Question

Calculate the degree of dissociation of acetic acid in its 0.1 m solution

Answer 1

Answer:

Answer

\displaystyle pK_a = -log K_a = 4.74pK

a

=−logK

a

=4.74

\displaystyle K_a = 10^{-pKa} = 10^{-4.74}= 1.8 \times 10^{-5}K

a

=10

−pKa

=10

−4.74

=1.8×10

−5

Let x be the degree of dissociation. The concentration of acetic acid solution, C = 0.05 M

The degree of dissociation, \displaystyle x = \sqrt {\frac {K_a}{C}} = \sqrt {\frac {1.8 \times 10^{-5}}{0.05}} = 0.019x=

C

K

a

=

0.05

1.8×10

−5

=0.019

(a) The solution is also 0.01 M in HCl.

Let x M be the hydrogen ion concentration from ionization of acetic acid. The hydrogen ion concentration from ionization of HCl is 0.01 M. The total hydrogen ion concentration

\displaystyle [H^+] = 0.01+x[H

+

] =0.01+x

The acetate ion concentration is equal to the hydrogen ion concentration from ionization of acetic acid. This is also equal to the concentration of acetic acid that has dissociated.

\displaystyle [CH_3COO^-] = x[CH

3

COO

−

]=x

\displaystyle [CH_3COOH] = 0.05-x[CH

3

COOH]=0.05−x

\displaystyle K_a = \frac {[H^+][CH_3COO^-]}{[CH_3COOH]}K

a

=

[CH

3

COOH]

[H

+

][CH

3

COO

−

]

\displaystyle 1.8 \times 10^{-5} = \frac {(0.01+x)x}{(0.05-x)}1.8×10

−5

=

(0.05−x)

(0.01+x)x

.....(1)

As x is very small, \displaystyle 0.01+x \approx 0.010.01+x≈0.01

\displaystyle 0.05-x \approx 0.050.05−x≈0.05

Hence, the equation (i) becomes

\displaystyle 1.8 \times 10^{-5} = \frac {0.01x}{0.05}1.8×10

−5

=

0.05

0.01x

\displaystyle x = 9.0 \times 10^{-5} Mx=9.0×10

−5

M

The degree of ionization is \displaystyle \dfrac {[CH_3COO^-]}{[CH_3COOH]}= \frac {x}{c} = \frac {9.0 \times 10^{-5}}{0.05} = 1.8 \times 10^{-3} = 0.0018

[CH

3

COOH]

[CH

3

COO

−

]

=

c

x

=

0.05

9.0×10

−5

=1.8×10

−3

=0.0018.

(b) The solution is also 0.1 M in HCl.

Let x M be the hydrogen ion concentration from ionization of acetic acid. The hydrogen ion concentration from ionization of HCl is 0.01 M. The total hydrogen ion concentration

\displaystyle [H^+] = 0.1+x[H

+

] =0.1+x

The acetate ion concentration is equal to the hydrogen ion concentration from ionization of acetic acid. This is also equal to the concentration of acetic acid that has dissociated.

\displaystyle [CH_3COO^-] = x[CH

3

COO

−

]=x

\displaystyle [CH_3COOH] = 0.05-x[CH

3

COOH]=0.05−x

\displaystyle K_a = \frac {[H^+][CH_3COO^-]}{[CH_3COOH]}K

a

=

[CH

3

COOH]

[H

+

][CH

3

COO

−

]

\displaystyle 1.8 \times 10^{-5} = \frac {(0.1+x)x}{(0.05-x)}1.8×10

−5

=

(0.05−x)

(0.1+x)x

.....(1)

As x is very small, \displaystyle 0.1+x \approx 0.10.1+x≈0.1

\displaystyle 0.05-x \approx 0.050.05−x≈0.05

Hence, the equation (i) becomes

\displaystyle 1.8 \times 10^{-5} = \frac {0.1x}{0.05}1.8×10

−5

=

0.05

0.1x

\displaystyle x = 9.0 \times 10^{-6} Mx=9.0×10

−6

M

The degree of ionization is \displaystyle \dfrac {[CH_3COO^-]}{[CH_3COOH]}= \frac {x}{c} = \frac {9.0 \times 10^{-6}}{0.05} = 1.8 \times 10^{-4} = 0.00018

[CH

3

COOH]

[CH

3

COO

−

]

=

c

x

=

0.05

9.0×10

−6

=1.8×10

−4

=0.00018

Answer 2

Given :

Molarity of acetic acid , C = 0.1 M .

To Find :

Degree of dissociation of acetic acid .

Solution :

We know , degree of dissociation of weak acid is given by :

$\alpha=\sqrt{\dfrac{K_a}{C}}$

Here , $K_a$ is equilibrium constant .

$K_a$ for acetic acid is $1.8 \times 10^{-5}$ .

Putting it in above equation .

We get :

$\alpha=\sqrt{\dfrac{1.8 \times 10^{-5}}{0.1}}$

$\alpha=1.34 \times 10^{-2}$

Therefore ,  the degree of dissociation of acetic acid in its 0.1 M solution

is $\alpha=1.34 \times 10^{-2}$ .

Learn More :

Equilibrium

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