Transcript Document 7724214
Ch. 15: Applications of Aqueous Equilibria
15.2 Buffers
Buffered Solutions
• • a solution that resists pH change when H+ or OH- is added made from: – weak acid and its salt – weak base and its salt
Example 4
• A buffered solution contains 0.5 M HC 2 H 3 O 2 (K a =1.8x10
-5 ) and NaC 2 H 3 O 2 . Find pH of the solution – major species: HC 2 H 3 O 2 , Na + , C 2 H 3 O 2 , H 2 O I C E HC 2 H 3 O 2 0.5
-x 0.5-x ⇄ H + 0 +x x + C 2 H 3 O 2 0.5
+x 0.5+x
Example 4
1 .
8 10 5 ( 0 .
5 ( 0 .
5
x
)(
x
)
x
) (
x
)( 0 .
5 ) ( 0 .
5 )
x
1 .
8 10 5 [
H
]
pH
4 .
74 • What would happen to the pH if we added 0.010 mol of solid NaOH was added to 1.0 L of this buffered solution?
Example 4
• • major species: HC 2 H 3 O 2 , Na + , OH-, C 2 H 3 O 2 -, H 2 O HC 2 H 3 O 2 + OH – C 2 H 3 O 2 + H goes to completion b/c of SB 2 O E HC 2 H 3 O 2 + OH 0.49 mol 0 C 2 H 3 O 2 + H 2 O I (1.0L)(0.050M) =0.50 mol 0.010 mol (1.0L)(0.050M) =0.50 mol C -x=-0.010 mol +x=+0.010 mol +x=0.010 mol 0.51 mol
Example 4
• Now that we know the amount of acetic acid left, we must find out how much H + is created from it.
I C E HC 2 H 3 O 0.49
2 -x 0.49-x ⇄ H + 0 +x x + C 2 H 3 O 2 0.51
+x 0.51+x 1 .
8 10 5 ( 0 .
51
x
)(
x
) ( 0 .
49
x
) (
x
)( 0 .
51 ) ( 0 .
49 )
x
1 .
8 10 5
pH
0 .
02 [
H
]
pH
4 .
76
How does it do that?
• if it is a WA and salt – it contains large amounts of weak acid, HA, and its conjugate base, A .
– when OH OH is added, the HA donates H does not buildup + to it so the – when H + is added, A not buildup either accepts the H + so H+ does – as long as the change is small compared to the original conc, pH won’t change much
How does it do that?
• if it is a WB and salt – it contains large amounts of weak base, B, and its conjugate base, BH + .
– when OH OH is added, the BH does not buildup + donates H + to it so the – when H + is added, B accepts the H + buildup either so H + does not – as long as the change is small compared to the original conc, pH won’t change much
How does it do that?
Henderson-Hasselbalch equation
Ka
[
H
][
A
] [
HA
] [
H
]
Ka
[
HA
] [
A
]
pH
pKa
log [ [
A HA
] ]
pKa
log [
base
] [
acid
] • • useful for calculating the pH of a solution when the concentrations of acid and base are known and the change (x) is negligible for a certain acid-base pair, if they have the same ratio, they have the same pH
Example 5
• Find the pH of 0.75 M HC 3 H 5 O 3 and 0.25 M NaC 3 H 5 O 3 – (Ka=1.4x10
major species: HC 3 H 5 O 3 , Na + , C 3 H 5 O 3 , H 2 O -4 )
Ka
[
x
][ 0 .
25 [ 0 .
75
x
]
x
]
x
( 1 .
4 10 4 [ 0 .
75 ] ) [ 0 .
25 ]
pH
3 .
38
OR pH
pKa
log [ [
A HA
] ] 3 .
85 log 0 .
25 0 .
75 3 .
38
Ch. 15: Applications of Aqueous Equilibria
• 15.3 Buffer Capacity
Buffer Capacity
• • • • amount of H + or OH the solution can absorb without a significant change in pH determined by the size of [HA] and [A ] if the amounts are equal (ratio of one) and they have large concentrations, it will have the largest buffer capacity the pKa of weak acid used should be very close to the pH desired for the buffer
Buffer Capacity
• • because the [HA] is so low in solution B, the buffer capacity is much lower this causes the change in pH after adding acid to be much larger
Example
I C E • A buffered solution contains 0.25 M NH 3 (K b =1.8x10
-5 ) and 0.40 M NH 4 Cl. Calculate the pH of this solution.
NH 3 + H 2 O ⇄ 0.25
-x 0.25-x NH 4 + 0.40
+x 0.40+x + OH 0 +x x
Example
1 .
8 10 5 ( 0 .
40 ( 0 .
25
x
)(
x
)
x
) (
x
)( 0 .
40 ) ( 0 .
25 )
x
1 .
8 10 5 [
OH
]
pOH
4 .
95
pH
9 .
05 I C E What would the pH be after 0.020 mol of OH- is added to 500.0 mL of the solution?
NH 4 + 0.200mol
-0.020
0.180
+ OH 0.020
-0.020
0 ⇄ NH 3 + H 0.125
+0.020
0.145
2 O
I C E NH 4 + 0.360 M -x 0.360-x
Example
⇄ NH 3 + H + 0.290
0 +x +x 0.290+x x 1 10 14 1 .
8 10 5 5 .
6 10 10
x
6 .
9 10 10 [
H
] ( 0 .
290 ( 0 .
360
pH
9
x
)(
x
) .
16
x
) (
x
)( 0 .
290 ) ( 0 .
360 )