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The molar solubility of calcium sulfite in a 0.218 M ammonium sulfite solution is ______ M.

The molar solubility of calcium sulfite in a 0.218 M ammonium sulfite solution is ______ M.

Chemistry
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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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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  1. The molar solubility of calcium sulfite in a 0.218 M ammonium sulfite solution is ______ M.
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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
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
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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