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thc +/- √b²-4ac 2a 4. 0.275 M CH3COOH; Ka = 1.8 x 10-5 Equation: EXPERIMENT 15 ACIDS AND BASES: pH, MOLARITY, K₁ AND Kb 1 × ICE Table: CH3COOH CH3 CDO + H+ 0.275 C E 0,275-x Ka= 0.275-X + X + X X 1.8x10-5 : 2 10-14/1.8x10-5 = 0.275 -x -5 4.95x10-6 1.8x10 5x - x² = 0 = k₁ the percent ionization of the acids in questions 1-4: [H*]: pH: POH: [OH-]: Kb: 5.55 × 10-10

thc +/- √b²-4ac 2a 4. 0.275 M CH3COOH; Ka = 1.8 x 10-5 Equation: EXPERIMENT 15 ACIDS AND BASES: pH, MOLARITY, K₁ AND Kb 1 × ICE Table: CH3COOH CH3 CDO + H+ 0.275 C E 0,275-x Ka= 0.275-X + X + X X 1.8x10-5 : 2 10-14/1.8x10-5 = 0.275 -x -5 4.95x10-6 1.8x10 5x - x² = 0 = k₁ the percent ionization of the acids in questions 1-4: [H*]: pH: POH: [OH-]: Kb: 5.55 × 10-10

Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter13: Acids And Bases
Section: Chapter Questions
Problem 94QAP: Consider a weak organic base (nonelectrolyte) with molar mass 281 g/mol. An aqueous solution of the...
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How do I solve question 4? This is a part of the non graded practice problems. Please help with a step by step solution.
quadratic : X = -b +/- √b²-4ac. Za EXPERIMENT 15 ACIDS AND BASES: pH, MOLARITY, Ka AND Kb 1×10-14/1.8x10-5 1.5 = kb 2 4.95x10-61.8x105x - x² = 0 4. 0.275 M CH3COOH; Ka = 1.8 x 10-5 Equation: ICE Table: CH3COOH CH3 C00 + H+ 0.275 C - X E 0,275-x + X 0.275 M HCOOH X + X X -5 1.8×10.5 = x 2 0.275 -x Ka=0.275-x 5. Calculate the percent ionization of the acids in questions 1-4: 0.275 M C6H5COOH [H*]: pH: POH: [OH-]: Kb: 5.55 x 10-10 M CH3COOH 0.275 M CH3COOH For Questions 6-9, refer to Table 15.1 for the "Situation" when each is applicable. 6. Three acids in 1-4 can be ranked based on pH. Rank them in order of increasing acidity (list the molarity and identity of the acid) and state the specific situation when this can be
Transcribed Image Text:quadratic : X = -b +/- √b²-4ac. Za EXPERIMENT 15 ACIDS AND BASES: pH, MOLARITY, Ka AND Kb 1×10-14/1.8x10-5 1.5 = kb 2 4.95x10-61.8x105x - x² = 0 4. 0.275 M CH3COOH; Ka = 1.8 x 10-5 Equation: ICE Table: CH3COOH CH3 C00 + H+ 0.275 C - X E 0,275-x + X 0.275 M HCOOH X + X X -5 1.8×10.5 = x 2 0.275 -x Ka=0.275-x 5. Calculate the percent ionization of the acids in questions 1-4: 0.275 M C6H5COOH [H*]: pH: POH: [OH-]: Kb: 5.55 x 10-10 M CH3COOH 0.275 M CH3COOH For Questions 6-9, refer to Table 15.1 for the "Situation" when each is applicable. 6. Three acids in 1-4 can be ranked based on pH. Rank them in order of increasing acidity (list the molarity and identity of the acid) and state the specific situation when this can be
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