Question

a buffer solution contain 0.1 mole per litres of acetic acid and 0.001 moles perlitre of sodium acetate.what will be its pH?(k=1.8×10-5)

Answer 1

Answer: The pH of the solution is 2.74.

Step-by-step explanation:

To calculate the pH of the acidic buffer, the equation for Henderson-Hasselbalch is used:

`pH=pK_a+\log (\frac{\text{[conjugate base]}}{\text{[acid]}})` .......(1)

The power of the acid dissociation constant is the negative logarithm of the acid dissociation constant. The equation for it is:

`pK_a=-log K_a` ......(2)

The chemical equation for the reaction of acetic acid and NaOH follows:

`CH_3COOH+NaOH\rightleftharpoons CH_3COONa+H-2O`

Given values:

`K_a=1.8* 10^(-5)`

Putting values in equation 2:

`pK_a=-log (1.8* 10^(-5))\\\\pK_a=4.74`

We are given:

`CH_3COOH` = 0.1 M

`CH_3COONa` = 0.001 M

Plugging values in equation 1:

`pH=4.74+\log ((0.001)/(0.1))\\\\pH=4.74-2\\\\pH=2.74`

Hence, the pH of the solution is 2.74.