TOPIC- ACIDS, BASES AND SALTS

                  SUB TOPIC- BUFFER SOLUTION

Objectives of subtopic- By the end of this subtopic you have to understand the following,

1. To know the concept of buffer solutions,

2. To describe the properties of buffer solution,

3. To know how to prepare buffer solution and

4. To know how to carry out calculation based on buffer solution

Before discussing the third objective, try to answer the following questions

i) By using examples, define the following terms

ii) What are the main components of buffer solution?

iii) Mention any two properties of buffer solution

iv) Identify buffer solutions from the list below.

a) Conjugate acid    b) Conjugate base   c) Weak acid    d)Weak base

1. 0.13 M sodium hydroxide + 0.27 M sodium bromide

2. 0.13 M nitrous acid + 0.14 M sodium nitrite

3. 0.24 M nitric acid + 0.17 M sodium nitrate

4. 0.31 M calcium chloride + 0.25 M calcium bromide

5. 0.34 M ammonia + 0.38 M ammonium bromide

PREPARATION OF BUFFER SOLUTION

The preparation of a buffer solution with a known pH is a two-step process.

1. A weak acid/conjugate base pair is chosen for which the weak acid pKa is within about

1 pH unit of the desired pH. The buffer is most effective when the ratio of component

concentration is close to 1, in which case pH: pKa of the acid.

2. The desired pH and the weak acid pKa are used to determine the relative concentrations of weak acid and conjugate base needed to give the desired pH.

Once the desired weak acid and conjugate base concentrations are known, the solution is

prepared in one of two ways,

1. Direct addition, where the correct amounts of the weak acid and conjugate base are added to water.

2. Acid-base reaction, where, for example, a conjugate base is created by reacting a weak

acid with enough strong base to produce a solution containing the correct weak acid and

conjugate base concentrations.

 EXAMPLE

1: Preparation of Buffers by Direct Addition

Describe how to prepare 500. mL of a buffer solution with pH = 9.85 using one of the weak acid/conjugate base systems shown below.

Solution

Step 1. Choose a weak acid/conjugate base pair. The bicarbonate ion/carbonate ion buffer

system is the best choice here because the desired pH is close to the pKa of the weak acid(Desired pH is 9.85 which is close to 10.32, pKa  of a weak acid)

Write the balanced equation for the acid hydrolysis reaction as follows

Step 2. Determine the necessary weak acid/conjugate base ratio using the rearranged  Ka

expression for the weak acid.)

Ka= [H3O+ to[H3O+(aq) ]= Ka[HCO3

expression

[H3O+(aq) ]/Ka =[HCO3   ratio of [HCO3

From the data given, pH =9.85 and Ka= 4.8 ×10

from this expression  pH =  -log[H3O+

Finding the ratio can be worked out as shown below.

(aq) ][ CO32-(aq]/ [HCO3-(aq)], then make [H3O+(aq) ]

 as the subject, this will results

(aq)]/ [ CO32-(aq], divide by Ka both sides, this will results to this

2-(aq], therefore the ratio of [H3O+(aq)]/ [ CO3-(aq)]/ [ CO3-2-(aq]–11

 the value of [H3O+

(aq) ]/Ka is equal to the (aq) ] can be obtained (aq) ]

Notice that the volume of buffer is cancelled in the ratio. The required amounts of weak acid

and conjugate base are independent of the solution volume, so the volume of a buffer has no effect on the buffer pH.

Step 3.

 Determine the amount of weak acid and conjugate base that must be combined to produce the buffer solution. Mixing 2.9 mol of HCO3–of this ratio) will result in a buffer with a pH of 9.85.and 1.0 mol of CO32– (or any multiple

Exercise

How would you prepare 10mL of a 0.01M phosphate buffer, pH 7.40, from stock solutions of  0.10M KH and 0.25M KPO24HPO? pKa of KH24PO= 7.20.24

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