12. If 50.0 g of silver phosphate (Kp = 1.8 x 1018) are placed in enough water to generate 500.0 ml of solution, calculate the following: A. The molar solubility of the silver phosphate solution. B. The molar solubility of the silver ion if 30.0 mL of 0.400 M silver nitrate solution is added to the original solution.

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12. If 50.0 g of silver phosphate (Kp = 1.8 x 10-18) are placed in enough water to generate
500.0 mL of solution, calculate the following:
A. The molar solubility of the silver phosphate solution.
B. The molar solubility of the silver ion if 30.0 mL of 0.400 M silver nitrate solution
is added to the original solution.
Transcribed Image Text:12. If 50.0 g of silver phosphate (Kp = 1.8 x 10-18) are placed in enough water to generate 500.0 mL of solution, calculate the following: A. The molar solubility of the silver phosphate solution. B. The molar solubility of the silver ion if 30.0 mL of 0.400 M silver nitrate solution is added to the original solution.
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When sparingly soluble salts are placed in water, they exist in equilibrium with its ions. The equilibrium constant for this condition is called the solubility product. If ions from outside are mixed in the solution, the equilibrium gets disturbed. To undo this change, equilibrium shifts in any desired direction.

 

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