The mass of lead chromate that is dissolved in 250 mL of a saturated solution is _____ grams. (Ksp = 1.8 x 10-14)The attached photo is an example of how to solve the problem The mass of iron(III) hydroxide that is dissolved in 225 mL of a saturated solution is 2.36x10^-9grams.IncorrectSTEP 1 Determine the number of moles of Fe(OH)3(s) that can dissolve:Let 's' = number of moles per liter that dissolve.%3DFe(OH)3 (s) Fe3+ (aq) + 30H¯ (aq)InitialsomeChangeEquilibrium+ 3s+ 3s+ ssome STEP 2 Set up the Ksp expression and solve for s:Ksp =[Fe3+] [OH13 = (s)(3s)³ = 27s1 = 38 (from the table)6.3x10%3D%3D%3D2.20x10 10 moles Fe(OH)3S = (Ksp /27)1/42.20x10 10M =%3DL solutionWe'll carry 3 significant figures and round to 2 at the end.STEP 3 In 225 mL of solution:2.20×10¯10 mol Fe(OH)31 L# moles Fe(0H)34.95x10 11 mol Fe%3D225 mLL1000 mL106.9 g Fe(0H)3# grams Fe(OH)34.95×10 11 mol Fe(OH)35.3x109 g Fe(0H)3 di%3D%3Dmol Fe(OH)3

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The mass of lead chromate that is dissolved in 250 mL of a saturated solution is _____ grams. (Ksp = 1.8 x 10-14)

Introductory Chemistry: An Active Learning Approach

6th Edition

ISBN:9781305079250

Author:Mark S. Cracolice, Ed Peters

Publisher:Mark S. Cracolice, Ed Peters

Chapter18: Chemical Equilibrium

Section: Chapter Questions

Problem 3E

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Question

100%

The mass of lead chromate that is dissolved in 250 mL of a saturated solution is _____ grams. (Ksp = 1.8 x 10^-14)
The attached photo is an example of how to solve the problem

The mass of iron(III) hydroxide that is dissolved in 225 mL of a saturated solution is 2.36x10^-9
grams.
Incorrect
STEP 1 Determine the number of moles of Fe(OH)3(s) that can dissolve:
Let 's' = number of moles per liter that dissolve.
%3D
Fe(OH)3 (s) Fe3+ (aq) + 30H¯ (aq)
Initial
some
Change
Equilibrium
+ 3s
+ 3s
+ s
some

STEP 2 Set up the Ksp expression and solve for s:
Ksp =[Fe3+] [OH13 = (s)(3s)³ = 27s1 = 38 (from the table)
6.3x10
%3D
%3D
%3D
2.20x10 10 moles Fe(OH)3
S = (Ksp /27)1/4
2.20x10 10M =
%3D
L solution
We'll carry 3 significant figures and round to 2 at the end.
STEP 3 In 225 mL of solution:
2.20×10¯10 mol Fe(OH)3
1 L
# moles Fe(0H)3
4.95x10 11 mol Fe
%3D
225 mL
L
1000 mL
106.9 g Fe(0H)3
# grams Fe(OH)3
4.95×10 11 mol Fe(OH)3
5.3x109 g Fe(0H)3 di
%3D
%3D
mol Fe(OH)3

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