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Whelan Whelan
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Posts: 3158
6 years ago
Phenobarbital is an antiepileptic drug with a water solubility of 4.3 × 10-3 M and pKa = 7.4. What is the pH and percent ionization of 4.3 × 10-3 M phenobarbital?
Textbook 

Introductory Chemistry Essentials


Edition: 5th
Author:
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hbutler2hbutler2
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6 years ago
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More questions for this book are available here
  4.88, 0.30%

GIVEN P^Ka = 7.4

           -logKa = 7.4

             Ka = 10^-7.4

          Ka = 3.98*10^-8

P^H = (1/2)P^Ka -(1/2) log C

      = (1/2)(7.4) -(1/2)log(4.3*10^-3)

     = 4.88

pecent of ionisation - (Ka/C)^1/2 *100

                            = (3.98*10^-8/4.3*10^-3)^1/2*100

                          = 0.30%
I'm a chemical engineer! Slight Smile

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