Q. 53

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

Answer :

Given:

Concentration of acetic acid = 0.05M

pK_{a} = 4.74

As we know that, pk_{a} = -logK_{a}

We can also write,

∴ pK_{a} = -logK_{a}

⇒4.74 = -logK_{a}

By taking antilog of both the sides, we get

Antilog -4.74 = K_{a}

⇒K_{a} = 1.82 × 10-5

By applying the formula, k_{a} = cα^{2}

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

We can also write,

α =√

K_{a} = 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 =

K_{a} is the ionization constant

c is the concentration

K_{a} = 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 =

K_{a} is the ionization constant

c is the concentration

K_{a} = 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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