Determine the value of Ksp for Zn (PO4)2 by constructing an ICE table, writing the solubility constant expression, and solving the expression. The molar solubility of Zn (PO4)2 is 5.6 × 105 M at a certain temperature. Complete Parts 1-2 before submitting your answer. NEXT > 1 2 Fill in the ICE table with the appropriate value for each involved species to determine concentrations of all reactants and products at this temperature. Initial (M) Change (M) Equilibrium (M) Zn(PO)₂(s) = 3 Zn2+(aq) + 2 PO3(aq) Determine the value of Ksp for Zn(PO4)2 by constructing an ICE table, writing the solubility constant expression, and solving the expression. The molar solubility of Zn (PO4)2 is 5.6 × 105 M at a certain temperature. Complete Parts 1-2 before submitting your answer. PREV 1 2 Using the values from the ICE table, construct the expression for the solubility constant. Each reaction participant must be represented by one tile. Do not combine terms Once the expression is constructed, solve for Ksp. Ksp = RESET RESET [0] [5.6 × 105] [1.12 × 104] [1.68 × 104] [5.6 × 10-5]² [1.12 x 104]2 [1.68 x 10413 [1.7 × 10-13] 0 5.6 × 10-5 -5.6 x 10-5 1.12 x 10+ -1.12 x 104 1.68 x 104 -1.68 × 104 [1.9 × 108] 5.9 × 10-20 [1.7 × 10-13]2 [1.9 × 108]2 3.5× 10-39 [1.7 × 10-13]3 [1.9 × 10-813 2.1 x 10-58 1.7 × 10-13 1.9 × 10-8 5.9 × 10-20 -5.6 × 10-20 -1.7 × 10-13 -1.9 × 10-8 -5.9 x 10-20 1.12 x 10-5 [1.12 × 105] [1.68 × 105] [1.12 × 105]² 2.6 × 10-3 3.8 × 10*4 -1.12 x 10-5 1.68 x 10-5 -1.68 × 10%

Introductory Chemistry: A Foundation
9th Edition
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Donald J. DeCoste
Chapter17: Equilibrium
Section: Chapter Questions
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Determine the value of Ksp for Zn (PO4)2 by constructing an ICE table, writing
the solubility constant expression, and solving the expression. The molar
solubility of Zn (PO4)2 is 5.6 × 105 M at a certain temperature. Complete Parts
1-2 before submitting your answer.
NEXT >
1
2
Fill in the ICE table with the appropriate value for each involved species to determine
concentrations of all reactants and products at this temperature.
Initial (M)
Change (M)
Equilibrium (M)
Zn(PO)₂(s)
=
3 Zn2+(aq)
+
2 PO3(aq)
Determine the value of Ksp for Zn(PO4)2 by constructing an ICE table, writing
the solubility constant expression, and solving the expression. The molar
solubility of Zn (PO4)2 is 5.6 × 105 M at a certain temperature. Complete Parts
1-2 before submitting your answer.
PREV
1
2
Using the values from the ICE table, construct the expression for the solubility constant.
Each reaction participant must be represented by one tile. Do not combine terms Once the
expression is constructed, solve for Ksp.
Ksp
=
RESET
RESET
[0]
[5.6 × 105]
[1.12 × 104]
[1.68 × 104]
[5.6 × 10-5]²
[1.12 x 104]2
[1.68 x 10413
[1.7 × 10-13]
0
5.6 × 10-5
-5.6 x 10-5
1.12 x 10+
-1.12 x 104
1.68 x 104
-1.68 × 104
[1.9 × 108]
5.9 × 10-20
[1.7 × 10-13]2
[1.9 × 108]2
3.5× 10-39
[1.7 × 10-13]3
[1.9 × 10-813
2.1 x 10-58
1.7 × 10-13
1.9 × 10-8
5.9 × 10-20
-5.6 × 10-20
-1.7 × 10-13
-1.9 × 10-8
-5.9 x 10-20
1.12 x 10-5
[1.12 × 105]
[1.68 × 105]
[1.12 × 105]²
2.6 × 10-3
3.8 × 10*4
-1.12 x 10-5
1.68 x 10-5
-1.68 × 10%
Transcribed Image Text:Determine the value of Ksp for Zn (PO4)2 by constructing an ICE table, writing the solubility constant expression, and solving the expression. The molar solubility of Zn (PO4)2 is 5.6 × 105 M at a certain temperature. Complete Parts 1-2 before submitting your answer. NEXT > 1 2 Fill in the ICE table with the appropriate value for each involved species to determine concentrations of all reactants and products at this temperature. Initial (M) Change (M) Equilibrium (M) Zn(PO)₂(s) = 3 Zn2+(aq) + 2 PO3(aq) Determine the value of Ksp for Zn(PO4)2 by constructing an ICE table, writing the solubility constant expression, and solving the expression. The molar solubility of Zn (PO4)2 is 5.6 × 105 M at a certain temperature. Complete Parts 1-2 before submitting your answer. PREV 1 2 Using the values from the ICE table, construct the expression for the solubility constant. Each reaction participant must be represented by one tile. Do not combine terms Once the expression is constructed, solve for Ksp. Ksp = RESET RESET [0] [5.6 × 105] [1.12 × 104] [1.68 × 104] [5.6 × 10-5]² [1.12 x 104]2 [1.68 x 10413 [1.7 × 10-13] 0 5.6 × 10-5 -5.6 x 10-5 1.12 x 10+ -1.12 x 104 1.68 x 104 -1.68 × 104 [1.9 × 108] 5.9 × 10-20 [1.7 × 10-13]2 [1.9 × 108]2 3.5× 10-39 [1.7 × 10-13]3 [1.9 × 10-813 2.1 x 10-58 1.7 × 10-13 1.9 × 10-8 5.9 × 10-20 -5.6 × 10-20 -1.7 × 10-13 -1.9 × 10-8 -5.9 x 10-20 1.12 x 10-5 [1.12 × 105] [1.68 × 105] [1.12 × 105]² 2.6 × 10-3 3.8 × 10*4 -1.12 x 10-5 1.68 x 10-5 -1.68 × 10%
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