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.)
Chapter16: Solubility And Complex Ion Equilibria
Section: Chapter Questions
Problem 10RQ: Figure 15-3 outlines the classic scheme for separating a mixture of insoluble chloride salts from...
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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.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc9fdbc9b-c7a4-4813-8fa6-3d75f8e76ceb%2F1ffa185f-75eb-49f9-88a5-3df23104edba%2Fqyyl8lm.png&w=3840&q=75)
Transcribed Image Text: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](https://content.bartleby.com/qna-images/answer/c9fdbc9b-c7a4-4813-8fa6-3d75f8e76ceb/d4884c4c-e526-4b7e-92b9-d3f0d9227098/lb5fi58.png)
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](https://content.bartleby.com/qna-images/answer/c9fdbc9b-c7a4-4813-8fa6-3d75f8e76ceb/d4884c4c-e526-4b7e-92b9-d3f0d9227098/ne3724.png)
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