The molar solubility of Zn3(PO4)2 is 5.6 × 10-6 M at a certain temperature. Determine the value of Ksp for Zn3(PO4)2.

Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter14: Chemical Equilibrium
Section: Chapter Questions
Problem 14.67QE
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10:46 1
Question 24 of 25
Submit
The molar solubility of Zn3(PO4)2 is 5.6 x 10-5
M at a certain temperature. Determine the
value of Ksp for Zn3(PO4)2.
1
2
Based on the given values, fill in the ICE table
to determine concentrations of all reactants
and products.
Zn:(PO.)2(s)= 3 Zn²*(aq) + 2 PO,8-(aq)
Initial (M)
5.6 x 10-5
Change (M)
-X
+3x
+2x
Equilibrium (M)
5.6 x 10-5 - x
+3x
+2x
5 RESET
5.6 x 10-5
-5.6 x 10-5
1.12 х 10-4
-1.12 x 10-4
1.68 x 10-4
-1.68 x 10-4
+x
+2x
+3x
-X
-2x
-3x
5.6 x 10-5 + x
5.6 x 10-5 - x
1.12 x 10-4 + 2x
1.12 x 10-4 - 2x
1.68 х 10-4 + Зx
1.68 x 10-4 - 3x
Transcribed Image Text:10:46 1 Question 24 of 25 Submit The molar solubility of Zn3(PO4)2 is 5.6 x 10-5 M at a certain temperature. Determine the value of Ksp for Zn3(PO4)2. 1 2 Based on the given values, fill in the ICE table to determine concentrations of all reactants and products. Zn:(PO.)2(s)= 3 Zn²*(aq) + 2 PO,8-(aq) Initial (M) 5.6 x 10-5 Change (M) -X +3x +2x Equilibrium (M) 5.6 x 10-5 - x +3x +2x 5 RESET 5.6 x 10-5 -5.6 x 10-5 1.12 х 10-4 -1.12 x 10-4 1.68 x 10-4 -1.68 x 10-4 +x +2x +3x -X -2x -3x 5.6 x 10-5 + x 5.6 x 10-5 - x 1.12 x 10-4 + 2x 1.12 x 10-4 - 2x 1.68 х 10-4 + Зx 1.68 x 10-4 - 3x
10:46 1
Question 24 of 25
Submit
The molar solubility of Zn3(PO4)2 is 5.6 x 10-5
M at a certain temperature. Determine the
value of Ksp for Zn3(PO4)2.
1
2
Based on the set up of your ICE table,
construct the expression for Ksp and then
evaluate it. Do not combine or simplify terms.
Ksp =
%3D
5 RESET
[0]
[5.6 x 10-]
[1.12 x 10-4]
[1.68 x 10-4]
[5.6 x 10-5]?
[1.12 x 10-4]?
[1.68 x 10-4]³
[x]
[2x]?
[3x]°
[5.6 x 10-5 + х]
[5.6 х 10-5 - х]
[1.12 x 10-4 + 2x] [1.12 x 10-4 - 2x] [1.68 × 10-4 + 3x] [1.68 × 10-4 - 3x]
5.6 x 10-5
1.7 x 10-13
1.9 x 10-8
5.9 x 10-20
Transcribed Image Text:10:46 1 Question 24 of 25 Submit The molar solubility of Zn3(PO4)2 is 5.6 x 10-5 M at a certain temperature. Determine the value of Ksp for Zn3(PO4)2. 1 2 Based on the set up of your ICE table, construct the expression for Ksp and then evaluate it. Do not combine or simplify terms. Ksp = %3D 5 RESET [0] [5.6 x 10-] [1.12 x 10-4] [1.68 x 10-4] [5.6 x 10-5]? [1.12 x 10-4]? [1.68 x 10-4]³ [x] [2x]? [3x]° [5.6 x 10-5 + х] [5.6 х 10-5 - х] [1.12 x 10-4 + 2x] [1.12 x 10-4 - 2x] [1.68 × 10-4 + 3x] [1.68 × 10-4 - 3x] 5.6 x 10-5 1.7 x 10-13 1.9 x 10-8 5.9 x 10-20
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