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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

General, Organic, and Biological Chemistry
General, Organic, and Biological Chemistry
7th Edition
ISBN:9781285853918
Author:H. Stephen Stoker
Publisher:H. Stephen Stoker
Chapter10: Acids, Bases, And Salts
Section: Chapter Questions
Problem 10.118EP: Using the diagrams shown in Problem 10-117, which of the solutions would have the greatest buffer...
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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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