If 2.00 mL of 0.0040 M NaOH are added to 50.0 mL of 0.100 M CdCl2, will a precipitate form? The ions present in solution are Na+, OH-, Cd 2+, CH-. The Only possible precipitate is: Cd(OH)2 (solubility rules). Is Qsp > Ksp for Cd(OH)2? Cd(OH) 2 (s) Ksp Cd(OH)2 = 2.50 x 10-14 EQ (M): Cd2+ (aq) + 2 OH- (aq) X mol Cd 2+ = 0.0500 L x 0.100 mol/L = 0.00500 mol [Cd2+] = 0.00500 mol/0.0520 L = 0.09615 M mol OH = 0.0020 L x 0.0040 M = 8.0 x 10-6 mol [OH-] 8.0 x 10-6 mol/0.052 L = 1.538 x 10-4 M = Qsp = [Cd 2+ ] [OH-] 2 2x = 0.09615 x (1.538 x 10-4)² = 2.27 x 10-9 NO precipitate will form Ksp = [Cd2+] [OH-]² = 2.50 x 10-14 Qsp < Ks sp
If 2.00 mL of 0.0040 M NaOH are added to 50.0 mL of 0.100 M CdCl2, will a precipitate form? The ions present in solution are Na+, OH-, Cd 2+, CH-. The Only possible precipitate is: Cd(OH)2 (solubility rules). Is Qsp > Ksp for Cd(OH)2? Cd(OH) 2 (s) Ksp Cd(OH)2 = 2.50 x 10-14 EQ (M): Cd2+ (aq) + 2 OH- (aq) X mol Cd 2+ = 0.0500 L x 0.100 mol/L = 0.00500 mol [Cd2+] = 0.00500 mol/0.0520 L = 0.09615 M mol OH = 0.0020 L x 0.0040 M = 8.0 x 10-6 mol [OH-] 8.0 x 10-6 mol/0.052 L = 1.538 x 10-4 M = Qsp = [Cd 2+ ] [OH-] 2 2x = 0.09615 x (1.538 x 10-4)² = 2.27 x 10-9 NO precipitate will form Ksp = [Cd2+] [OH-]² = 2.50 x 10-14 Qsp < Ks sp
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter15: Complex Ion And Precipitation Equilibria
Section: Chapter Questions
Problem 37QAP
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Can you explain to me how to find the ions from the chemical equation and when to know the possible precipitate
![If 2.00 mL of 0.0040 M NaOH are added to 50.0 mL of 0.100 M
CdCl2, will a precipitate form?
The ions present in solution are Na+, OH-, Cd 2+, CH-.
The Only possible precipitate is: Cd(OH)2 (solubility rules).
Is Qsp > Ksp for Cd(OH)2?
Cd(OH) 2 (s)
Ksp Cd(OH)2 = 2.50 x 10-14
EQ (M):
Cd2+ (aq) + 2 OH- (aq)
X
mol Cd 2+ = 0.0500 L x 0.100 mol/L = 0.00500 mol
[Cd2+] = 0.00500 mol/0.0520 L = 0.09615 M
mol OH = 0.0020 L x 0.0040 M = 8.0 x 10-6 mol
[OH-] 8.0 x 10-6 mol/0.052 L = 1.538 x 10-4 M
=
Qsp = [Cd 2+ ] [OH-] 2
2x
= 0.09615 x (1.538 x 10-4)² = 2.27 x 10-9
NO precipitate will form
Ksp = [Cd2+] [OH-]² = 2.50 x 10-14
Qsp < Ks
sp](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F637a8243-98d2-45ca-8dd1-4e9d7d0f97ca%2Fff4896a1-8a11-439c-a097-96115218aa36%2F5ifqfm.jpeg&w=3840&q=75)
Transcribed Image Text:If 2.00 mL of 0.0040 M NaOH are added to 50.0 mL of 0.100 M
CdCl2, will a precipitate form?
The ions present in solution are Na+, OH-, Cd 2+, CH-.
The Only possible precipitate is: Cd(OH)2 (solubility rules).
Is Qsp > Ksp for Cd(OH)2?
Cd(OH) 2 (s)
Ksp Cd(OH)2 = 2.50 x 10-14
EQ (M):
Cd2+ (aq) + 2 OH- (aq)
X
mol Cd 2+ = 0.0500 L x 0.100 mol/L = 0.00500 mol
[Cd2+] = 0.00500 mol/0.0520 L = 0.09615 M
mol OH = 0.0020 L x 0.0040 M = 8.0 x 10-6 mol
[OH-] 8.0 x 10-6 mol/0.052 L = 1.538 x 10-4 M
=
Qsp = [Cd 2+ ] [OH-] 2
2x
= 0.09615 x (1.538 x 10-4)² = 2.27 x 10-9
NO precipitate will form
Ksp = [Cd2+] [OH-]² = 2.50 x 10-14
Qsp < Ks
sp
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