I.
Acid-Base Properties of Salts
A) Neutral Salts
1) Salt = ionic compound = one that completely ionizes in water
2) Cations of strong bases have no effect on pH
1) Na+, K+, etc…
2) These cations have no affinity for OH- in water
3) Anions of strong acids have no effect on pH
1) Cl-, NO3-, etc…
2) These anions have no affinity for H+ in water
4) Solutions of these combined ion salts have pH = 7.00
B. Basic Salts
1) Salts containing the conjugate base of a weak acid produce basic solutions
2) The conjugate base must be strong if the acid is weak, so it must have a strong
affinity for H+, which will affect the pH of a solution
3) Sodium Acetate Example
NaC2H3O2
Na+ + C2H3O2[HC2 H3O2 ][OH ]
C2H3O2- + H2O
HC2H3O2 + OHK 
 ???
b
-
[C2 H3O2 ]
4) For any weak acid and its conjugate base, Ka x Kb = KW
Kb = KW / Ka = 1 x 10-14 / 1.8 x 10-5 = 5.6 x 10-10 for acetate
C.
Acidic Salts
1) Salts having the conjugate acid of a weak base produce acidic solutions
2) Ammonium chloride = NH4Cl
NH4+ + ClNH4+
NH3 + H+
3)
Since ammonia is a weak base, ammonium is “strong” acid and will effect pH
4)
Highly charged metal ions can also be acidic
a) AlCl3 + 6 H2O
Al(H2O)63+ + 3 Clb) Al(H2O)63+ + H2O
Al(H2O)5(OH)2+ + H3O+
c) The higher the metal’s charge the more acidic
Example: AlCl3 Ka = 1.4 x 10-5
II.
Salt Solution Procedure
A.
Add ~0.5 g of eight different salts to 25 ml water and test with pH paper
1) Use “pH = 7” water and graduated cylinder
2) Don’t have to be very careful with amount; just seeing if acidic/basic/neutral
3) Hydrolysis Reaction = reaction with water (neutral salt: no hydrolysis)
NaF: F- + H2O -------> HF + OH- (pH is basic)
B.
Carefully add 0.500 g of NH4Cl and NaC2H3O2 to 25 ml water; test with pH meter
1) Use “pH = 7” water and Volumetric Flask
2) Need to know mass to 3 decimal places and pH to 2 decimal places
3) Use ICE tables to calculate theoretical pH of your solutions
III. Buffers
A.
A Buffer is a solution containing a weak acid and the salt of its conjugate base (or
a weak base and the salt of its conjugate acid) which resist changes to its pH.
1) Example: Blood stays at pH = 7.4, even when various reactions in the body
produce much H+ and OH2) The buffer system in blood is H2CO3/HCO33) Buffer Capacity = amount of H+/OH- a buffer can absorb with only a small
pH change
B.
How a Buffer Works
1) Buffers have a large concentration of both HA and A2) When OH- is added, it reacts with HA and is replaced in solution by A3) When we add H+, it is used up by reaction with A- and replaced in solution by HA
4)
HA
H+ + A-
[H ][A- ]
[HA]
Ka 
 [H ]  K a
[HA]
[A- ]
pH depends on [H ], which depends on
C.
The Henderson-Hasselbalch Equation for Buffers:
[HA]
[A- ]
 [A  ] 

pH  pK a  log
 [HA] 
a)
We can use this simple equation instead of doing the longer equilibrium
problem approach for most buffer problems
b)
Any concentration of buffer with the same [A-]/[HA] ratio will have the same
pH
c)
The Henderson-Hasselbalch Equation assumes that [A-] and [HA] are the
same as [A-]0 and [HA]0. If we keep the buffer concentration high, this is a
valid approximation.
IV. Procedure and Calculations for Buffers
A. Make 50ml of a buffer with pH = 5.00 from 1.00 M Acetic Acid and 5.00 g of
Sodium Acetate
1) Example calculation for pH = 5.20
2) Ka = 1.8 x 10-5 so pKa = 4.74
 [A ] 
 [A ] 
  5.20  4.74  log

pH  pKa  log
 [HA]
 [HA]

 [A ] 

[A
]
  100.46  2.88  

0.46  log
 [HA]
 [HA]
 1mol 
0.061mol
  0.061molNaC2 H 3O 2 
(5.00g)
 1.22M
0.050L
 82.034g
[A ] 1.22M
2.88 

 [HA]  0.424M
[HA] [HA]
M V (0.424M)(5
0ml)
V1  2 2 
 21.2ml
M1
1.00M
3)
To make buffer: Add 5.00g Sodium Acetate and 21.2ml 1.00M Acetic Acid to a
50ml Volumetric Flask and dilute with water to the mark.
B.
Test the pH of your buffer with the pH meter, and calculate %Error
%Error
Experimental - Assigned
Assigned
x 100%
C.
Add 10ml of your buffer to a 100ml Volumetric and dilute to the line. Find pH
1) Original buffer: [HA] = 0.424M [A-] = 1.22M
2) Diluted buffer: [HA] = 0.0424M [A-] = 0.122M
D.
What is the pH when we add 5.00ml of 0.10M HCl to our diluted buffer?
1) 100ml + 5ml = 105ml = new volume
(0.122mol/L)(0.100L) = 0.0122mol A(0.0424mol/L)(0.100L) = 0.00424 mol HA
2) HCl will completely dissociate to H+ (0.10mol/L)(0.005L) = 0.0005mol H+
3)
C2H3O2+
H+
HC2H3O2
Initial
After H+
0.0122mol
0.0117mol
0.0005mol
0
0.00424mol
0.00474mol
 [A ] 
0.105L 
 0.0117mol/
  4.74  log
pH  pKa  log
  5.13
/0.105L
 0.00474mol
 [HA]
E.
What is the pH when we add 5.00ml of 0.10M NaOH to our diluted buffer?
1) 100ml + 5ml = 105ml = new volume
(0.122mol/L)(0.100L) = 0.0122mol A(0.0424mol/L)(0.100L) = 0.00424 mol HA
2) NaOH will completely dissociate to OH- (0.10mol/L)(0.005L) = 0.0005mol OH3)
HC2H3O2
+
OHC2H3O2-
Initial
After OH-
0.00424mol
0.00374mol
0.0005mol
0
0.0122mol
0.0127mol
 [A ] 
0.105L 
 0.0127mol/
  4.74  log
pH  pKa  log
  5.27
/0.105L
 0.00374mol
 [HA]