Gen Chem 3

Question 0

Acid/CB and Base/CA

Answer 0

CB of an acid is the acid MINUS its HYDROGEN.
CA of a base is the base PLUS a HYDROGEN

Amphoteric substances can act as EITHER (H2O)

Question 1

Acid Base Chemistry

Answer 1

Arrhenius: Acids produce H+ ions in solution; bases OH- in solution
Bronsted-Lowry: acids DONATE protons (H+), bases ACCEPT protons
Lewis: acids accept a pair of electrons; bases donate a pair of electrons (AlCl3, BF3, electrophiles in o. chem are Lewis acids, NH3 and OH- are Lewis bases since they have electron pairs to donate).

Question 2

pH Scale

Answer 2

WATER ALWAYS PRESENT! pH scale ranks solutions based not so much on the acids and bases themselves, but on HOW THOSE ACIS OR BASES INFLUENCE the equilibrium for the ionization of WATER

Question 3

pH Formula

Answer 3

pH=-log[H+]

pOH=-log[OH-]

pH+pOH=14

pKa=-log(Ka)

Question 4

Acid Base Equilibria

Answer 4

ALL EQ CONSTANTS (Keq, Ka, Kb, Kw or Ksp) ARE WRITTEN VIA THE LAW OF MASS ACTION, with pure liquids (l) and solids (s) omitted. Large Ka means strong acid (lots of dissociation to form lots of product than reactants). Same with Kb.

Question 5

Ionization of Water

H2O + HA –> OH- + H3O+

Answer 5

Kw=[H3O+][OH-]= 10^-14 (at 25 C)

pKw= pH + pOH = 14
pKa+pKb= 14

Question 6

Acid Dissociation

HA + H2O -> H3O+ + A-

Answer 6

Ka=[H+][A-]/[HA]

Question 7

Base Dissociation

A- + H2O -> HA + OH-

Answer 7

Kb=[OH-][HA]/[A-]

Ka x Kb = Kw = 10^-14

Question 8

Summary of Acid and bases

Answer 8

WHEN WE ADD AN ACID OR BASE TO WATER, THE EQUILIBRIUM OF THE ACID OR BASE WILL DIRECTLY IMPACT THE EQUILIBRIUM FOR THE IONIZATION OF WATER ACCORDING TO Le Chatelier’s Principle.
The addition of either acid or base shifts the eq. for the ionization of water to the LEFT.

Question 9

Calculating pH for STRONG ACIDS/BASES

Answer 9

pH or pOH=-log[strong acid or base]

*Must assume at 25 C

Question 10

Calculating pH for WEAK ACIDS

Answer 10

1) Write out the equilibrium equation (HA-> H+ + A-)
2) Use x to represent the concentration of each of the two products (2x, 3x, etc. depending on the coefficient of the balanced equation)
3) Use “[HA]-x” for the concentration of the ORIGINAL ACID
4) If this results in a QUADRATIC equation, assume X is much smaller than [HA] and omit it.
5) Solve for x from Ka=(x)(x)/[HA-X]
6) Use -log[H+] (-log[x] to find pH)

Ka should be given right?

Question 11

Acid/Base Strength & Dissociation

Answer 11

Strong Acids/bases: all strong acids and bases dissociate 100% in water. Good electrolytes.

Strong ACIDS: HI, HBr, HCl, HNO3, HClO4, HClO3, H2SO4, H3O+
*H3O+ borderline strong, just barely pka<0 (-1.7). HF NOT A STRONG ACID.
Strong BASES: Group IA hydroxides (NaOH, KOH, etc), NH2-, H-, Ca(OH)2, Sr(OH)2, Ba(OH)2, Na2O, CaO

Question 12

Weak ACIDS/BASES

Answer 12

DON’T DISSOCIATE READILY IN SOLUTION. AS GENERAL RULE, pka > 0, or Ka<1, can be considered a weak acid. (Same with pkb or Kb)
EXAMPLES OF WA: anything not on the strong acid list. H2O, H2S, NH4+, HF, HCN, H2CO3, H3PO4, Acetic acid, benzoic acid, etc.

EXAMPLES OF WB: anything not on the SB list. H2O, NH3, R3N, pyridine, Mg(OH)2, ETC.

Question 13

Impact of SALTS ON THE DISSOLUTION OF WA and WB

Answer 13

• % dissociation of benzoic acid (weak acid) DECREASES in a sodium BENZOATE solution
• % dissociation of ammonium hydroxide (weak base) DECREASES in an ammonium chloride solution.
• This is due that the salts dissociate to the respective CA/CB, shifting the equilibrium to the left.

Question 14

Hydrolysis of SALTS

Answer 14

When the salts of weak acids or weak bases dissolve in water, one of the ions will undergo hydrolysis to REFORM the weak acid or weak base.
Example:
NaNO2-> Na+ + NO2- —> NO2- + H2O—> HNO2 + OH- (inc. pH)
NH4Cl-> NH4+ + Cl- –> NH4+ + H2O –> NH3 + H3O+ (dec. pH)

Question 15

TITRATION

Answer 15

Drop by drop mixing of an acid and base with an indicator.

ONE “EQUIVALENT”= the amount of acid or base necessary to produce or consume ONE MOLE of [H+] ions.

Question 16

TITRATION CURVES

Answer 16

x-axis: VOLUME OF TITRANT (mL)

y-axis: pH

Question 17

TITRATION OF SA w/ SB or SB w/ SA

Answer 17

EQUIVALENCE POINT/STOICHIOMETRIC POINT=midpoint of the nearly vertical section of the graph. [tirtant]=[analyte].
[H+]=[OH-] at the equivalence point, or pH=7

Question 18

Titration of a WA w/ SB or WB w/ SA

Answer 18

EQUIVALENCE POINT=midpoint of the nearly vertical section of the graph. Again, [titrant]=[analyte], however since acid or base not as strong, the [H+] does NOT EQUAL [OH-]. pH NOT 7

For WB with SA: pH7
For SA with SB: pH=7
For WA with WB: pH=7 (assuming equal strength; rarely attempted)

Question 19

Half-Equivalence Point

Answer 19

MIDPOINT OF THE NEARLY HORIZONTAL SECTION OF THE GRAPH.
Here the pH=pKa. Also, we can say that [HA]=[A-] AT THE HALF-EQUIVALENCE POINT. HA will continue to be deprotonated until at the equivalence point the solution contains 100% A- and 0% HA.
SA/SB titrations don’t have half equivalence points because they readily dissociate 100%.

Question 20

Indicators

Answer 20

Weak acids that change color as they dissociate from HA into H+ and A-. To set up a titration, you must know the approximate pH of your equivalence point; you then select an indicator that will change color at that pH. It has no impact on pH.

Question 21

BUFFERS

Answer 21

A buffer solution CONTAINS A WEAK ACID AND WEAK BASE, often the conjugates of each other. In a buffer there is an equilibrium between the WA and its CB, or between a WB and its CA.

The NEARLY HORIZONTAL AREA SURROUNDING THE HALF-EQUIVALENCE POINT IS CALLED THE “BUFFER REGION”

pH = pKa + log[A-]/[HA] (Henderson Hasselbach equation) or pH=pKA-log[HA]/[A-]

Question 22

How to recognize buffer problems?

Answer 22

1) Watch for equimolar amounts. The maximum buffering strength occurs when [HA] = [A-]. This is the ration one would start with if making a buffer in the lab. Therefore, be on the lookout for equimolar amounts of a weak acid and its CB, or a WB and its CA.
2) Watch for conjugates. To be a buffer, the two equimolar substances are often conjugates of each other such as: NH3 and NH4+, CH3COO- and CH3COOH, HCO3- and CO32-

3) Watch for WEAK acids/bases
4) Watch for resistance to pH change. BUFFER!!
5) Watch for half-equivalence point.
6) Watch for pH= pKa (at the midpoint of the buffer region)
7) Watch for the ration of [HA]/[A-] or [A-]/[HA].

Question 23

Electrochemistry

Answer 23

Oxidation-Reduction (REDOX) Reactions: Any reaction where one or more electrons are transferred from one atom to another. The atom that loses electrons is OXIDIZED and the atom that gains electrons is REDUCED. A REDUCING AGENT is an atom or molecule that donates electrons to another atom or molecule and is itself oxidized in the process. An OXIDIZING AGENT is an atom or molecule that accepts electrons and is itself reduced in the process.

Question 24

Oxidation State

Answer 24

Apparent charge that an atom takes on while in a molecule. The sum of the oxidation states for all of the atoms in a molecule must equal the charge on that molecule, or equal to zero fi the molecule is neutral.

Any elemental atom: 0
F: -1
H: +1
Hydrogen w/ a metal: -1
Oxygen (except peroxides): -2
Alkali metals: +1
Alkaline earth metals: +2
Group V: -3
Group VI: -2
Group VII: -1

Question 25

Electrical Potential

Answer 25

Tells us the degree to which a species “wants electrons” or “wants to be reduced”. Given in volts and are presented as HALF REACTIONS.

A species with a + E is more likely to GAIN ELECTRONS (be reduced) than are hydrogen ions.
A species with a - E is LESS likely to GAIN ELECTRONS than are hydrogen ions.

CATIONS GET REDUCED TO FORM SOLIDS. SOLIDS GET OXIDIZED TO FORM CATIONS.

Question 26

Cell Potential

Answer 26

Sum of the electrical potentials for the two half reactions that make up an electrochemical cell.
Can’t add two values directly from a table. MUST REVERSE HALF REACTION OF SPECIES WITH THE LOWEST REDUCTION POTENTIAL AND TAKE THE NEGATIVE OF ITS E value. NO STOICHIOMETRY.

Question 27

Electrochemical Cells: The GALVANIC CELL

Answer 27

Convert chemical energy into electrical energy. By taking advantage of the difference in reduction potentials between two metals, a current can be spontaneously generated along a wire that connects two metal electrodes submerged in solutions that contain metal ions.

Question 28

Purpose of SALT BRIDGE

Answer 28

It allows for the circuit to continue running. Sodium ions can float towards the Copper side and the nitrate ions can flow to the zing side, neutralizing the build up of charge and allowing for electron flow to continue.

Question 29

Galvanic Cell CONT

Answer 29

REDUCTION ALWAYS AT THE CATHODE AND OXIDATION ALWAYS AT THE ANODE (FOR ALL ELECTROCHEMICAL CELLS).

Cathode = (+); Anode = (-) True of only GALVANIC CELLS
CELL POTENTIAL ALWAYS POSITIVE. Galvanic cells only.
A functioning galvanic cell can be created using ANY TWO METALS, regardless of their reduction potentials. Electrons will automatically flow from the species with the LOWER REDUCTION potential to the species with the HIGHER reduction potential.

Question 30

Electrolytic Cell

Answer 30

Galvanic cell to which external voltage is applied, forcing electrons to flow in the opposite direction.
OXIDATION STILL AT THE ANODE AND REDUCTION AT CATHODE.
Species with LOWER reduction potential will be REDUCED!
Cell potential is ALWAYS NEGATIVE.
Sum of externally applied voltage (Vbattery) and the negative cell potential MUST BE POSITIVE.

Cathode=(-), Anode=(+)

Question 31

Concentration Cell

Answer 31

THE SAME ELECTRODES and SOLUTIONS ARE USED IN BOTH BEAKERS. In one beaker, a metal is oxidized via its oxidation half-reaction, and in the other beaker it is reduced via its reduction half reaction. E cell = 0.00 V.

NERNST EQUATION: E= Enot - (0.06/n) * log[lower]/[higher]
n=number of moles of electrons tranferred, Enot will ALWAYS=0 for concentration cell.

Question 32

Relationship Between FREE ENERGY AND CHEMICAL ENERGY

Answer 32

ΔGo= -nFEo, n is the number of moles of electrons transferred in the BALANCED REDOX REACTION, F is FARADAY'S CONSTANT
\+Eo= NEGATIVE ΔG = spontaneous reaction

Faraday’s constant: 9.6e4 C/mol (charge on one mole of electrons)