Student Reader's Chemistry Notes

Elimination Reaction Type 1 (E1)

Levi Clancy — Fri, 21 Nov 2008 02:02:57 +0000

In the E1 mechanism, a leaving group leaves. The resulting carbocation has a R-C⁺-C-H arrangement. A nucleophile attacks the hydrogen, and the hydrogen donates it’s e^-‘s to create a new π-bond. There are 3 fundamental requirements:

Moderate or better leaving group.
Stable carbocation.
Polar solvent.

Aqueous Strong Acid or Base

Levi Clancy — Mon, 21 Jul 2008 19:57:02 +0000

Q · Find the pH of .2M HCl_aq.
A · Since HCl is a strong acid, it completely dissociates into H⁺ and Cl^-. The concentration of H⁺ is .2M. Thefore, the pH is: - log (.2) = .7

Q · Find the pH of .2M NaOH_aq.
A · Since NaOH is a strong base, it completely dissociates into OH^- and Na⁺. The concentration of OH^- is .2M. Therefore, the pOH is: - log (.2) = .7 & the pH will be: 14-.7 = 13.3

Titrations

Levi Clancy — Mon, 21 Jul 2008 19:56:46 +0000

When pH=pK_a, dissociation occurs. Half protonated and half deprotonated.

Titrations

1. pH of an unknown.
2. K_a/b of an unknown.
3. Identity of an unknown.

   |                    ______________________
   |                   /
   |                  /
pH |                  |
   |                  ∗ ← equivalence point
   |⌈ buffer region ⌉ |
   |                  /
   |________∗________/
   |        ↑
   | ½ equivalence point
   |__________________________________________
                 volume of base

mol acid = mol base (aded) at equivalence point

How much 3.0 M NH3 is needed to fully tirate 40mL 2.0M HCl

3.0Vb = 2.0*40

Vb = 26.3 mL

How much of 4.0M CA(OH)2 need to reach equivalence point w/50 mL of 3.0 M HC2H3O2 (acetic acid)

polyprotic base

3.0*50=2*4.0Vb Vb=18.75 mL

w/ 50mL of 3.0M H3PO4

3*3.0*50 = 2*4.0Vb

Vb=450/8 = 56.25

Henderson-Hasselbach:

pH = pK_a – log ([rons base]/[acid])

BUFFER REGION…SOMETHING WHICH PREVENTS CHANGE IN PH

H2CO3/NaHCO3

weak acid or base/conjugate salt

Indicator…

Weak acid that changes color at some pH.

Rule: choose indicator with a pK_a Â± 1 away from pH of equivalence point.

HA (color 1) → H+ + A- (color 2)

Given that pK_a of acetic acid is 4.82, which form (protonated or deprotonated) will be favored at a pH of 6.3?

Since pH > pK_a, deprotonated form is favored.

HC2H3O2 &rarr H+ + C2H3O2- (in solution of pH > pK_a, protons are removed, shifting equilibrium to right; that means deprotonated is favored).

pH of a strong acid:

Fully dissociates:

HCl → H⁺ + Cl^-

Calculate the pH of .1M HCl solution at 25ºC equilibrium.

.1M = [H⁺] → pH = 1

H₂SO₄ is strong
HSO₄ is amphoteric (both basic and acidic)

Calculate pH of .02M H₂SO₄.

HSO₄^- is so weak that it does not matter. [H⁺] = .02

pH OF WEAK ACIDS

Acidity ∝ K_a, so ∴ K_a tell the strength of an acid.

H₂CO₃ → HCO₃^- → CO₃^2-

K_a of H₂CO₃ is 1.0. K_b of HCO₃^- is 10^-14

Calculate pH of 2.0M aqueous acetic acid (HC₂H₃O₂). The K_a = 1.8 x 10^-5

K_a = [H+][C2H3O2-]
[HC2H3O2]

1.5 x 10^-5 = xx
2.0

1.5 x 10^-5 = x²
2.0

conjugates…

HCl → H⁺ + Cl^- (conjugate base)

HCN (Ka=10-10) → H+ + CN-Kb=10-4

+ formerly atached to stron base = neutral
- attached to strong acid = neutral
+ attached to weak base = acidic
- weak acid = basic

KCN → K+(neutral) + CN-(basic)

CN- + H2O → HCN + OH-

HCl + NaOH → H2O + NaCl (N)

HCl + NH3 → NH4+ + Cl- (A)

G-Chem 6

bmik32@yahoo.com

k is equilibrium constant.

k=p/r

Ka HA → H⁺ + A^- (A^- is conjugate base)

Strong acid if H+A-/HA >1 … fully dissociates …

HCl, HBr, H2SO4, HI, HNO3, HClO4, HClO3

Kb BOH → B+ + OH- If kb OH-B+/BOH >1 then strong base …

group i metal hydroxides

arrhenius
donates H+
donates OH-NaOH

bronsted-lowry
donates H+
accepts H+ NH3

lewis
accepts e^-
donates e^- (nucleophiles, ligands, chelator)

Dentate: NH3 is a monodentate lewis base because it donates one lone pair to iron. a bidentate offers two lone pairs. 3…tri. 4…poly. FeN4 is a polydentate famous molecule called hemoglobin.

      __
     /  \
    N    N
   / \  / \
  |   Fe   |   ← hemoglobin molecule
   \ /  \ /
    N    N
     \__/

H+…4.2×10-6

10-4.2 =5.8

.58

-(-6) -1 = 5

10-4.2 =5.8/10

5.58

this is only at 25ºC.

oh- of 7.2×10-12

poh = 12-1 =11 10-7.2 =2.8 11.28
ph=2.72
h+=2.8*10-3

ph=9.36

poh=4.64
[h+] = 6.4*10-10
[oh-]=3.6*10-5

Properties of Water (dH2O)

Levi Clancy — Mon, 21 Jul 2008 19:47:09 +0000

Water molecules interact with eachother via hydrogen bonding. This hydrogen bonding also gives water three important properties:

Water is an excellent solvent.
Water is cohesive, meaning the molecules stick together.
Frozen water is less dence than liquid water.

Atoms & Elements

Levi Clancy — Thu, 02 Feb 2006 02:06:21 +0000

Atoms contain a nucleus. This nucleus contains positively charged protons and neutral neutrons. It is surrounded by a smear of negatively charged electrons (e^-‘s). Atoms have volume and mass, which are both properties of matter. Mass measures the amount of matter present (not to be confused with weight, which measures the gravitational force on mass). Atoms have equal numbers of electrons and protons. Therefore, an atom is electrically neutral. Electrons determine the chemical properties of an atom. They swarm about the nucleus in concentric regions called electron shells or energy levels. Each electron shell holds a limited number of electrons. The one closest to the nucleus holds a maximum of 2 electrons, the second level can hold up to 8 electrons, and the third holds a maximum of 18 electrons. The electrons of the outermost shell, called the valence electrons, determine the binding properties of an atom. Atoms react to complete the outermost shell. Atoms with a single shell, react to fill this shell with 2 electrons. All other atoms react to produce eight electrons in the valence shell. This tendency is referred to as the octet rule.