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# The Ionization Constant of Phenol is 1.0 × 10–10. What is the Concentration of Phenolate Ion in 0.05 M Solution of Phenol? What Will Be Its Degree of Ionization If the Solution is Also 0.01m in Sodium Phenolate? - CBSE (Science) Class 11 - Chemistry

ConceptIonization of Acids and Bases Ionization Constants of Weak Acids

#### Question

The ionization constant of phenol is 1.0 × 10–10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate?

#### Solution 1

Ionization of phenol:

C_6H_5OH + H_2O  ↔C_6H_5O^(-) + H_3O^(+)

Initial conc                    0.05                             0               0

At equilibrium           0.05 - x                           x                x

K_a = ([C_6H_5O^(-)][H_3O^(+)])/[C_6H_5OH]

K_a = (x xx x)/(0.05 - x)

As the value of the ionization constant is very less, x  will be very small. Thus we can ignore x in the denominator

:. x = sqrt(1xx10^(-10) xx 0.05)

= sqrt(5xx10^(-12))

= 2.2 xx 10^(-6) M = [H_3O^(+)]

Since {H_3O^+] = [C_6H_5O^(-)]

[C+6H_5O^(-)] = 2.2 xx 10^(-6) M

Now, let ∝ be the degree of ionization of phenol in the presence of 0.01 M C6H5ONa.

C_6H_5ONa -> C_6H_5O^(-) + Na^(+)

Conc                               0.01

Also

C_6H_5OH   + H_2O ↔ C_6H_5O^(-) + H_3O^(+)

Conc   0.05-0.05alpha                    0.05alpha     0.05alpha

[C_6H_5OH] = 0.05 - 0.05alpha ; 0.05 M

[C_6H_5O^(-)] = 0.01 + 0.05 alpha 0.01 M

H_3O^+] = 0.05 alpha

K_a = ([C_6H_5O^(-)][H_3O^(+)])/[C_6H_5OH]

K_a = ((0.01)(0.05alpha))/0.05

1.0 xx 10^(-10) = .01 alpha

alpha = 1 xx 10^(-8)

#### Solution 2

C_6H_5OH ⇌ C_6H_5O^(-) + H^(+)

Initial                  0.05 N

After disso         0.05-x          xx

:. K_a = (x xx x)/(0.05 - x) = 1.0 xx 10^(-10) (Given) or x^2/0.05 = 1.0 xx   10^(-10)

or x^2 = 5 xx 10^(-12) or x = 2.2 xx 10^(-6) M

In presence of 0.01 C_6H_5ONa suppose is the amount of phenol dissociate, then at equilibrium

[C_6H_5OH] = 0.05 - y ~~ 0.05

[C_6H_5O^(-)] = 0.01 + y ~~ 0.01 M, [H^+] = yM

:. K_a = ((0.01)(y))/0.05 = 1.0 xx 10^(-10) Given or y = 5 xx 10^(-10)

:. alpha = y/c = (5xx10^(-10))/(5xx10^(-2)) = 10^(-8)

Is there an error in this question or solution?

#### APPEARS IN

Solution The Ionization Constant of Phenol is 1.0 × 10–10. What is the Concentration of Phenolate Ion in 0.05 M Solution of Phenol? What Will Be Its Degree of Ionization If the Solution is Also 0.01m in Sodium Phenolate? Concept: Ionization of Acids and Bases - Ionization Constants of Weak Acids.
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