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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
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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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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