CHEM 102 1nd Edition Lecture 23Outline of Last Lecture I. ArrheniusII. Bronsted-LowryIII. Dissociation vs IonizationIV. Hydronium IonsV. Strong and Weak AcidsVI. Strong and Weak BasesVII. Strong ElectrolytesVIII. Acid and Bond StrengthIX. Binary AcidsX. Acid Strength and KOutline of Current Lecture I. pH and pOHII. KaIII. KbIV. Determining pH and pOHV. Determining % IonizationCurrent LectureI. PH and pOH - PH and pOH add up to be 14- PH = log[H]- pOH = log[OH]II. Ka- the equilibrium constant for the dissociation of an acid- or for the acid hydrolysis reaction- hydrolysis is the reaction of a substance with water- Ka = [products]/[reactants]- The larger the Ka, the greater the ionization- The stronger the acid the larger [H]- The greater the [H], the smaller the pH and pKaIII. KbThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.- The quilibrium constant for the dissociation of a base- Or for the base hydrolysis reaction- Kb = [products]/[reactants]IV. Determine the pH and pOH- What is the pH and the pOH of a 0.050 M HNO3 solution HNO3 + H20 H3O+ + NO3- Initial 0.050 Change-0.50 0.050 0.050 –log(0.050) = -(-1.30) = 1.30 14 – pH = 14 – 1.30 = 12.7 = pOH- What is the pH of a 0.10 M HC2H3O2; Ka = 1.8e-5 HAc + H20 H3O+ + Ac- 0.10 -x +x +x 0.10-x x x Ka = 1.e-5 = [H+][Ac-]/[HAc] = x x/ (10-x) = 1.8e-5 1.8e-5 = x x /10 X = 1.34 e-3 pH = - log[H] pH = - log(1.34e-3) = 2.87- What is the pH of a 0.10 M NaOH solution [OH] = 0.10M pOH = -log(0.10) = 1 pH = 14 – pOH = 14 – 1 = 13V. Determining % Ionization- % ionization = concentration ionized/original concentration x 100- 3.9e-5/1.0M x 100 -
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