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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

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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**Titration Curve Explanation** **Description:** The image presents a titration curve graph for the titration of a weak monoprotic acid with a strong base. **Graph Details:** - **Axes:** - The y-axis represents pH, ranging from 0 to 14. - The x-axis represents the volume of base added, in milliliters (mL), ranging from 0 to 160 mL. - **Curve Analysis:** - Initially, the curve rises gradually from a pH of 2.3 at 0 mL to a pH of 5.8 at 49 mL of base added. - A steep increase occurs, reaching a pH of 11.8 at 51 mL. - Following this steep rise, the curve continues to increase more gradually, reaching a pH of 12.2 at 160 mL. **Problem Description:** - **Part D:** - Question: At what volume of added base does pH = pKₐ? - Answer: 30 mL - Explanation: The point where pH equals pKₐ, known as the half-equivalence point, is where the concentration of the acid equals the concentration of its conjugate base. This occurs halfway to the equivalence point. - **Part E:** - The exercise asks to calculate the volume where the pH is determined by an equilibrium problem involving the conjugate base of the weak acid. Understanding this curve is essential for analyzing the titration process and determining key properties such as the equivalence and half-equivalence points in acid-base reactions.
Transcribed Image Text:**Titration Curve Explanation** **Description:** The image presents a titration curve graph for the titration of a weak monoprotic acid with a strong base. **Graph Details:** - **Axes:** - The y-axis represents pH, ranging from 0 to 14. - The x-axis represents the volume of base added, in milliliters (mL), ranging from 0 to 160 mL. - **Curve Analysis:** - Initially, the curve rises gradually from a pH of 2.3 at 0 mL to a pH of 5.8 at 49 mL of base added. - A steep increase occurs, reaching a pH of 11.8 at 51 mL. - Following this steep rise, the curve continues to increase more gradually, reaching a pH of 12.2 at 160 mL. **Problem Description:** - **Part D:** - Question: At what volume of added base does pH = pKₐ? - Answer: 30 mL - Explanation: The point where pH equals pKₐ, known as the half-equivalence point, is where the concentration of the acid equals the concentration of its conjugate base. This occurs halfway to the equivalence point. - **Part E:** - The exercise asks to calculate the volume where the pH is determined by an equilibrium problem involving the conjugate base of the weak acid. Understanding this curve is essential for analyzing the titration process and determining key properties such as the equivalence and half-equivalence points in acid-base reactions.
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