Reaction !3D • PbClh(s) 2 Pb2*(aq) +2 CI(aq) Fe(OH),(s) = Fe2*(aq)+ 2 OH (aq) CaCO3(s) 2 Ca"(aq) + CO,2 (aq) Hgal,(s) Hg,"(ag) + 21 (aq) Fe(OH);(s) = Fe"(aq) + 3 OH (aq) = 1.6 x 105 Initial Conditions Q=[1.0 x 10[1.0 x 10 = 1.00 x 10-27 Equilibrium Conditions [Pb2") = 1.0 x 10° M [CT] = 1.0 x 10 M Mass of PbCly(s) =0.00 g Initial Concentrations D (Pb?*] = 1.0 x 10 M D (CI) = 1.0 x 10 M Adjust the concentrations of each ion up and down, paying attention to the value of Q and whether a precipitate forms. What is the highest value Q can be without forming a precipitate?
Reaction !3D • PbClh(s) 2 Pb2*(aq) +2 CI(aq) Fe(OH),(s) = Fe2*(aq)+ 2 OH (aq) CaCO3(s) 2 Ca"(aq) + CO,2 (aq) Hgal,(s) Hg,"(ag) + 21 (aq) Fe(OH);(s) = Fe"(aq) + 3 OH (aq) = 1.6 x 105 Initial Conditions Q=[1.0 x 10[1.0 x 10 = 1.00 x 10-27 Equilibrium Conditions [Pb2") = 1.0 x 10° M [CT] = 1.0 x 10 M Mass of PbCly(s) =0.00 g Initial Concentrations D (Pb?*] = 1.0 x 10 M D (CI) = 1.0 x 10 M Adjust the concentrations of each ion up and down, paying attention to the value of Q and whether a precipitate forms. What is the highest value Q can be without forming a precipitate?
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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Question
The value of Q is 1.0*10^-4
![Reaction
Kap = [Pb][CI]?
= 1.6 x 105
PbCl,(s) 2 Pb"(aq)+2 CF(aq)
Fe(OH),(s) = Fe"(aq) + 2 OH (aq)
O CaCO;(s) 2 Ca2"(aq) + Co,2 (aq)
Hgal,(s) 2 Hg,"(aq) + 21 (aq)
Fe(OH);(s) = Fe"(aq) + 3 OH (aq)
Initial Conditions
Q=[1.0 x 10 [1.0x 10 2
= 1.00 x 10-27
!3!
Equilibrium Conditions
[Pb"] = 1.0 x 10° M
[Cr] = 1.0 x 10°M
Mass of PbCly(s) =0.00 g
Initial Concentrations
D (Pb?*] = 1.0 x 10 M
D (CI] = 1.0 x 10 M
Adjust the concentrations of each ion up and down, paying attention to the value of Q and whether a precipitate forms.
What is the highest value Q can be without forming a precipitate?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa7fedcfb-33e5-442f-8597-a591190de8f9%2F9829a09e-1dc0-4e35-b1ae-3b87a7aa8d2c%2F3x6ok3a_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Reaction
Kap = [Pb][CI]?
= 1.6 x 105
PbCl,(s) 2 Pb"(aq)+2 CF(aq)
Fe(OH),(s) = Fe"(aq) + 2 OH (aq)
O CaCO;(s) 2 Ca2"(aq) + Co,2 (aq)
Hgal,(s) 2 Hg,"(aq) + 21 (aq)
Fe(OH);(s) = Fe"(aq) + 3 OH (aq)
Initial Conditions
Q=[1.0 x 10 [1.0x 10 2
= 1.00 x 10-27
!3!
Equilibrium Conditions
[Pb"] = 1.0 x 10° M
[Cr] = 1.0 x 10°M
Mass of PbCly(s) =0.00 g
Initial Concentrations
D (Pb?*] = 1.0 x 10 M
D (CI] = 1.0 x 10 M
Adjust the concentrations of each ion up and down, paying attention to the value of Q and whether a precipitate forms.
What is the highest value Q can be without forming a precipitate?
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