Chem 120 Exams

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unsaturated solution

solution that contains less solute than it has the capacity to dissolve

saturated solution

solution that contains the maximum amount of solute that will dissolve in a solvent at a specific temp

supersaturated solution

solution that contains more dissolved solute than it present in a saturated solution

solution formation

solute dissolves in a solvent to form a solution
to form a liquid solution, the solute needs to separate into individual components, overcome IMF to make room and expand, than allow the solute and solvent to form a solution

predicting solubility for solids

increases with increasing temperature and not affected by increasing pressure

predicting solubility for gases

decreases with increasing temperature and increases with increasing pressure

intermolecular forces

London Dispersal < Dipole-Dipole < Hydrogen Bonding < Ion-Dipole

entropy factors

increasing temperature, volume, number of molecules, and complexity of molecules all increases the entropy of the substance due to microstates

When ∆H is (-) and ∆S is (+)

∆G is (-) so always spontaneous

When ∆H is (+) and ∆S is (-)

∆G is (+) so never spontaneous

When ∆H is (-) and ∆S is (-)

∆G is (-) at low temperatures but (+) at high temperatures

When ∆H is (+) and ∆S is (+)

∆G is (+) at low temperatures but (-) at high temperatures

vapor pressure

proportional to mole faction so it is higher for pure compounds compared to solutions

First Order Reactions

k[A]

k = 1 / s so graph is 1 / s vs T

t_1/2 = ln(2) / k

ln[A] = -kt + ln[A]_o

Second Order Reactions

k[A]²

k = 1 / Ms so graph is 1 / Ms vs T

t_1/2 = 1 / k[A]_o

1 / [A] = kt + 1 / [A]_o

Zeroth Order Reactions

k

k = M / s so graph is M / s vs T

t_1/2 = [A]_o / 2k

[A] = -kt + [A]_o

what step determines the rate of a reaction

slowest step

collision theory

molecules must collide with sufficient energy and in the correct orientation; more concentration, less volume, and higher temperature increases collision

chemical equilibrium

K = products / reactants

ignore solids and pure liquids

Q > K

reaction proceeds spontaneously in the reverse direction or shift left or products to reactants

Q < K

reaction proceeds spontaneously in the forward direction or shift right or reactants to products

Approximation of Reactant Favored

[A]o / Kc > 100

Le Chatelier’s Principle

if a system at equilibrium is distributed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance

Therefore, K only changes with temperature

ΔH > 0 (endothermic)

they have a direct relationship so temperature and K increases

ΔH < 0 (exothermic)

they have an inverse relationship so temperature increases and K decreases

Increasing reactant concentrations/pressures

denominator increases so Q decreases so Q < K

Increasing product concentrations/pressures

numerator increases so Q increases so Q > K

Increasing overall pressure with an inert gas

Q = K so no shift

Decrease Volume of Container

shift towards the side with less moles

Increase Volume of Container

shift towards the side with more moles

K >> 1

∆G_o < 0 and spontaneous and product favored

Q >> 1

Q is most likely > K and is non-spontaneous in the reverse direction and ∆G > 0

K << 1

∆G_o > 0 and non-spontaneous and reactant favored

Q << 1

Q is most likely < K and is spontaneous in the forward direction and ∆G < 0

Common Ion Effect

the shift in an equilibrium position caused by the addition or presence of an ion involved in the equilibrium reaction

Q < Ksp

solution is unsaturated (more salt will dissolve)

Q > Ksp

The solution is supersaturated (precipitation will occur)

Q = Ksp

The solution is saturated (at equilibrium)

Acid

Produces hydrogen ions (H+) in an aqueous solution and H+ donor

Base

Produces hydroxide ions (OH-) in an aqueous solution and H+ acceptor

Conjugate Base

What remains of an molecule after a proton is lost

Conjugate Acid

The species formed when a proton is transferred to

Strong acids and bases

K = 10⁶ to 10¹⁰

Weak acids and bases

K = 10^-3 to 10^-10

Polyprotic acids

Acids that have more than one dissociable proton

salt of a strong base and a strong acid

no hydrolysable ions and gives neutral aqueous solutions

salt of a strong base and a weak acid

hydrolyze and a basic solution

salt of a weak base and a strong acid

hydrolyze and a acidic solution

salt of a weak base and a weak acid

hydrolyze and will depend on the relative acid-base strengths of the two ions

Buffer solution

A solution that resists a change in its pH when either hydroxide ions or protons are added and resist drastic change in pH

weak acid/base and its conjugate base/acid

can use Henderson-Hasselbalch equation

Buffering capacity

the ability of a buffered solution to absorb protons or hydroxide ions without a significant change in pH; determined by the magnitude of [HA] and [A-] in the solution

0.1 < [A-]/[HA] < 10

strong acid + strong base titration

1. The initial pH (at about 12 which is a strong base at first)

2. Between the Initial pH and equivalence point (midway point between the middle and beginning)

3. The equivalence point (7 for strong acid strong base and the mols of acid = mols base)

4. After the equivalence point (at about 0 which is a strong acid now)

• You can also start with a strong acid and add a strong base, which gives an analogous titration curve

strong acid + weak base titration

1. The initial pH

2. Buffer Region

3. Equivalence point (not 7, but below)

4. After the equivalence point

• equivalence point happens when second derivative = 0

• when volume added = 1/2 volume at equivalence point, pH = pKa

weak acid + strong base titration

1. The initial pH

2. Buffer Region

3. Equivalence point (not 7, but above)

4. After the equivalence point

• equivalence point happens when second derivative = 0

• when volume added = 1/2 volume at equivalence point, pH = pKa

Reduction

gaining the electron → oxidizing agent

Oxidation

losing the electron → reducing agent

Anode

where oxidation occurs

Cathode

where reduction occurs

Line notation

oxidation anode (s) | oxidation anode (aq) || reduction cathode (aq) | reduction cathode (s)

Electrolysis

find the moles of electrons transferred then:

current and time → quantity of charge in coulombs → moles of electrons → moles of element → grams of element

oxidation anode

Zn(s) → Zn2+(aq) + 2e-

reduction cathode

2e- + 2Ag+(aq) → 2Ag(s)