Calculate the pH at each of the following points for the titration of 100 mL of 0. 6 M dl-Histidine with the following volumes of 1.20 M sodium hydroxide. H3His ↔ H+ + H2His- Ka1 = 3.98x10-3 H2His- ↔ H+ + HHis2- Ka2 = 9.12x10-7 HHis2- ↔ H+ + His3- Ka3 = 4.67x10-10 Vol. of NaOH Calculated pH 0 mL 30 mL 50 mL 75 mL 100 mL 125 mL 135 mL 150 mL 175 mL Please provide a sketch of the titration curve and label both the x & y axis and applicable equations used at each data point (e.g., pH = - log[H+] or pH = pKa or pH = pKa + log )
Calculate the pH at each of the following points for the titration of 100 mL of 0. 6 M dl-Histidine with the following volumes of 1.20 M sodium hydroxide. H3His ↔ H+ + H2His- Ka1 = 3.98x10-3 H2His- ↔ H+ + HHis2- Ka2 = 9.12x10-7 HHis2- ↔ H+ + His3- Ka3 = 4.67x10-10 Vol. of NaOH Calculated pH 0 mL 30 mL 50 mL 75 mL 100 mL 125 mL 135 mL 150 mL 175 mL Please provide a sketch of the titration curve and label both the x & y axis and applicable equations used at each data point (e.g., pH = - log[H+] or pH = pKa or pH = pKa + log )
Chapter14: Principles Of Neutralization Titrations
Section: Chapter Questions
Problem 14.9QAP
Related questions
Question
Calculate the pH at each of the following points for the titration of 100 mL of 0. 6 M dl-Histidine with the following volumes of 1.20 M sodium hydroxide.
H3His ↔ H+ + H2His- Ka1 = 3.98x10-3
H2His- ↔ H+ + HHis2- Ka2 = 9.12x10-7
HHis2- ↔ H+ + His3- Ka3 = 4.67x10-10
Vol. of NaOH |
Calculated pH |
0 mL |
|
30 mL |
|
50 mL |
|
75 mL |
|
100 mL |
|
125 mL |
|
135 mL |
|
150 mL |
|
175 mL |
|
Please provide a sketch of the titration curve and label both the x & y axis and applicable equations used at each data point (e.g., pH = - log[H+] or pH = pKa or pH = pKa + log )
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 6 steps with 1 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Recommended textbooks for you