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.

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:

$K_a=1.8\times 10^{-5}$

Putting values in equation 2:

$pK_a=-log (1.8\times 10^{-5})\\\\pK_a=4.74$

We are given:

$CH_3COOH$ = 0.1 M

$CH_3COONa$ = 0.001 M

Plugging values in equation 1:

$pH=4.74+\log (\frac{0.001}{0.1})\\\\pH=4.74-2\\\\pH=2.74$

Hence, the pH of the solution is 2.74.