HOW DOES A BUFFER MAINTAIN PH?, Schemes and Mind Maps of Chemistry

The buffer capacity is the amount of acid or base that can be added before the pH begins to change significantly. It can be.

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1Updated 2/16/2019
HOW DOES A BUFFER MAINTAIN PH?
A buffer is a special solution that stops massive changes in pH levels. Every buffer that is made has a certain buffer capacity, and
buffer range. The buffer capacity is the amount of acid or base that can be added before the pH begins to change significantly. It can be
also defined as the quantity of strong acid or base that must be added to change the pH of one liter of solution by one pH unit. The
buffer range is the pH range where a buffer effectively neutralizes added acids and bases, while maintaining a relatively constant pH.
INTRODUCTION
The equation for pH also shows why pH does not change by much in buffers.
Where,
is the concentration of the conjugate base
is the concentration of the acid
When the ratio between the conjugate base/ acid is equal to 1, the pH = pK . If the ratio between the two is 0.10, the pH drops by 1 unit
from pK since log (0.10) = -1. If a ratio increases to a value of 10, then the pH increases by 1 unit since \log (10) = 1. The buffer
capacity has a range of about 2. This means when a buffer is created, the pH can be changed by -1 by acid or +1 by base before the pH
begins to change substantially. After the addition of base to raise the pH by 1 or more, most of the conjugate acid will have been
depleted to try to maintain a certain pH, so the pH will be free to increase faster without the restraint of the conjugate acid. The same
goes for the addition of acid, once the conjugate base has been used up, the pH will drop faster since most of the conjugate base has
been used up.
EXAMPLE 1
What is the effect on the pH of adding 0.006 mol HCL to 0.3L of a buffer solution that is 0.250M HC H O and 0.560 M
NaC H O ? pK = 4.74
Calculate the starting amount of C H O
Calculate the starting amount of HC H O
=
K
a
[ ][ ]
H
+
A
[
HA
]
(1)
pH
=
p
+ log
K
a
[ ]
A
[
HA
]
(2)
A
HA
a
a
232
2 3 2 a
pH
= 4.74+ log = 4.74+ 0.35= 5.09
0.560
0.250
(1)
+
H
+
O
C
2
H
3
O
2
H
3
O
+
C
2
H
3
O
2
H
2
(2)
232-
0.300
L
× 0.560
M
= 0.168
mol
C
2
H
3
O
2
(3)
232
pf3

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HOW DOES A BUFFER MAINTAIN PH?

A buffer is a special solution that stops massive changes in pH levels. Every buffer that is made has a certain buffer capacity, and buffer range. The buffer capacity is the amount of acid or base that can be added before the pH begins to change significantly. It can be also defined as the quantity of strong acid or base that must be added to change the pH of one liter of solution by one pH unit. The buffer range is the pH range where a buffer effectively neutralizes added acids and bases, while maintaining a relatively constant pH.

INTRODUCTION

The equation for pH also shows why pH does not change by much in buffers.

Where,

is the concentration of the conjugate base is the concentration of the acid

When the ratio between the conjugate base/ acid is equal to 1, the pH = pK. If the ratio between the two is 0.10, the pH drops by 1 unit from pK since log (0.10) = -1. If a ratio increases to a value of 10, then the pH increases by 1 unit since \log (10) = 1. The buffer capacity has a range of about 2. This means when a buffer is created, the pH can be changed by -1 by acid or +1 by base before the pH begins to change substantially. After the addition of base to raise the pH by 1 or more, most of the conjugate acid will have been depleted to try to maintain a certain pH, so the pH will be free to increase faster without the restraint of the conjugate acid. The same goes for the addition of acid, once the conjugate base has been used up, the pH will drop faster since most of the conjugate base has been used up.

EXAMPLE 1

What is the effect on the pH of adding 0.006 mol HCL to 0.3L of a buffer solution that is 0.250M HC H O and 0.560 M NaC H O? pK = 4.

Calculate the starting amount of C H O

Calculate the starting amount of HC H O

Ka =

[ H +^ ][ A −]

[ HA ]

pH = p Ka + log

[ A −]

[ HA ]

A −

HA

a a

2 3 2 2 3 2 a

pH = 4.74 + log = 4.74 + 0.35 = 5.

C 2 H 3 O − 2 + H 3 O +^ ⇌ H C 2 H 3 O 2 + H 2 O (2)

2 3 2 -

0.300 L × 0.560 M = 0.168 mol C 2 H 3 O − 2 (3)

2 3 2

C H O H O HC H O

Original Buffer 0.168 mol 0.075 mol

Add 0.006 mol

Change -0.006 mol -0.006 mol +0.006 mol

Final Amount 0.162 mol 0.081 mol

Now calculate the new concentrations of C H O and HC H O :

Using the new concentrations, we can calculate the new pH:

Calculate the pH change:

Therefore, the pH dropped by 0.05 pH units.

EXAMPLE 2

What is the effect on the pH of adding 0.006 mol NaOH to 0.3L of a buffer solution that is 0.250M HC H O and 0.560 M NaC H O? pK = 4.

Calculate the starting amount of HC H O

Calculate the starting amount of C H O

0.300 L × 0.250 M = 0.075 mol HC 2 H 3 O 2 (4)

2 3 2 -^3 -^2 3

2 3 2 -^2 3

= 0.540 M 0.162 mol 0.300 L

C 2 H 3 O − 2 (5)

= 0.540 M H

0.081 mol 0.300 L

C 2 H 3 O 2 (6)

pH = 4.74 + log = 4.74 + 0.30 = 5.

p Hfinalp Hinitial = 5.04 − 5.09 = −0.05 (8)

2 3 2 2 3 2 a

pH = 4.74 + log = 4.74 + 0.35 = 5.

H C 2 H 3 O 2 + O H −^ ⇌ C 2 H 3 O − 2 + H 2 O (10)

2 3 2

0.300 L × 0.250 M = 0.075 mol HC 2 H 3 O 2 (11)

2 3 2 -

0.300 L × 0.560 M = 0.168 mol C 2 H 3 O − 2 (12)