Buffer Solution Formula

 

Buffer solution formula

 

A buffer solution is an aqueous solution comprising a mixture of a weak acid and its conjugate base, or vice versa. There is a minute change in its pH when a small or moderate amount of strong acid or base is added to it and that’s the reason it is used to prevent changes in the pH of a solution.

The pH value of 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 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.

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 = -logKa

= -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 100ml 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

 

 


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When Sulphur is burnt in air, the major product is Sulphur trioxide – True or False?