Another sample problem: calculate the pH of 1.0x10-8 M KOH. Answer: -log[OH-] = 8 pOH pH = 14pOH = 6 Wait...is it even possible for a KOH solution to have pH = 6? The autoionization of water can no longer be neglect when the electrolyte's concentration is low (-10-7). . == Answer: use the systematic treatment of equilibrium [H][OH] = Kw [K] + [H] = [OH] [K] 1.0 × 108 M = Eq1 Eq2 Eq3 ī 125 11- 12 3 KOH Hör -2-3455 Systematic treatment required here

Another sample problem: calculate the pH of 1.0x10-8 M KOH. Answer: -log[OH-] = 8 pOH pH = 14pOH = 6 Wait...is it even possible for a KOH solution to have pH = 6? The autoionization of water can no longer be neglect when the electrolyte's concentration is low (-10-7). . == Answer: use the systematic treatment of equilibrium [H][OH] = Kw [K] + [H] = [OH] [K] 1.0 × 108 M = Eq1 Eq2 Eq3 ī 125 11- 12 3 KOH Hör -2-3455 Systematic treatment required here

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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