Question

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A 0.10 mol/L solution of phenol has a pH of 5.43 at 25°C. What is Ka, for phenol?
What is the degree of ionization of the phenol in this solution?
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Answer 1

Explanation:

pH =-log[H+]

5.43=-log[H+]

[H+]= 5*10^-6

[H+]= C alpha

5*10^-6=10*10^-2 alpha

Alpha=5*10^-5

Ka=C (alpha) ^2

Ka=0.1*5*10^-5

Ka=5*10^-6

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