When pH=pKa, dissociation occurs. Half protonated and half deprotonated.
Titrations
1. pH of an unknown.
2. Ka/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 = pKa – 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 pKa ± 1 away from pH of equivalence point.
HA (color 1) → H+ + A- (color 2)
Given that pKa of acetic acid is 4.82, which form (protonated or deprotonated) will be favored at a pH of 6.3?
Since pH > pKa, deprotonated form is favored.
HC2H3O2 &rarr H+ + C2H3O2- (in solution of pH > pKa, 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 ∝ Ka, so ∴ Ka tell the strength of an acid.
H2CO3 → HCO3- → CO32-
Ka of H2CO3 is 1.0. Kb of HCO3- is 10-14
Calculate pH of 2.0M aqueous acetic acid (HC2H3O2). The Ka = 1.8 x 10-5
Ka = [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
Q · Find the pH of .2M HClaq.
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 NaOHaq.
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
|
|