I need help completing this page mier moTitration and Equilibria of Diprotic Acids Hbin dgu ais y olsomalor niog nolesi Section: brue (1lo dossName:bacqanPre-Lab Problemseldat steb uovBelow is a student's titration curve of the weak acid H2SO3.Doy solav qg bua q oili mo1.218 ni ben1. 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 2o151413.Cbr soy biou sdi 1012 -11-10 -ntao muindilhp orls otin W.CTo sulav ordi esob woH.EH SAil bioo gnone a109.Ao lo 2 oz vlolit tai 1210305040iay19ne vileimad to loodbmflmL NaOH2. If the pKal point has a value of 1.85, what is Kal?opoigotuda d3. What is the pH at pKal?

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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

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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?

Expert Solution

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