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Example • Since only 0.09 mol of HBr were added to the solution, and there were (0.6M) × (1.0 L) = 0.6 mol of CH3CO2 already present, all the added HBr will be used up, leaving us with these species at equilibrium: HCH3CO, Na+ CH3CO2 Br H₂O Since at this point there's no strong acid, there's no need to consider Rule 4. Now you can complete the table. ANSWER 0.7 mol of NaOH is added to 1.0 L of a 0.7M HCH3CO2 solution. 0.09 mol of HBr is added to 1.0 L of a solution that is 0.6M in both HCH3CO2 and NaCH3CO2. acids: bases: CH3CO2 other: Na + acids: HCH3CO2 bases: CH3CO2 + other: Na+, Br ©2024 McGraw HilLLLC. All Rights Reserved. Terms of Use. Privacy Center Accessibilit The preparations of two aqueous solutions are described in the table below. For each solution, write the chemical formulas of the major species present at equilibrium. You can leave out water itself. Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases' row, and the formulas of the species that will act as neither acids nor bases in the 'other' row. You will find it useful to keep in mind that NH3 is a weak base. acids: H', , Cl, NH 0.7 mol of HCl is added to 1.0 L of a 0.4M NH3 solution. bases: ☐ 0.46 mol of HNO3 is added to 1.0 L of a solution that is 1.0M in both NH3 and Try again other: ☐ acids: bases: NO3, Br, NH other: ☐ 4 0,0,... ☑ ⑤

Example • Since only 0.09 mol of HBr were added to the solution, and there were (0.6M) × (1.0 L) = 0.6 mol of CH3CO2 already present, all the added HBr will be used up, leaving us with these species at equilibrium: HCH3CO, Na+ CH3CO2 Br H₂O Since at this point there's no strong acid, there's no need to consider Rule 4. Now you can complete the table. ANSWER 0.7 mol of NaOH is added to 1.0 L of a 0.7M HCH3CO2 solution. 0.09 mol of HBr is added to 1.0 L of a solution that is 0.6M in both HCH3CO2 and NaCH3CO2. acids: bases: CH3CO2 other: Na + acids: HCH3CO2 bases: CH3CO2 + other: Na+, Br ©2024 McGraw HilLLLC. All Rights Reserved. Terms of Use. Privacy Center Accessibilit The preparations of two aqueous solutions are described in the table below. For each solution, write the chemical formulas of the major species present at equilibrium. You can leave out water itself. Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases' row, and the formulas of the species that will act as neither acids nor bases in the 'other' row. You will find it useful to keep in mind that NH3 is a weak base. acids: H', , Cl, NH 0.7 mol of HCl is added to 1.0 L of a 0.4M NH3 solution. bases: ☐ 0.46 mol of HNO3 is added to 1.0 L of a solution that is 1.0M in both NH3 and Try again other: ☐ acids: bases: NO3, Br, NH other: ☐ 4 0,0,... ☑ ⑤

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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Example • Since only 0.09 mol of HBr were added to the solution, and there were (0.6M) × (1.0 L) = 0.6 mol of CH3CO2 already present, all the added HBr will be used up, leaving us with these species at equilibrium: HCH3CO, Na+ CH3CO2 Br H₂O Since at this point there's no strong acid, there's no need to consider Rule 4. Now you can complete the table. ANSWER 0.7 mol of NaOH is added to 1.0 L of a 0.7M HCH3CO2 solution. 0.09 mol of HBr is added to 1.0 L of a solution that is 0.6M in both HCH3CO2 and NaCH3CO2. acids: bases: CH3CO2 other: Na + acids: HCH3CO2 bases: CH3CO2 + other: Na+, Br ©2024 McGraw HilLLLC. All Rights Reserved. Terms of Use. Privacy Center Accessibilit
Transcribed Image Text:Example • Since only 0.09 mol of HBr were added to the solution, and there were (0.6M) × (1.0 L) = 0.6 mol of CH3CO2 already present, all the added HBr will be used up, leaving us with these species at equilibrium: HCH3CO, Na+ CH3CO2 Br H₂O Since at this point there's no strong acid, there's no need to consider Rule 4. Now you can complete the table. ANSWER 0.7 mol of NaOH is added to 1.0 L of a 0.7M HCH3CO2 solution. 0.09 mol of HBr is added to 1.0 L of a solution that is 0.6M in both HCH3CO2 and NaCH3CO2. acids: bases: CH3CO2 other: Na + acids: HCH3CO2 bases: CH3CO2 + other: Na+, Br ©2024 McGraw HilLLLC. All Rights Reserved. Terms of Use. Privacy Center Accessibilit
The preparations of two aqueous solutions are described in the table below. For each solution, write the chemical formulas of the major species present at equilibrium. You can leave out water itself. Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases' row, and the formulas of the species that will act as neither acids nor bases in the 'other' row. You will find it useful to keep in mind that NH3 is a weak base. acids: H', , Cl, NH 0.7 mol of HCl is added to 1.0 L of a 0.4M NH3 solution. bases: ☐ 0.46 mol of HNO3 is added to 1.0 L of a solution that is 1.0M in both NH3 and Try again other: ☐ acids: bases: NO3, Br, NH other: ☐ 4 0,0,... ☑ ⑤
Transcribed Image Text:The preparations of two aqueous solutions are described in the table below. For each solution, write the chemical formulas of the major species present at equilibrium. You can leave out water itself. Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases' row, and the formulas of the species that will act as neither acids nor bases in the 'other' row. You will find it useful to keep in mind that NH3 is a weak base. acids: H', , Cl, NH 0.7 mol of HCl is added to 1.0 L of a 0.4M NH3 solution. bases: ☐ 0.46 mol of HNO3 is added to 1.0 L of a solution that is 1.0M in both NH3 and Try again other: ☐ acids: bases: NO3, Br, NH other: ☐ 4 0,0,... ☑ ⑤
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