Titrations

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 ± 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

Filed Under: Monoprotic Rxns I
Tagged With: Aqueous Strong Acid or Base

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

Filed Under: Monoprotic Rxns I
Tagged With: