Question 44 of 69intermediate🔧 ApplyNumerical3 marks

The ionization constant of phenol is 1.0×10−101.0 \times 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 0.01 M in sodium phenol?

Correct Answer

(i) Concentration of phenolate ion = 2.24×10−62.24 \times 10^{-6} M

(ii) Degree of ionization = 1.0×10−81.0 \times 10^{-8}

Exercise: EXERCISES | Q: 6.44 | (Page 45)
For More Understanding

Explanation

The ionization of phenol is calculated using its ionization constant KaK_a. In a pure solution, the approximation Ka=cα2K_a = c\alpha^2 is used. When a common ion (phenolate from sodium phenolate) is added, the common ion effect suppresses the ionization, and the degree of ionization is calculated using Ka=[C6H5O−][H+][C6H5OH]≈[C6H5O−]⋅cαcK_a = \frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]} \approx \frac{[C_6H_5O^-] \cdot c\alpha}{c}.

Solution Steps

  1. Part 1: Concentration of phenolate ion in 0.05 M phenol Given: Ka=1.0×10−10K_a = 1.0 \times 10^{-10}, c=0.05c = 0.05 M For phenol C6H5OH⇌C6H5O−+H+C_6H_5OH \rightleftharpoons C_6H_5O^- + H^+

    Ka=cα2K_a = c\alpha^2 1.0×10−10=0.05×α21.0 \times 10^{-10} = 0.05 \times \alpha^2 α2=1.0×10−100.05=2×10−9\alpha^2 = \frac{1.0 \times 10^{-10}}{0.05} = 2 \times 10^{-9} α=20×10−10=4.47×10−5\alpha = \sqrt{20 \times 10^{-10}} = 4.47 \times 10^{-5}

    [C6H5O−]=cα=0.05×4.47×10−5=2.235×10−6[C_6H_5O^-] = c\alpha = 0.05 \times 4.47 \times 10^{-5} = 2.235 \times 10^{-6} M

  2. Part 2: Degree of ionization in presence of 0.01 M sodium phenolate Sodium phenolate provides common ion C6H5O−C_6H_5O^-, so [C6H5O−]=0.01[C_6H_5O^-] = 0.01 M

    Ka=[C6H5O−][H+][C6H5OH]K_a = \frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}

    Since [H+]=cα[H^+] = c\alpha and [C6H5OH]≈c=0.05[C_6H_5OH] \approx c = 0.05 M

    1.0×10−10=0.01×(0.05×α)0.051.0 \times 10^{-10} = \frac{0.01 \times (0.05 \times \alpha)}{0.05} 1.0×10−10=0.01×α1.0 \times 10^{-10} = 0.01 \times \alpha α=1.0×10−100.01=1.0×10−8\alpha = \frac{1.0 \times 10^{-10}}{0.01} = 1.0 \times 10^{-8}
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