Ksp / Molar Solubility - TBR #5 PS

[sugarbabee0]

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The molar solubility of CdCO3(s) is 2.2 x 10-6 M, and the molar solubility of NiCO3(s) is 3.8 x 10-4 M. What is the voltage in Trial 3 when the solutions in both half-cells are completely saturated?



A. +0.37 V

B. +0.24 V

B is the best answer. When the concentrations of Cd2+(aq) and Ni2+(aq) are equal, the cell voltage for the electrochemical cell made up of their respective half-reactions is 0.17 volts (as shown in Trial 3). The concentration of cadmium cation in the anode half-cell is the molar solubility of CdCO3, 2.2 x 10-6 M. The concentration of nickel cation in the cathode half-cell is the molar solubility of NiCO3, 3.8 x 104 M. Because the cathode has a greater cation concentration than the anode, where the cathode cation is a reactant and the anode cation is a product, there is more reactant than product. This means that the reaction is more favorable than the standard reaction. Thus, the cell voltage must be greater than the standard value of 0.17 volts. This eliminates choices C and D. To determine the correct answer, the Nernst equation must be used as shown below....


I know there are some data missing in the question i copied above, but my question is about the molar solubility. If the molar solubility of CdCO3(s) is 2.2 x 10-6 M, wouldn't the concentration of Cd++ be the square root of 2.2 x 10-6 M since CdCO3(s) ---> Cd(++) AND CO3(-2) ???

 

[fizzgig]

i think the molar solubility is the ACTUAL amount that can dissolve in these conditions. so 2.2E-6 M of CdCO3 is dissolved, and so you have 2.2E-6mols of EACH Cd2+ and CO3 2-

if they'd given the solubility constant and you had to calculate molar solubility that'd be different.