General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter17: Solubility And Complex-ion Equilibria
Section: Chapter Questions
Problem 17.97QP: A saturated solution of copper(II) iodate in pure water has a copper-ion concentration of 2.7 103...
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Question
![The molar solubility of silver carbonate in a 0.165 M potassium carbonate solution is 7.78x10^-7
М.
Incorrect
K2CO3 is a soluble salt. Ag,C03 is not.
The solubility of Ag,CO3 in pure water is 1.3x10 4 M. The solubility of Ag,CO3 will be less than this
in the presence of potassium carbonate, because there is already CO3²- ion present in the solution.
This is known as the "Common Ion Effect".
Initial [CO2 ] = 0.165 M from the K2CO3 solution.
Step 1: Set up the equilibrium using the ICE method
Let 's' = number of moles of Ag2CO3 (s) per liter that dissolves = molar solubility
%3D
Ag2C03 (s) = 2Ag* (aq) +
Co3² (aq)
Initial
some
0.165 M
Change
Equilibrium
+ 2s
some
+ 2s
0.165 +s](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0a4e7880-3731-4736-9087-0e959d4f3aa8%2F05266efe-8946-494d-97fd-b3aea67dd5c1%2Fatc5iz_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The molar solubility of silver carbonate in a 0.165 M potassium carbonate solution is 7.78x10^-7
М.
Incorrect
K2CO3 is a soluble salt. Ag,C03 is not.
The solubility of Ag,CO3 in pure water is 1.3x10 4 M. The solubility of Ag,CO3 will be less than this
in the presence of potassium carbonate, because there is already CO3²- ion present in the solution.
This is known as the "Common Ion Effect".
Initial [CO2 ] = 0.165 M from the K2CO3 solution.
Step 1: Set up the equilibrium using the ICE method
Let 's' = number of moles of Ag2CO3 (s) per liter that dissolves = molar solubility
%3D
Ag2C03 (s) = 2Ag* (aq) +
Co3² (aq)
Initial
some
0.165 M
Change
Equilibrium
+ 2s
some
+ 2s
0.165 +s
![A92CO3 (s) = 2Ag* (aq) +
Co,² (aq)
Initial
some
0.165 M
Change
Equilibrium
S
+ 2s
some
+ 2s
0.165 + s
Setp 2: Substitute the equilibrium values into the the Ksp expression
Ksp =[Ag*]? [CO3²] =(2s)?(0.165 + s) = (from the table)
8.1×10 12
%3D
Step 3: Assume that s is small relative to 0.165 M, so that the approximation 0.165 + s 0.165 can
be made. This is reasonable because the solubility is low without the common ion and it will be even
lower in the presense of added CO3?. Then:
Ksp = (2s)2(0.165) = 8.1×10 12
%3D
%3D
Step 4: Solve for s
Ksp
(1/2)
(1/2)
:(
8.1x10-12
s= (1/2)
(1/2)
3.5×10 6 M
%3D
(0.165)
(0.165)
Step 5: Check the approximation:
Next
0.165 + S
0.165 + (3.5×10-6)
0.165
0.165
OK
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0a4e7880-3731-4736-9087-0e959d4f3aa8%2F05266efe-8946-494d-97fd-b3aea67dd5c1%2F7ofm7z_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A92CO3 (s) = 2Ag* (aq) +
Co,² (aq)
Initial
some
0.165 M
Change
Equilibrium
S
+ 2s
some
+ 2s
0.165 + s
Setp 2: Substitute the equilibrium values into the the Ksp expression
Ksp =[Ag*]? [CO3²] =(2s)?(0.165 + s) = (from the table)
8.1×10 12
%3D
Step 3: Assume that s is small relative to 0.165 M, so that the approximation 0.165 + s 0.165 can
be made. This is reasonable because the solubility is low without the common ion and it will be even
lower in the presense of added CO3?. Then:
Ksp = (2s)2(0.165) = 8.1×10 12
%3D
%3D
Step 4: Solve for s
Ksp
(1/2)
(1/2)
:(
8.1x10-12
s= (1/2)
(1/2)
3.5×10 6 M
%3D
(0.165)
(0.165)
Step 5: Check the approximation:
Next
0.165 + S
0.165 + (3.5×10-6)
0.165
0.165
OK
%3D
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