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MISSED THIS? Read Section 18.4 (Page); Watch KCV 18.48. Consider the curve shown here for the titration of a weak monoprotic acid with a strong base. Hd 14 12 10- 6- 4 2- 0 0 40 80 120 160 Volume of base added (mL) The figure shows pH as a function of volume of base added. The pH is measured from 0 to 14 on the y-axis, while the volume of base added is measured from 0 to 160 milliliters on the x-axis. The curve of the plot goes up gradually from pH 2.3 at 0 milliliters to pH 5.8 at 49 milliliters, next the plot goes up steeply to pH 11.8 at 51 milliliters, and finally the plot goes up gradually to pH 12.2 at 160 milliliters. Part D Correct An equilibrium problem based on the initial concentration and Ka of a weak acid should be solved f aqueous solution of the weak acid. This is true for the initial state of titration, when no base has bee addition of a small amount of a strong base turns this solution into a buffer whose pH is calculated way. At what volume of added base does pH = pK₂? Express your answer in milliliters as an integer. V = 30 mL Submit ✓ Part E V = Correct The point at which pH = pK (the half-equivalence point) corresponds to the state when the buffer components are exactly equal. This point is reached halfway to the equivalence point, na 60 mL/2= 30 mL. Previous Answers At what volume of added base is the pH calculated by working an equilibrium problem based on the Kb of the conjugate base? Express your answer in milliliters as an integer. 15| ΑΣΦ Submit G d Previous Answers Request Answer ? mL

MISSED THIS? Read Section 18.4 (Page); Watch KCV 18.48. Consider the curve shown here for the titration of a weak monoprotic acid with a strong base. Hd 14 12 10- 6- 4 2- 0 0 40 80 120 160 Volume of base added (mL) The figure shows pH as a function of volume of base added. The pH is measured from 0 to 14 on the y-axis, while the volume of base added is measured from 0 to 160 milliliters on the x-axis. The curve of the plot goes up gradually from pH 2.3 at 0 milliliters to pH 5.8 at 49 milliliters, next the plot goes up steeply to pH 11.8 at 51 milliliters, and finally the plot goes up gradually to pH 12.2 at 160 milliliters. Part D Correct An equilibrium problem based on the initial concentration and Ka of a weak acid should be solved f aqueous solution of the weak acid. This is true for the initial state of titration, when no base has bee addition of a small amount of a strong base turns this solution into a buffer whose pH is calculated way. At what volume of added base does pH = pK₂? Express your answer in milliliters as an integer. V = 30 mL Submit ✓ Part E V = Correct The point at which pH = pK (the half-equivalence point) corresponds to the state when the buffer components are exactly equal. This point is reached halfway to the equivalence point, na 60 mL/2= 30 mL. Previous Answers At what volume of added base is the pH calculated by working an equilibrium problem based on the Kb of the conjugate base? Express your answer in milliliters as an integer. 15| ΑΣΦ Submit G d Previous Answers Request Answer ? mL

General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter16: Acid-base Equilibria
Section: Chapter Questions
Problem 16.149QP: A solution of weak base is titrated to the equivalence point with a strong acid. Which one of the...
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MISSED THIS? Read Section 18.4 (Page); Watch KCV 18.4B. Consider the curve shown here for the titration of a weak monoprotic acid with a strong base. Hd 14 12 10 8 6420 80 120 160 40 Volume of base added (mL) The figure shows pH as a function of volume of base added. The pH is measured from 0 to 14 on the y-axis, while the volume of base added is measured from 0 to 160 milliliters on the x-axis. The curve of the plot goes up gradually from pH 2.3 at 0 milliliters to pH 5.8 at 49 milliliters, next the plot goes up steeply to pH 11.8 at 51 milliliters, and finally the plot goes up gradually to pH 12.2 at 160 milliliters. ✓ Correct An equilibrium problem based on the initial concentration and Ka of a weak acid should be solved for aqueous solution of the weak acid. This is true for the initial state of titration, when no base has been addition of a small amount of a strong base turns this solution into a buffer whose pH is calculated in way. Part D At what volume of added base does pH = pka? Express your answer in milliliters as an integer. V = 30 mL Submit ✓ Correct The point at which pH = pKa (the half-equivalence point) corresponds to the state when the co buffer components are exactly equal. This point is reached halfway to the equivalence point, nan 60 mL/2 = 30 mL. Part E Previous Answers At what volume of added base is the pH calculated by working an equilibrium problem based on the Kb of the conjugate base? Express your answer in milliliters as an integer. V = [VE ΑΣΦ VO Submit ว Previous Answers Request Answer ? X Incorrect; Try Again; 6 attempts remaining mL
Transcribed Image Text:MISSED THIS? Read Section 18.4 (Page); Watch KCV 18.4B. Consider the curve shown here for the titration of a weak monoprotic acid with a strong base. Hd 14 12 10 8 6420 80 120 160 40 Volume of base added (mL) The figure shows pH as a function of volume of base added. The pH is measured from 0 to 14 on the y-axis, while the volume of base added is measured from 0 to 160 milliliters on the x-axis. The curve of the plot goes up gradually from pH 2.3 at 0 milliliters to pH 5.8 at 49 milliliters, next the plot goes up steeply to pH 11.8 at 51 milliliters, and finally the plot goes up gradually to pH 12.2 at 160 milliliters. ✓ Correct An equilibrium problem based on the initial concentration and Ka of a weak acid should be solved for aqueous solution of the weak acid. This is true for the initial state of titration, when no base has been addition of a small amount of a strong base turns this solution into a buffer whose pH is calculated in way. Part D At what volume of added base does pH = pka? Express your answer in milliliters as an integer. V = 30 mL Submit ✓ Correct The point at which pH = pKa (the half-equivalence point) corresponds to the state when the co buffer components are exactly equal. This point is reached halfway to the equivalence point, nan 60 mL/2 = 30 mL. Part E Previous Answers At what volume of added base is the pH calculated by working an equilibrium problem based on the Kb of the conjugate base? Express your answer in milliliters as an integer. V = [VE ΑΣΦ VO Submit ว Previous Answers Request Answer ? X Incorrect; Try Again; 6 attempts remaining mL
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