Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
Bartleby Related Questions Icon

Related questions

Question
I need help completing this page
mier moTitration and Equilibria of Diprotic Acids H
bin dg
u ais y ols
omalor niog nolesi Section: brue (1
lo doss
Name:
bacqan
Pre-Lab Problems
eldat steb uov
Below is a student's titration curve of the weak acid H2SO3.
Doy solav qg bua q oili mo1.21
8 ni ben
1. Identify the regions on the graph corresponding to equivalence point #1, equivalence point
#2, the approximate pKal and pK2 points.
Con dul uo Titration of 0.10 M H2SO,
brte t 2o
15
14
13.
Cbr soy biou sdi 10
12 -
11-
10 -
ntao muindilhp orls otin W.C
To sulav ordi esob woH.E
H SAil bioo gnone a109.A
o lo 2 oz vlolit tai 12
10
30
50
40
iay19ne vileimad to loodbmfl
mL NaOH
2. If the pKal point has a value of 1.85, what is Kal?
opoigotuda d
3. What is the pH at pKal?
expand button
Transcribed Image Text:mier moTitration and Equilibria of Diprotic Acids H bin dg u ais y ols omalor niog nolesi Section: brue (1 lo doss Name: bacqan Pre-Lab Problems eldat steb uov Below is a student's titration curve of the weak acid H2SO3. Doy solav qg bua q oili mo1.21 8 ni ben 1. Identify the regions on the graph corresponding to equivalence point #1, equivalence point #2, the approximate pKal and pK2 points. Con dul uo Titration of 0.10 M H2SO, brte t 2o 15 14 13. Cbr soy biou sdi 10 12 - 11- 10 - ntao muindilhp orls otin W.C To sulav ordi esob woH.E H SAil bioo gnone a109.A o lo 2 oz vlolit tai 12 10 30 50 40 iay19ne vileimad to loodbmfl mL NaOH 2. If the pKal point has a value of 1.85, what is Kal? opoigotuda d 3. What is the pH at pKal?
Expert Solution
Check Mark
Step 1

Equivalence point: point in titration at which the amount of titrant added is just enough to completely neutralize the analyte solution. At the equivalence point in an acid-base titration, moles of base = moles of acid and the solution only contains salt and water. 

For diprotic acid two equivalence point is possible due to presence of two types of protons 

From the graph we got

1. equivalance point 1 is at 20 ml and pKa1 is 1.85 

equivalance point 2 is at 40 ml and pKa2 is 7.20

2. pka1 = 1.85

We know

pka1 = -logka1 

1.85 = -logKa1,

Ka1= antilog(- 1.85)

= 1.41x 10^-2 

3. pH at pka1 is obtained from the graph that is 4.5

Knowledge Booster
Background pattern image
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY