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

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

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 112.41 O 16.00 H 1.008

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?

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.

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.

Explanation / Answer

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

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