis base.

is a Lewis acid, and the

According to the Lewis theory, substances other than proton donors (e.g., BF₃) behave as acids:

Copyright ©2016 John Wiley & Sons, Inc.

Summary of the Acid/Base Theories

The theory that best explains the reaction of interest is used.

Table 15.1 Summary of Acid–Base Definitions

Reactions of Acids (1 _{of 2)}

1. Reactivity with Metals

Acids react with any metals above hydrogen in the activity series.

2 HCl(aq) + Mg(s)→ MgCl₂ (s)+ H₂ (g) In general: acid + metal → salt + hydrogen

2. Reactivity with Bases

Also called a neutralization reaction.

2 HCl(aq) + Ca OH( )₂ (aq)→ CaCl₂ (aq)+ 2 H O ₂ (l)

In general: acid + base → salt + water

Reactions of Acids (2 _{of 2)}

3. Reactivity with Metal Oxides

2 HCl(aq) + Na O₂ (s)→ 2 NaCl(aq)+ H₂O (l)

In general:

acid + metal Oxide →salt + water

(base)

4. Reactivity with Metal Carbonates

2 HCl(aq) + Na CO_{2 3}(aq) → 2 NaCl(aq)+ H O ₂ (l) + CO₂ (g) In general:

acid + carbonate → salt + water + carbon dioxide (base)

Base Reactions

Bases can be amphoteric (act as either Brönsted acids or bases) As a base:

Zn OH( )₂ (aq) + 2 HBr aq( )→ ZnBr₂ (aq)+ 2 H O ₂ (l) As an acid:

Zn OH( )₂ (aq) + 2 NaOH(aq)→ Na Zn OH₂ ( )₄ (aq) NaOH and KOH can also react with metals.

2 NaOH(aq)+ 2 Al(s)+6 H O₂ (l)→ 2 NaAl OH( )₄ (aq)+3 H₂ (g)

In general: base + metal + water → salt + hydrogen

Salts (1 of 2)

Salts: products from acid-base reactions.

HCl(aq)+ NaOH(aq)→ NaCl(aq) + H O ₂ (l)

Sodium chloride (table salt)

Salts are ionic compounds.

Salts contain a cation (a metal or ammonium ion) derived from the base and an anion (excluding oxide or hydroxide ions) derived from the acid.

Salts are generally crystalline compounds with high melting and boiling points.

Salts Practice (1 _{of 4)}

Which salt forms in the reaction of aluminum oxide and hydrobromic acid?

3 Al O_{2 3}(s) + 18 HBr(aq) → 6 AlBr₃(aq)+ 9 H O ₂ (l)

1. AlBr
2. AlBr₃
3. Al₂Br
4. Al₂Br₃

Salts Practice (2 of 4)

Which salt forms in the reaction of aluminum oxide and hydrobromic acid?

3 Al O_{2 3}(s) + 18 HBr(aq) → 6 AlBr₃(aq)+ 9 H O ₂ (l)

1. AlBr
2. Answer: AlBr₃
3. Al₂Br
4. Al₂Br₃