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

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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