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# The Ionization Constant of Acetic Acid is 1.74 × 10–5. Calculate the Degree of Dissociation of Acetic Acid in Its 0.05 M Solution. Calculate the Concentration of Acetate Ion in the Solution and Its Ph. - CBSE (Science) Class 11 - Chemistry

ConceptIonization of Acids and Bases Ionization Constants of Weak Acids

#### Question

The ionization constant of acetic acid is 1.74 × 10–5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.

#### Solution 1

CH_3COOH ⇌ CH_3COO^- + H^+

K_a = ([CH_3COO^-][H^+])/[CH_3COOH] = ([H^+]^2)/[[CH_3COOH]]

or [H^+] =  sqrt(K_a[CH_3COOH]) = sqrt((1.74xx10^(-5))(5xx10^(-2))) = 9.33 xx 10^(-4) M

[CH_3COO^-] = [H^+] = 9.33 xx 10^(-4) M

pH = -log(9.33 xx 10^(-4)) = 4 - 0.9699 = 4 - 0.97 = 3.03

#### Solution 2

Method 1

1) CH_3COOH ↔ CH_3COO^(-) + H^(-)   K_a = 1.74 xx 10^(-5)

2) H_2O + H_2O ↔ H_3O^(+) +  OH^(-) K_w = 1.0 xx 10^(-14)

Since Ka >> Kw, :

CH_3COOH + H_2O ↔  CH_3COO^(-) +H_3O^(+)

C_i = 0.05                   0               0

0.05 - 05alpha          0.05alpha    0.05alpha

K_a =((.05alpha) (.05 alpha))/((.05 - 0.05alpha))

= ((.05alpha)(0.05alpha))/(.05(1-alpha))

= (.05alpha^2)/(1-alpha)

1.74 xx 10^(-5) = (0.05alpha^2)/(1-alpha)

1.74  xx 10^(-5) - 1.74 xx 10^(-5) alpha = 0.05alpha^2

0.05 alpha^2 + 1.74  xx 10^(-5) alpha - 1.74 xx 10^(-5)

D = b^2 - 4c

= (1.74 xx 10^(-5))^2 - 4(.05)(1.74 xx 10^(-5))

= 3.02 xx 10^(-23) + .348  xx 10^(-5)

alpha =  sqrt(K_a)/c

alpha = sqrt(((1.74 xx 10^(-5)))/.05

= sqrt((34.8 xx 10^(-5) xx 10)/10)

= sqrt(3.48 xx 10^(-6))

= CH_3COOH ↔ CH_3COO^(-) + H^(+)

= (0.93xx 10^(-3))/1000

= .000093

Method 2

Degree of dissociation,

alpha = sqrt((K_a)/c)

c = 0.05 M

Ka = 1.74 xx 10^(-5)

Then alpha = sqrt((1.74 xx 10^(-5))/.05)

alpha = sqrt(34.8 xx 10^(-5))

alpha = 1.8610^(-2)

CH_3COOH ↔ CH_3COO^(-) + H^(+)

Thus, concentration of CH3COO– = c.α

= .05 xx 1.86 xx 10^(-2)

=.093 xx 10^(-2)

= .00093  M

Since [oAc^(-)] = [H^+]

[H^+] = .00093 = .093 xx 10^(-2)

pH = -log[H^+]

= -log(.093 xx 10^(-2))

∴pH = 3.03

Hence, the concentration of acetate ion in the solution is 0.00093 M and its Ph is 3.03.

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#### APPEARS IN

Solution The Ionization Constant of Acetic Acid is 1.74 × 10–5. Calculate the Degree of Dissociation of Acetic Acid in Its 0.05 M Solution. Calculate the Concentration of Acetate Ion in the Solution and Its Ph. Concept: Ionization of Acids and Bases - Ionization Constants of Weak Acids.
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