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Concentration of standard NaOH (base) .200 M Sample V of acid V of base [HP] (X) Solubility (-X) [K*] (X or .5 +X) Ksp KHP + H20 4.00 ml 9.38 ml KHP + 5.00 ml 7.25 ml .50M KCI The Common lon Effect uses Le Châteler's Principle so less solid dissolves: KHP (s) K* + HP- More Calculate the % decrease in solubility = diff. in sol x 100 sol. In H20 %3D

Concentration of standard NaOH (base) .200 M Sample V of acid V of base [HP] (X) Solubility (-X) [K*] (X or .5 +X) Ksp KHP + H20 4.00 ml 9.38 ml KHP + 5.00 ml 7.25 ml .50M KCI The Common lon Effect uses Le Châteler's Principle so less solid dissolves: KHP (s) K* + HP- More Calculate the % decrease in solubility = diff. in sol x 100 sol. In H20 %3D

Chemistry
Chemistry
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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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Hello, can you please complete these problems and show all your work. Please complete the rest of the table on the second page and calculate the % decrease in solubility. Thank you.

The Solubility Product of KHP and the Common lon Effect Benzene Benzoic Acid Phthalic Acid Potassium Hydrogen Phthalate Acidic Hydrogen ОН H OH HO. HO H. Abbreviation = KHP The Reaction: KHP(s) K* + HP Ksp = [K*[HP] (X)( X) %3D The Chemistry: KHP(s) K* + HP- Start & Completion Change Equilibrium y -X ++ + How solve for X? X = HP is an acid. (-X = the solubiity) Will do an A-B titration. (H* + OH_H2O) Sample numbers: Acid-Base titration 4.00 ml KHP solution in flask "Phenolphthalein indicator" .200M NAOH in the buret 10.00 ml of NaOH added to reach the endpoint. molesacid = molesbase molesAXI = molesB XI 250ml End Point (MA)(VA) = (MB)(VB) (X)(4 ml) = (.2M)(10ml) X= .500M Overshot %3D %3D
Transcribed Image Text:The Solubility Product of KHP and the Common lon Effect Benzene Benzoic Acid Phthalic Acid Potassium Hydrogen Phthalate Acidic Hydrogen ОН H OH HO. HO H. Abbreviation = KHP The Reaction: KHP(s) K* + HP Ksp = [K*[HP] (X)( X) %3D The Chemistry: KHP(s) K* + HP- Start & Completion Change Equilibrium y -X ++ + How solve for X? X = HP is an acid. (-X = the solubiity) Will do an A-B titration. (H* + OH_H2O) Sample numbers: Acid-Base titration 4.00 ml KHP solution in flask "Phenolphthalein indicator" .200M NAOH in the buret 10.00 ml of NaOH added to reach the endpoint. molesacid = molesbase molesAXI = molesB XI 250ml End Point (MA)(VA) = (MB)(VB) (X)(4 ml) = (.2M)(10ml) X= .500M Overshot %3D %3D
The Common lon Effect KHP(s) is placed in a solution of .50OM KNO3. K* is the common ion. The Chemistry: KHP (s) EK* + HP- Start & Completion Change Equilibrium .5 y -X- + + .5 + X X Solve for X by doing an acid-base titration on this solution. Then, Ksp = [K*][HP] = (.5 +X)(X) %3D %3D Data and Calculations Concentration of standard NaOH (base) 200 M Sample V of acid V of base [HP] (X) Solubility (-X) [K*] (X or .5 +X) Ksp KHP + H20 4.00 ml 9.38 ml KHP + 5.00 ml 7.25 ml .50M KCI The Common lon Effect uses Le Châteler's Principle so less solid dissolves: KHP (s) K* + HP- -More Calculate the % decrease in solubility = diff. in sol x 100 sol. In H20
Transcribed Image Text:The Common lon Effect KHP(s) is placed in a solution of .50OM KNO3. K* is the common ion. The Chemistry: KHP (s) EK* + HP- Start & Completion Change Equilibrium .5 y -X- + + .5 + X X Solve for X by doing an acid-base titration on this solution. Then, Ksp = [K*][HP] = (.5 +X)(X) %3D %3D Data and Calculations Concentration of standard NaOH (base) 200 M Sample V of acid V of base [HP] (X) Solubility (-X) [K*] (X or .5 +X) Ksp KHP + H20 4.00 ml 9.38 ml KHP + 5.00 ml 7.25 ml .50M KCI The Common lon Effect uses Le Châteler's Principle so less solid dissolves: KHP (s) K* + HP- -More Calculate the % decrease in solubility = diff. in sol x 100 sol. In H20
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