Q. 53

# Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01M (b) 0.1M in HCl?

Given:

Concentration of acetic acid = 0.05M

pKa = 4.74

As we know that, pka = -logKa

We can also write,

pKa = -logKa

4.74 = -logKa

By taking antilog of both the sides, we get

Antilog -4.74 = Ka

Ka = 1.82 × 10-5

By applying the formula, ka = cα2

Where c is the concentration and α is the degree of ionization

We can also write,

α =√

Ka = 1.82 × 10-5(given)

C = 0.05M (given)

α =√

α = 1.908 × 10-2

In the presence of HCl due to the high concentration of H+ , the dissociation equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.

a) In the presence of HCl, let x be the amount of acetic acid dissociated after the addition of HCl.

Ionization of acetic acid

We can assume, 0.05-x≈ 0.05

0.01+x ≈ 0.01

To calculate the degree of ionization, we apply the formula

α=

Where α =

Ka is the ionization constant

c is the concentration

Ka = 1.82 × 10-5(given)

C = 0.01M (given)

α =

α =1.82 × 10-3

Thus, the degree of ionization in the presence of 0.01M HCl is 1.82 × 10-3

b) In the presence of 0.1 M HCl, let y be the amount of acetic acid dissociated after the addition of HCl.

Ionization of acetic acid

We can assume, 0.05-x≈ 0.05

0.1+x ≈ 0.1

To calculate the degree of ionization, we apply the formula

α=

Where α =

Ka is the ionization constant

c is the concentration

Ka = 1.82 × 10-5(given)

C = 0.1M (given)

α =

α =1.82 × 10-4

Thus, the degree of ionization in the presence of 0.1M HCl is 1.82 × 10-4

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