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Answer to Question #77232 in General Chemistry for eliza
Question #77232
Propanoic acid, CH3CH2COOH, has a pKa of 4.89 at 25 °C. Which of the following is the correct Ka?
Expert's answer
pKa = -lgKa
Ka = 10-pKa
Ka = 10-4.89 = 1.29·10-5
Ka = 10-pKa
Ka = 10-4.89 = 1.29·10-5
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