7.44. The ionization constant of phenol is 1.0 x 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 phenolate?
7.44. The ionization constant of phenol is 1.0 x 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 phenolate?
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1 Answer
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C6H5OH? C6H5O- + H+
C6H5OH
C6H5O-
H+
Initial
0.05 M
0
0
After dissociation
0.05 –x
x
x
Ka = x2 / (0.05 - x) = 1.0 x 10-10
=> x2 / 0.05 = 1.0 x 10-10
=> x2 = 5 x 10-12
=> x= 2.2 x 10-6 M
In presence of 0.01 C6H5Na, suppose y is the amount of phenol dissociated, then at equilibrium
[C6H5OH] = 0.05 – y ≈ 0.05,
[C6H5O-] = 0.01 + y ≈ 0.01 M, [H+] = yM
Therefore, Ka = (0.01) (y) / (0.05) = 1.0 x 10-10
=> y = 5 x 10-10
and α = y/c = (5 x 10-10) / (5 x 10-2)
= 10-8
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