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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

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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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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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