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The maximum amount of calcium fluoride that will dissolve in a 0.152 M calcium nitrate solution is ______ M.

The maximum amount of calcium fluoride that will dissolve in a 0.152 M calcium nitrate 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
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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  1. The maximum amount of calcium fluoride that will dissolve in a 0.152 M calcium nitrate solution is ______ M.
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The maximum amount of silver carbonate that will dissolve in a 0.125 M ammonium carbonate solution is x M. Incorrect (NH4)2CO3 is a soluble salt, Ag,CO3 is not. The solubility of Ag2CO3 in pure water is 1.3×10-4 M. The amount of silver carbonate that can dissolve in a solution of ammonium carbonate is less than in plain water, because there is already carbonate ion present in the solution. This is known as the "Common Ion Effect". Initial [CO3?-] = 0.125 M from the (NH4)2CO3 solution. Step 1: Set up the equilibrium using the ICE method Let 's'= number of moles of Ag2CO3 (s) per liter that dissolves. Ag2CO3 (s) = 2Ag+ (aq) + Co32 (aq) Initial some 0.125 M Change Equilibrium + 2s + s some + 2s 0.125 +s Setp 21 Substitute the equilibrium values into the Ksp expression Previous Next
Transcribed Image Text:The maximum amount of silver carbonate that will dissolve in a 0.125 M ammonium carbonate solution is x M. Incorrect (NH4)2CO3 is a soluble salt, Ag,CO3 is not. The solubility of Ag2CO3 in pure water is 1.3×10-4 M. The amount of silver carbonate that can dissolve in a solution of ammonium carbonate is less than in plain water, because there is already carbonate ion present in the solution. This is known as the "Common Ion Effect". Initial [CO3?-] = 0.125 M from the (NH4)2CO3 solution. Step 1: Set up the equilibrium using the ICE method Let 's'= number of moles of Ag2CO3 (s) per liter that dissolves. Ag2CO3 (s) = 2Ag+ (aq) + Co32 (aq) Initial some 0.125 M Change Equilibrium + 2s + s some + 2s 0.125 +s Setp 21 Substitute the equilibrium values into the Ksp expression Previous Next
Setp 2: Substitute the equilibrium values into the Ksp expression Ksp =[Ag*]2 [CO,²] =(2s)²(0.125 + s) = 8.1×10-12 (from the table) %3D %3D Step 3: Assume that s is small relative to 0.125 M, so that the approximation 0.125 + s. 0.125 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 CO3²-. Then: Ksp = (2s)2(0.125) = 8.1×10-12 %3D Step 4: Solve for s 8.1x10 12 (1/2) = (1/2) = 4.0×10-6 M (0.125) Step 5: Check the approximation: 0.125 +s = 0.125 + (4.0×10 6) 0.125 0.125 OK %3D
Transcribed Image Text:Setp 2: Substitute the equilibrium values into the Ksp expression Ksp =[Ag*]2 [CO,²] =(2s)²(0.125 + s) = 8.1×10-12 (from the table) %3D %3D Step 3: Assume that s is small relative to 0.125 M, so that the approximation 0.125 + s. 0.125 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 CO3²-. Then: Ksp = (2s)2(0.125) = 8.1×10-12 %3D Step 4: Solve for s 8.1x10 12 (1/2) = (1/2) = 4.0×10-6 M (0.125) Step 5: Check the approximation: 0.125 +s = 0.125 + (4.0×10 6) 0.125 0.125 OK %3D
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