9) An unknown weak acid is titrated to its equivalence point using 10 mL of 1M NaOH. At the equivalence point, the pH of the 100 mL solution is 9. What is the pKa of the acid? Knowing that, what was the pH of the original 90 mL solution? mol NaOH = (10ML) (IM) = 10 mmol [x-] = 10mmol 100mL = 0.1M pka: pH
9) An unknown weak acid is titrated to its equivalence point using 10 mL of 1M NaOH. At the equivalence point, the pH of the 100 mL solution is 9. What is the pKa of the acid? Knowing that, what was the pH of the original 90 mL solution? mol NaOH = (10ML) (IM) = 10 mmol [x-] = 10mmol 100mL = 0.1M pka: pH
Chemistry: An Atoms First Approach
2nd Edition
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Steven S. Zumdahl, Susan A. Zumdahl
Chapter14: Acid- Base Equilibria
Section: Chapter Questions
Problem 110CP: A 0.400-M solution of ammonia was titrated with hydrochloric acid to the equivalence point, where...
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