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I.^
Acid-Base Properties of SaltsA)^
Neutral Salts 1)^
Salt = ionic compound = one that completely ionizes in water 2)^
Cations of strong bases have no effect on pH1)^
Na
+^ , K
+^ , etc…
2)^
These cations have no affinity for OH
-^ in water
- Anions of strong acids have no effect on pH
1)^
, NO
-^ , etc… 3
2)^
These anions have no affinity for H
+^ in water
4)^
Solutions of these combined ion salts have pH = 7. 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 NaC
H 2
O 3
2
Na
+^
+^
C^2
H^3
O^2
C^2
H^3
O^2
-^
+^
H^2
O^
HC
H 2
O 3
2
+^
OH
4)^
For any weak acid and its conjugate base, K
x Ka
= Kb^
W
K^ b
= K
/ KW^
= 1 x 10a^
-^
/ 1.8 x 10
-5^ = 5.6 x 10
-^
for acetate
]
O
H
[C
]
][OH
O
H
[HC
K^
2 3 2
b^
C.^
Acidic Salts1)^
Salts having the conjugate acid of a weak base produce acidic solutions 2)^
Ammonium chloride = NH
Cl 4
NH
+^4
+^
NH
+^4
NH
3
+^
+H
3)^
Since ammonia is a weak base, ammonium is “strong” acid and will effect pH 4)^
Highly charged metal ions can also be acidica)^
AlCl
3
+^
6 H
O 2
Al(H
O) 2
3+ 6
3 Cl
b)^
Al(H
O) 2
3+ 6
+ H
O 2
Al(H
O) 2
(OH) 5
2+^
+ H
O 3
c)^
The higher the metal’s charge the more acidic
Example
: AlCl
K 3
= 1.4 x 10a^
B.^
How a Buffer Works1)^
Buffers have a large concentration of both HA and A
2)^
When OH
-^ is added, it reacts with HA and is replaced in solution by A
3)^
When we add H
+^ , it is used up by reaction with A
-^ and replaced in solution by HA
4)^
HA
+H
A
C.^
The Henderson-Hasselbalch Equation for Buffers:
a)^
We can use this simple equation instead of doing the longer equilibriumproblem 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]
. If we keep the buffer concentration high, this is a 0
valid approximation.
]
[HA][A
on
depends
which],
[H
on
depends
pH
]
[HA][A
K
]
[H
[HA]
]
][A
[H
K
a
a
[HA]
]
[A
log
pK
pH
a
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IV.
Procedure and Calculations for BuffersA.
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. 2)^
K^ a
= 1.8 x 10
-5^ so pK
= 4.74a^
3)^
To make buffer: Add 5.00g Sodium Acetate and 21.2ml 1.00M Acetic Acid to a50ml Volumetric Flask and dilute with water to the mark.
21.2ml
1.00M
0ml)
(0.424M)(
VM M
V
0.424M
[HA]
1.22M[HA]
[HA]
]
[A
1.22M
0.061mol0.050L
O
H
NaC
0.061mol
1mol82.034g
(5.00g)
[HA]
]
[A
[HA]
]
[A
log
[HA]
]
[A
log
[HA]
]
[A
log
pK
pH
(^221)
1
2 3 2
a
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 OH
3)^
HC
H 2
O 3
2
+^
OH
-^
C^2
H^3
O^2
Initial
0.00424mol
0.0005mol
0.0122mol
After OH
-^
0.00374mol
0.0127mol
/0.105L
0.00374mol
0.105L
0.0127mol/
log
[HA]
]
[A
log
pK
pH
a^
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