6 What is the molar solubility of cadmium hydroxide, Cd(OH) 2 in pure water at 2

Question

6 What is the molar solubility of cadmium hydroxide, Cd(OH)2 in pure water at 25 oC? Note that KSP for Cd(OH)2 is 5.3 x 10-15 at 25 oC

IMPORTANT: Enter your answer in scientific notation to 2 significant figures. For example, if your answer is 9.8765432 x 10-1, you should enter 9.9E-1. Failure to adhere to this format may cause WebCT to mark your answer as incorrect.

7 In the previous problem, you calculated the solubility of Cd(OH)2 in mol/L. What is this solubility in mg/L? Note the following atomic weights: Cd 112.411 O 15.9994 H 1.0079

IMPORTANT: Enter your answer in floating point to 1 decimal place. For example, if your answer is 9.876543210, you should enter 9.9. Failure to adhere to this format may cause WebCT to mark your answer as incorrect <?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" /?>

8 In the previous two problems, you looked at the solubility of Cd(OH)2 in pure water. Now consider a 0.50 M Cd(NO3)2. Cadmium nitrate, Cd(NO3)2 is a freely soluble salt. When a 0.50 M Cd(NO3)2 is prepared, the cadmium nitrate dissolves without establishing an equilibrium:

Cd(NO3)2(s) ==========> Cd2+(aq) + 2NO3-(aq)

In a 0.50 M Cd(NO3)2 solution, the concentration of Cd2+ will be 0.50 mol/L. When the Cd(OH)2 is dissolved in this solution, it establishes its usual equilibrium between the solid and the ions in solution:

Cd(OH)2(s) <----------> Cd2+(aq) + 2OH-(aq)

However, the presence of Cd(NO3)2 in this solution means there will be Cd2+ ions in solution even before the Cd(OH)2 dissolves. What is the molar solubility of Cd(OH)2 in a 0.50 M Cd(NO3)2 solutioN

9. In problem 6, you calculated the molar solubility of Cd(OH)2 in distilled water, and in problem 8, you calculated the molar solubility of Cd(OH)2 in a 0.50 M Cd(NO3)2 solution. Now consider a 0.50 M NaCl solution. Sodium chloride, NaCl is a freely soluble salt. When a 0.50 M NaCl solution is prepared, the NaCl dissolves without establishing an equilibrium:

NaCl(s) ==========> Na+(aq) + Cl-(aq)

In a 0.50 M NaCl solution, the concentration of Cl- will be 0.50 mol/L. When the Cd(OH)2 is dissolved in this solution, it establishes its usual equilibrium between the solid and the ions in solution:

Cd(OH)2(s) <----------> Cd2+(aq) + 2OH-(aq)

The Cd2+ ions then react with the Cl- ions from NaCl to form a complex ion:

Cd2+(aq) + 4Cl-(aq) <----------> [CdCl4]2-(aq)

Note that the formation constant for [CdCl4]2-, Kf is 1.0 x 104.

IMPORTANT: Enter your answer in scientific notation to 2 significant figures. For example, if your answer is 9.8765432 x 10-1, you should enter 9.9E-1.

Explanation / Answer

4*s^3=5.3*10^-15 s=1.1*10^-5

Related Questions

Navigate

• Browse (All)
• Browse 6
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.