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You find a bottle of MCO3 (M is some metal) on the shelf in a chemistry lab, and you want to dissolve this in solution, so you try two methods. Note: the solubility of this compound in water is 3.9 x 10-4 M. Remember to check any assumptions you make in your calculations or your setup.

a) Method 1: Dissolving MCO3 in a solution of 1.00 molar M(NO3)2. Calculate the solubility in this M(NO3)2 solution and compare it to the solubility in pure water (i.e. state whether S is greater than, equal to, or less than S in pure water).

b) Method 2: Dissolving MCO3 in an aqueous solution of pH = 4. Does this affect the solubility of MCO3? Briefly answer qualitatively. Note: For H2CO3, Ka1 = 4.5 x 10-7 and Ka2 = 4.7 x 10-11

 

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Related questions



Confused on number 10 please.

6 What is the molar solubility of cadmium hydroxide,Cd(OH)2 in pure water at 25 oC? Note thatKSP for Cd(OH)2 is 5.3 x 10-15 at25 oC

IMPORTANT: Enter your answer in scientific notation to 2significant figures. For example, if your answer is 9.8765432 x10-1, you should enter 9.9E-1. Failure to adhere to thisformat may cause WebCT to mark your answer as incorrect.

7 In the previous problem, you calculated the solubility ofCd(OH)2 in mol/L. What is this solubility in mg/L? Notethe 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 youranswer as incorrect <?xml:namespace prefix = o ns ="urn:schemas-microsoft-com:office:office" /?>

8 In the previous two problems, you looked at the solubility ofCd(OH)2 in pure water. Now consider a 0.50 MCd(NO3)2. Cadmium nitrate,Cd(NO3)2 is a freely soluble salt. When a0.50 M Cd(NO3)2 is prepared, the cadmiumnitrate dissolves without establishing an equilibrium:

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

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

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

However, the presence of Cd(NO3)2 in thissolution means there will be Cd2+ ions in solution evenbefore the Cd(OH)2 dissolves. What is the molarsolubility of Cd(OH)2 in a 0.50 MCd(NO3)2 solutioN

9. In problem 6, you calculated the molar solubility ofCd(OH)2 in distilled water, and in problem 8, youcalculated the molar solubility of Cd(OH)2 in a 0.50 MCd(NO3)2 solution. Now consider a 0.50 M NaClsolution. Sodium chloride, NaCl is a freely soluble salt. When a0.50 M NaCl solution is prepared, the NaCl dissolves withoutestablishing an equilibrium:

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

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

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

The Cd2+ ions then react with the Cl- ionsfrom 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 x104.

IMPORTANT: Enter your answer in scientific notation to 2significant figures. For example, if your answer is 9.8765432 x10-1, you should enter 9.9E-1.

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