Transcript Acid-Base Properties

Acid-Base Properties
H O
1)
Arrhenius Definition
a) Acids produce H+ in water
b) Bases produce OH- in water
c) Applies only to aqueous solutions
d) Allows for only one kind of base (OH-)
2)
Bronsted-Lowery Definition
a) Acid is an H+ donor
b) Base is an H+ acceptor
+
H Cl
H
base
H O H
+
Cl
H
acid
conjugate
acid
conjugate
base
3) General Acid Equation
HA + H2O
H3O+ + Aa) Conjugate base = what is left after H+ leaves acid
b) Conjugate acid = base + H+
c) Conjugate acid-base pair are related by loss/gain of H+
4)
Ka = acid dissociation constant
[H 3O ][A  ] [H  ][A  ]
Ka 

[HA]
[HA]
5)
General Base Equation (Kb = base dissociation constant)
B: + H2O
BH+ + OH
6 ) Bases require N: or O: lone pair
H
H N
+
H O H
H
base
acid
[BH ][OH  ]
Kb 
[B]
H
H N H
H
conjugate acid
+
OH
conjugate base
Strong Acid = equilibrium lies far to the right ([H+] = [HA]0)
HCl
H+ + Cl- K = large
8) Strong Base = equilibrium lies far to the right ([OH-] ≈ [B]0)
NaOH
Na+ + OH- K = large
9) Weak Acid = equilibrium lies far to the left ([H+] << [HA]0)
HC2H3O2
H+ + C2H3O2- K = 1.8 x 10-5
10) Weak Base: equilibrium lies far to the left ([OH-] << [B]0)
NH3 + H2O
NH4+ + OH- K = 1.8 x 10-5
7)
Water as an Acid and Base
1) An amphoteric substance can behave as an acid or a base (water)
2) Autoionization of water (reaction with itself)
H2O + H2O
H3O+ + OH3)
Ionization constant for water = KW = [H3O+][OH-] = [H+][OH-]
a) For any water solution at 25 oC, [OH-] x [H+] = KW = 1 x 10-14
b) Neutral solutions (pure water) have [OH-] = [H+] = 1 x 10-7
c) Acidic solutions: [H+] > [OH-]
d) Basic solutions: [OH-] > [H+]
e) Example: If [H+] = 1 x 10-5, what is [OH-]?
4)
Relationship of pH and pOH
a) KW = [H+][OH-]
b) -logKW = -log[H+] – log[OH-]
c) -log(1 x 10-14) = pH + pOH = 14
d) Example: Find [H+], [OH-], and pOH for sample of pH = 7.41
The pH Scale
1) pH = -log[H+] (simplifies working with small numbers)
2) If [H+] = 1.0 x 10-7, pH = -log(1 x 10-7) = -(-7.00) = 7.00
3)
Number of decimal places in a log:
The number of sig. fig’s in the number is
how many decimal places you keep when
you take the log
4)
Other p Scales:
a) pOH = -log[OH-]
b) pKa = -logKa
5) pH changes by 1 unit for every power
of 10 change in [H+]
a) pH = 3 [H+] = 10 times the [H+] at pH = 4
b) pH decreases as [H+] increases
(pH = 2 more acidic than pH = 3)
Percent Dissociation = amount dissociated / initial conc. x 100%
a) For 1.00 M HF we found [H+] = 2.7 x 10-2
Percent Dissociated = (2.7 x 10-2 / 1.00) x 100% = 2.7%
b) Strong acids are 100% dissociated always
c) For weak acids, percent dissociation increases as acid is diluted
HA + H2O
H3O+ + A-
Le Chatelier’s Principle
Finding Ka or Kb from the pH
1) Example: an unknown acid HA at 0.40 M has a pH = 2.6. What is the Ka?
pH  log[H  ]  [H  ]  10-pH  102.6  2.5 x 103
[H  ][A  ] [2.5 x 10-3 ][2.5 x 10-3 ]
-5
Ka 


1
.
56
x
10
[HA]
[(0.4) - (2.5 x 10-3 )]
2)
Example: an unknown base at 0.5 M has a pH = 11.3. What is the Kb?
pOH  14 - pH  14 - 11.3  2.7
pOH  log[OH  ]  [OH  ]  10-pOH  102.7  2.0 x 103
[BH  ][OH  ] [2.0 x 10-3 ][2.0 x 10-3 ]
-6
Kb 


8
.
03
x
10
[B]
[(0.5) - (2.0 x 10-3 )]
Experimental Details
1) You can skip Procedure 1, the standardization of the pH meter (it has been done)
2) In Procedure 2, use clean pipet to blow breath through solution (broken glass box)
-get your own tap and distilled water
-vials of other solutions up front; take to your pH meter and then bring back
3)
In Procedure 3, take 10 ml of the stock solution and dilute it to 100 ml
-use volumetric pipet and volumetric flask
M1V1  M 2 V2

M1 V1 
V2 
 [HCl]
M 2 

0.001M 0.100L 

 0.010L
0.010M 
4)
Procedure 4 and 5: Use graduated cylinder and be as careful as you can
-use pH meter readings to find [H+], find other concentrations from it
-substitute concentrations into Ka/Kb expression to find Ka/Kb
5)
Skip the Indicator Procedure Part 6 and PreLab Question #4