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2. What is the mass of KHP that you will weigh? a. The instructions read "Weigh out accurately 700 to 800 mg of KHP". Why are you given a range in the instructions and why is this acceptable? + is acceptable because KH P is the excess reactant and therefore, you only need to hayo enough moles of this compound but a spečific nunber or mole's.

2. What is the mass of KHP that you will weigh? a. The instructions read "Weigh out accurately 700 to 800 mg of KHP". Why are you given a range in the instructions and why is this acceptable? + is acceptable because KH P is the excess reactant and therefore, you only need to hayo enough moles of this compound but a spečific nunber or mole's.

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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Prelab: Before coming to lab: Make an outline of procedures in your lab notebook as described in the Laboratory Notebook Requirements at the beginning of the manual. Prepare a table like that shown in the Experimental Procedures section of the experiment to record your titration data. Use a straight edge. Instructor use only Complete this prelab assignment (questions 1-8) on these sheets in the spaces provided. Tear out at the beginning of the lab period and stack on your instructor's bench along with your notebook. 1. Define the following terms: a point where the stoichiometry of the equation is satistied. This means that eyoush moles of tritrant hgve becn ndded to the analyte, in order to completely reaot and satisfu the solution according to the indicator. a. Equivalence point titration where the tritiation çonplete. there is in b. End point – +he poinnt in a is complete. One particularity end point is thnt color of the solution accordingto'the indicator. cte is a change c. Molarity - A COnceyrtration ynit that relates the moles if solute and the volume of solution in of. liters, according to the followins moley Molarity-Volune equation d. Units for molarity –the units of molaritu are moles/Liters 45
Transcribed Image Text:Prelab: Before coming to lab: Make an outline of procedures in your lab notebook as described in the Laboratory Notebook Requirements at the beginning of the manual. Prepare a table like that shown in the Experimental Procedures section of the experiment to record your titration data. Use a straight edge. Instructor use only Complete this prelab assignment (questions 1-8) on these sheets in the spaces provided. Tear out at the beginning of the lab period and stack on your instructor's bench along with your notebook. 1. Define the following terms: a point where the stoichiometry of the equation is satistied. This means that eyoush moles of tritrant hgve becn ndded to the analyte, in order to completely reaot and satisfu the solution according to the indicator. a. Equivalence point titration where the tritiation çonplete. there is in b. End point – +he poinnt in a is complete. One particularity end point is thnt color of the solution accordingto'the indicator. cte is a change c. Molarity - A COnceyrtration ynit that relates the moles if solute and the volume of solution in of. liters, according to the followins moley Molarity-Volune equation d. Units for molarity –the units of molaritu are moles/Liters 45
of titrant that can be writHen accordint he butet, 2. What is the mass of KHP that you will weigh? a. The instructions read "Weigh out accurately 700 to 800 mg of KHP". Why are you given a range in the instructions and why is this acceptable? t is acceptable because KH P is the excess reactant and therefore, you only need to haye enough, moles of this compound but OV not a spečific number or mole's. Why is it important to not have any bubbles in the tip of the buret? H is jmgortautnot to have huy bubbles inthe tie. of the buret becawe, of herwise the yolvne of titrant 3. nuy actually added will be differevrt from the yolume itant that can be written accoraing Tuet to the butet. If there are bubbles in the tip the volume of titraut that, actva ly cts in contact with the analyte, will be lower than the recorded volume. 4. How many milliliters of 6.0 M NaOH must be used to prepare 1.0 L of 0.10 M NAOH? Show calculation with units. Given: MiVi = M2 V2 Vi = MaV- 0.10(mol)xIL VI= Mi M = 0.10M V2 = 1L V = 0-01aelale L Viにle.07 mL 5. A student titrates a 25.00 mL aliquot of HCl (concentration unknown) with 33.10 mL of 0.1500 M NaOH. What is the concentration of HCl? Show calculation with units. MIVI Ma V2 M2V20.1500(ml)x 33.10x102 MI= VI HCl+ Na DH -> He0+NaCl Given: 25x10-5L Mi =0.1906M V = 25 mL Ma= 0.1500M Vz=3,3.10ML=33.10x/DL Concentrstion of Hei=D0.1904M 46
Transcribed Image Text:of titrant that can be writHen accordint he butet, 2. What is the mass of KHP that you will weigh? a. The instructions read "Weigh out accurately 700 to 800 mg of KHP". Why are you given a range in the instructions and why is this acceptable? t is acceptable because KH P is the excess reactant and therefore, you only need to haye enough, moles of this compound but OV not a spečific number or mole's. Why is it important to not have any bubbles in the tip of the buret? H is jmgortautnot to have huy bubbles inthe tie. of the buret becawe, of herwise the yolvne of titrant 3. nuy actually added will be differevrt from the yolume itant that can be written accoraing Tuet to the butet. If there are bubbles in the tip the volume of titraut that, actva ly cts in contact with the analyte, will be lower than the recorded volume. 4. How many milliliters of 6.0 M NaOH must be used to prepare 1.0 L of 0.10 M NAOH? Show calculation with units. Given: MiVi = M2 V2 Vi = MaV- 0.10(mol)xIL VI= Mi M = 0.10M V2 = 1L V = 0-01aelale L Viにle.07 mL 5. A student titrates a 25.00 mL aliquot of HCl (concentration unknown) with 33.10 mL of 0.1500 M NaOH. What is the concentration of HCl? Show calculation with units. MIVI Ma V2 M2V20.1500(ml)x 33.10x102 MI= VI HCl+ Na DH -> He0+NaCl Given: 25x10-5L Mi =0.1906M V = 25 mL Ma= 0.1500M Vz=3,3.10ML=33.10x/DL Concentrstion of Hei=D0.1904M 46
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