Ch 8.5 Solution Concentration Units % (m/m or w/w) = mass of solute x 100 total mass of solution mass of solution = mass solute + mass solvent

Ch 8.5 Solution Concentration Units

% (m/m or w/w) = mass of solute x 100 total mass of solution

mass of solution = mass solute + mass solvent

% (v/v) = volume of solute x 100 volume of solution

filled to a total volume of solution

% (m/v or w/v) = mass of solute (g) x 100 volume of solution (mL)

units are specified Parts-per-million

ppm = mass or volume of solute x 10

mass or volume of solution

Sample calculation: How many ounces of wine (12 % v/v) are in one standard drink?

12 mL pure alcohol in 100 mL wine Density of pure alcohol = 0.789 g/mL 1 ounce = 29.57 mL

g alcohol → mL alcohol → mL wine → ounces wine 14 g alcohol x 1 mL x 100 mL wine x 1 oz . 1 standard drink 0.789 g 12 mL alcohol 29.57 mL

= 5.0 oz of wine/standard drink

Molarity (M) = moles of solute L of solution

Drill problem: What volume of 2.0 M CaCl2 is needed to provide 1.94 g Ca²⁺?

g Ca²⁺ → mol Ca²⁺ → mol CaCl₂ → volume soln

1.94 g Ca²⁺ x 1 mol x 1 mol CaCl2 x 1 L . 40.1 g 1 mol Ca²⁺ 2.0 mol CaCl2

= 0.024 L or 24 mL

Ch 8.6 Dilution

C

x V

= C

x V

Practice problem

:

C

x V

= C

x V

0.300 M x 100 mL = 0.0500 M x V

V

= 600 mL = final volume (V

+ V

) V

= V

– V

= 600 mL – 100 mL = 500 mL

Q: How many-fold of a dilution has occurred?

Ch 8.7 Colloidal Dispersions and Suspensions

Table 8.4 Property comparison for solutions, colloidal dispersions, and suspensions.

Ch 8.8 Colligative Properties of Solutions

• reduction of vapor pressure

• elevation of boiling point

• depression of freezing point

• osmotic pressure

directly proportional to the total concentration of all the species formed when solutes dissolve

Ch 8.9 Osmosis and Osmotic Pressure

osmolarity = molarity x i

i = number of particles from one formula unit of solute

Example: What is the osmolarity of a solution that is 1 M in MgBr₂ and 2 M in glucose?

Osmolarity = 1 M x 3 + 2 M x 1 = 5 osmol MgBr2 glucose

The osmotic pressure of a solution is directly proportional to the number of solute particles present.

Solution A 0.30 osmol = 7.6 atm Solution B 0.15 osmol = 3.8 atm Solution C 1.5 osmol = ?

Isotonic, Hypotonic, and Hypertonic solutions.

Ch 8.10 Dialysis: a semipermeable membrane allows the passage of solvent, dissolved ions, and small molecules, but blocks the passage of larger particles. (separation technique)

Chapter 9 Chemical Reactions

Ch 9.1-9.3 covered in Review II (Chapter Medley) Ch 9.4 Collision Theory and Chemical Reactions

Activation energy (E

) is the minimum combined kinetic energy that colliding reactants must possess in order for their collision to result in a chemical reaction.

Higher activation energy → Slower reaction Lower activation energy → Faster reaction

Collision Orientation

Reaction rates are sometimes very slow because reactant molecules must be oriented in a certain way.

Ch 9.5 Exothermic Chemical Reactions

Figure 9.7 Energy diagram for an exothermic reaction.

Sometimes an initial input of energy may be needed but once it has started, an exothermic reaction is self-

sustaining.

Ch 9.5 Endothermic Chemical Reactions

Figure 9.7 Energy diagram for an endothermic reaction.

A continuous input of energy is needed for endothermic reactions to occur.

Ch 9.6 Factors That Influence Chemical Reaction Rates

• Physical nature of the reactants

• Reactant concentrations (surface area for solids)

• Reaction temperature (rate doubles every 10

C)

• Reaction pressure in the case of gases

• Presence of catalysts

Catalyst is not consumed.

It provides an alternative route for the reaction with a lower activation energy (E_a).

Catalysts DO NOT influence the amount of product formed, they only speed up the process.

Drill Problem: Will each of the changes listed increase or decrease the rate of the following reaction?

N2 + 3 H2 → 2 NH3

• Adding some N2

• Raising T

• Removing a catalyst

• Removing some H2

Drill Problem:

Reaction A ∆H = -20 kcal/mole E_a = 25 kcal/mole Reaction B ∆H = -15 kcal/mole E_a = 20 kcal/mole Which reaction occurs faster at the same temperature?

Which reaction releases more heat energy?

Ch 9.7 Chemical Equilibrium = Dynamic Equilibrium Ch 9.8 Equilibrium Constant

wA + xB yC + zD

[products] [C]

[D]

K

= =

[reactants] [A]

[B]

[ ] = molar concentrations (M = moles/L)

When [products] >> [reactants], K_eq = large, Equilibrium right When [reactants] >> [products], Keq = small, Equilibrium left

At values of Keq > 1000, [products] >> [reactants] and the reaction is usually considered complete.

Pure solids and pure liquids have constant concentrations, which are incorporated into the equilibrium constant itself.

2 KClO₃(s) 2 KCl(s) + 3 O₂(g)

K_eq = [O₂]³

catalyst

N2(g) + 3 H2(g) 2 NH3(g)

Sample calculation: Keq = 70 at 350^oC

[N2] = 0.100 M [H2] = 0.300 M [NH3] = ?

[NH₃]² [NH₃]²

K_eq = = = 70 [N₂][H₂]³ [0.100][0.300]³ [NH₃]² = 70 x 0.100 x (0.300)³

[NH₃]² = 0.189

√[NH₃]² = √0.189 [NH₃] = 0.435 M

Ch 9.9 Altering Equilibrium Conditions

• Concentration changes (Keq is unchanged)

• Pressure changes (Keq is unchanged)

• Temperature changes (Keq changes)

• Addition of catalyst has NO EFFECT on the equilibrium position, it merely allows the equilibrium to be reached more quickly. (Keqis unchanged)

N₂(g) + 3 H₂(g) 2 NH₃(g) Stress imposed Shift observed

Add N₂ right

Remove N₂ left

Add H2 right

Remove H2 left

Add NH3 left

Remove NH3 right

C hange in pressure

Drill Problem:

Rxn A: 2 NO2(g) + 7 H2(g) 2 NH3(g) + 4 H2O(g) Rxn B: H2(g) + I2(g) 2 HI(g)

What will be the effects of increasing pressure?

Rxn A:

Rxn B:

What will be the effects of decreasing pressure?

Rxn A:

Rxn B:

When the reaction temperature changes, Keq also changes!

H₂(g) + F₂(g) 2 HF(g) + Heat For exothermic reactions heat = product

Increase T, shift left - [HF] decreases, K_eq decreases Decrease T, shift right - [HF] increases, K_eq increases

2 CO₂(g) + Heat 2 CO(g) + O₂(g) For endothermic reactions heat = reactant

Increase T, shift right - K_eq increases Decrease T, shift left - Keq decreases

Supplemental material: Spontaneity and Chemical Reactions Enthaply (∆H) and Entropy (∆S) determine whether a reaction is spontaneous or nonspontaneous

Most spontaneous chemical reactions are exothermic.

Many endothermic reactions are nonspontaneous.

Some endothermic reactions are spontaneous because Entropy increases: ∆S = positive

Drill Problem:

PCl

(g) PCl

(g) + Cl

(g) ∆H = (+)

1. What changes would produce more PCl₅(g)?

[ ] ? add Cl₂ or PCl₃

P ? increase P to shift equilibrium left

T ? ∆H = (+) for endothermic reaction and Heat is on left side, decrease T to favor left side, K_eq changes

Catalyst? No effect on equilibrium, just reaction rate.

2. What is the sign of

∆S

1 mole of gas → 2 moles of gas = disorder increases,

∆S

0.60 mole PCl5

0.20 mole PCl3

1.0 mole Cl2

Calculate the Keq for the reaction.

Keq = [PCl3][Cl2] = (0.20mol/4L)(1.0mol/4L) = 0.083 [PCl₅] (0.60mol/4L)

Chapter 10 Acids, Bases, and Salts

Ch 10.1 Arrhenius Acid-Base Theory

Arrhenius Acids produce Arrhenius Bases produce H⁺ in water OH^- in water

HCl hydrochloric acid KOH HNO3 nitric acid Ba(OH)2

HClO₄ perchloric acid H₂SO₄ sulfuric acid H₃PO₄ phosphoric acid

Ch 10.2 Brønsted–Lowry Acid-Base Theory Brønsted–Lowry acid = proton (H

ion) donor Brønsted–Lowry base = proton (H

ion) acceptor

Hydronium Ion

Base Acid H⁺ acceptor H⁺ donor

H

O(l) + HCl(g) → H

O

(aq) + Cl

(aq)

Only acidic H atoms are donated:

Acidic

at can either donate or accept a H⁺ are called amphiprotic.

Acetic acid is monoprotic

Substances th

Conjugate Acid-Base pairs differ by a single H

Drill Problem. Write the chemical formula for:

1. The conjugate base of H

PO

2. The conjugate acid of H

PO

3. The conjugate base of H

O

4. The conjugate acid of H

O

Ch 10.3 Mono-, Di-, and Triprotic Acids

Monoprotic acids can transfer 1 H

to H

O or base Examples: HCl and HNO

Diprotic acids can transfer 2 H

Example: H

SO

+ H

O → H

O

+ HSO

HSO

+ H

O H

O

+ SO

Triprotic acids can transfer 3 H

Example: H

PO

+ H

O H

O

+ H

PO

H

PO

+ H

O H

O

+ HPO

HPO

+ H

O H

O

+ PO

A polyprotic acid supplies 2 or more H

Ch 10.4 Strengths of Acids and Bases

A strong acid donates all or nearly 100% of its H

to H

O

Table 10.1 Learn the name and formulas these commonly encountered strong acids, and then assume that all other acids you encounter are weak, unless you are told otherwise.

s of

These acids are strong even in dilute solution because in water they are all or mostly ionized.

A weak acid does not ionize completely.

CH

CO

H(l) + H

O(l) CH

CO

O

Strong bases are limited to the hydroxides of Group IA and IIA listed in Table 10.2.

Ammonia gas (NH

) is the most common weak base.

NH

(g) + H

O(l) NH

(aq) + OH

(aq)

Ch 10.5 Ionization Constants for Acids and Bases HA(aq) + H

O(l) H

O

(aq) + A

(aq)

acid ionization constant K

= [H

O

][A

] [HA]

Acid strength increases with increasing K

values.

B(aq) + H

O(l) BH

(aq) + OH

(aq) base ionization constant K

= [BH

][OH

]

[B]

Base strength increases with increasing K

values.

Acid-Base Neutralization Reactions

(Ch 10.1, 10.6 & 10.7 covered in Chapter Medley)

Acid + Base → Salt + Water HX + BOH → BX + HOH Net ionic equation: H⁺ + OH^- → H₂O

Ch 10.8 Self-Ionization of Water

H

O(l) + H

O(l)

H

O

(aq) + OH

(aq) K

= [H

O

][OH

] = 1.00 x 10

K

= Ion product constant for pure H

O at 25

C

Pure H₂O at 25^oC, [H₃O⁺] = [OH^-] = 1.00 x 10^-7 M Neutral

HCl is added to produce a 0.010 M solution:

acidic solution [H3O⁺] > 10^-7 and [H3O⁺] > [OH^-] K_w = [H₃O⁺][OH^-] = 1.00 x 10^-14

[OH^-] = K_w/[H₃O⁺] = 1.00 x 10^-14/1.0 x 10^-2 [OH^-] = 1.0 x 10^-12

NaOH is added to produce a 0.0010 M solution:

basic solution [OH^-] > 10^-7 and [OH^-] > [H₃O⁺]

[H3O⁺] = Kw/[OH^-] = 1.00 x 10^-14/1.0 x 10^-3 [H3O⁺] = 1.0 x 10^-11

Ch 10.9 The pH concept

pH is the negative logarithm of the molarity of the hydronium ion:

pH = -log[H

O

] [H O

] = 6.3 x 10

pH = 4.20

enter 6.3 x 10^-5 in your calculator, press the log key, then switch the sign

pH = 7 = neutral

pH < 7 = acidic

pH > 7 = basic

Ch 10.10 The pK

Method for Expressing Acid Strength

pK

= -logK

Acid strength increases with increasing K

increases with decreasing pK

acetic acid K_a = 1.8 x 10^-5 pK_a = 4.74 HF K_a = 6.8 x 10^-4 pK_a = 3.17

Ch 10.11 The pH of Aqueous Salt Solutions

Table 10.6 Examples of neutral, acidic, and basic salts.

Ch 10.12 Buffers

CH₃CO₂H/CH₃CO₂^- H₂PO₄^-/HPO₄^2-

H₂CO₃/HCO₃^- (buffer in human blood)

• Added base is absorbed by the acid OH^- + H2CO3 → HCO3- + H2O

• Added acid is absorbed by the conjugate base H3O⁺ + HCO3- → H2CO3 + H2O

Ch 10.14 Electrolytes

Strong Electrolyte = soluble ionic compounds and covalent molecules that ionize completely, such as strong acids

Weak electrolytes = molecules that ionize incompletely, such as weak acids and bases

Nonelectrolyte = covalent molecules that do not ionize

Drill Problem. Classify each of the following compounds as a strong electrolyte, weak electrolyte, or nonelectrolyte:

H₃PO₄ HCl Cl₂ HF KBr CH₃CH₂-OH CH₃COOH

Ch 10.15 Equivalents and Milliequivalents of Electrolytes One equivalent (Eq) supplies one mole of charge 1 mole Na⁺ = 1 equivalent

1 mole Ca²⁺ = 2 equivalents 1 mole PO43- = 3 equivalents

Sample Calculation. Human blood plasma contains 2.4 mg Mg²⁺ per dL. How many Eq or mEq are in 1.0 L of plasma?

Strategy: dL → L mg → g → mol → Eq → mEq

1.0 L x 2.4 mg Mg²⁺ x 10 dL x . 1 g . x 1 mol x 2 Eq Mg²⁺

1 dL 1 L 10³ mg 24.3 g mol Mg²⁺

= 2.0 x 10^-3 Eq Mg²⁺ or 2.0 mEq Mg²⁺

Ch 10.16 Acid-Base Titrations

endpoint detected with acid-base indicator:

HInd + H₂O H₃O⁺ + Ind^-

Sample calculation. In an acid-base titration, 13.07 mL of 0.100 M H₃PO₄ is needed to neutralize 25.0 mL of KOH of unknown concentration. Calculate the molarity of the KOH.

H₃PO₄ (aq) + 3 KOH(aq) → K₃PO₄(aq) + 3 H₂O(l)

13.07 mL x 0.100 mol H3PO4 x 3 mole KOH = 0.00392 mole KOH 1000 mL 1 mol H3PO4

M = mol/L = 0.00392 mole KOH/0.025 L = 0.157 M KOH