Be sure to answer all parts. Calculate the concentrations of H,C,0,, HC,0,,C,0,, and H* ions in a 0.301 M oxalic acid solution at 25°C. (K,, = 6.5 x 10-2 and K,, = 6.1 x 105 for oxalic acid.) [H,C,0,] = |м [HC,0,]= [C,0,]= [H'] = M x 10 M (Enter your answer in scientific notation.)

Be sure to answer all parts. Calculate the concentrations of H,C,0,, HC,0,,C,0,, and H* ions in a 0.301 M oxalic acid solution at 25°C. (K,, = 6.5 x 10-2 and K,, = 6.1 x 105 for oxalic acid.) [H,C,0,] = |м [HC,0,]= [C,0,]= [H'] = M x 10 M (Enter your answer in scientific notation.)

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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Question

Be sure to answer all parts.
Calculate the concentrations of H,C,0,, HC,0,,C,0,, and H* ions in a 0.301 M oxalic acid solution
at 25°C. (K,, = 6.5 x 10-2 and K,, = 6.1 x 105 for oxalic acid.)
[H,C,0,] =
|м
[HC,0,]=
[C,0,]=
[H'] = M
x 10 M
(Enter
your answer in scientific notation.)

Expert Solution

Step 1

Dissociation reactions of oxalic acid is given below,

Н.С,О, + Н,о
НС,О, + Н,о"
K,
— 6.5х10-2
al
С,о +Н,о"
6.1x10-5
НС,О, + Н,О
K, =
Here,
[нC о: [н,о"]
[Н.С.О,]
[со: Тно]
нC о:]
K,2

Step 2

ICE table is constructed for the first dissociation as follows,

Н.О (ад
H,O(1) +
H,C,O,
НС 0, +
Initial
0.301
Change
+x
-X
+X
(0.301- x)
Equilibrium
X
х
K, can be given as:
(x -X)
Kal
(0.301 – x)
х?
6.5x10-2
(0.301– x
Solving for x,
X = 0.111
[H,C,0,]=(0.301– x) = 0.19 M
Equilibrium concentration of H,C,0, is 0.19 M
[H,0*]=[HC,0;]= 0.111 M
= Concentration after 1* dissociation

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