Buffer Solution Formula

Buffer Solution

A buffer solution is an aqueous solution consisting of a weak acid and its conjugated base mixture or vice verse. There is a minute change in its pH when a little or medium amount of strong base or acid is added to it and that is why it is used to avoid changes in the pH of solution.

The pH value of the water solvent is 7, but if we add a few drops of HCl or NaOH solution, its pH decreases or increases respectively. Therefore it is necessary to have solutions whose pH does not change even on the addition of strong alkalies or strong acids. Such solutions are called buffer solutions.

Buffer capacity is the capacity of a buffer solution to resist change in its pH. The equation is given by,

pH = pKa + log [Salt] / [Acid]

The pH of any acidic buffer solution is always less than 7 and the pH of any basic buffer solution is always greater than 7.

Solved Examples

Example 1

Determine the ratio of concentrations of formate ion and formic acid in a buffer solution so that its pH is required to be 4? Identify the pH of this buffer to have the maximum buffer capacity? Ka of formic acid is 1.8 × 10-4.

Solution

We have the equation

pH = pKa + log[Salt] / [Acid]

4 = – log(1.8 × 10-4) + log[Formate] / [Formic acid]

4 = 3.74 + log [Formate] / [Formic acid]

log[Formate] / [Formic acid] = 4 – 3.74 = 0.26

[Formate] / [Formic acid]= 1.8

The buffer capacity would be maximum near the pKa of the acid.

For maximum buffer capacity

pH = pKa = -log Ka

= -log (1.8 × 10-4)

= 3.74

buffer capacity = 3.74

Example 2

Calculate the volume of 0.2M solution of acetic acid that needs to be added to 100 ml of 0.2M solution of sodium acetate to obtain a buffer solution of pH 5.00. pKa of acetic acid is 4.74.

Solution

We have the equation,

pH = pKa + log[Salt] / [Acid]

log[Salt] / [Acid] = pH – pKa

= 5 – 4.74

= 0.26

[Salt] / [Acid] = 1.82

Number of moles of sodium acetate

= 100×0.2 / 1000

= 0.02 mol

Let volume of 0.2M acetic acid added = V mL

Number of moles of acetic acid = V×0.2 / 1000

0.02 / V× 0.2/1000 = 1.82

Therefore,

V = 0.02×1000 / 0.2×1.82

= 55 mL

55 mL of acetic acid is added

Stay tuned with BYJU’S for more such interesting articles.

Leave a Comment

Your email address will not be published. Required fields are marked *