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Masteringchem - Print View CHEM162S08DC Assignment 5 Due at 3:00am on Friday View Grading Details Introduction to Solubility and the Solubility

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[ Print View ] CHEM162S08DC Assignment 5 Due at 3:00am on Friday, March 28, 2008 View Grading Details Introduction to Solubility and the Solubility Product Constant Learning Goal: To learn how to calculate the solubility from and vice versa. Consider the following equilibrium between a solid salt and its dissolved form (ions) in a saturated solution: At equilibrium, the ion concentrations remain constant because the rate of dissolution of solid equals the rate of the ion crystallization. The equilibrium constant for the dissolution reaction is is called the solubility product and can be determined experimentally by measuring the solubility , which is the amount of compound that dissolves per unit volume of saturated solution.
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Part A A saturated solution of lead(II) fluoride, , was prepared by dissolving solid in water. The concentration of ion in the solution was found to be . Calculate for . Part A.1 Find the concentration of F using stoichiometry If solid dissolves to produce , what concentration of ion is produced? Hint A.1.a How to approach the problem Hint not displayed Express the molar concentration numerically. ANSWER: = 4.16×10 3 Now you can plug the concentrations into the expression. Part A.2 Identify the K sp expression Part not displayed Express your answer numerically. ANSWER: = 3.60×10 8
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Part B The value of for silver sulfate, , is . Calculate the solubility of in grams per liter. Part B.1 Identify the K sp expression The balanced equation for the complete dissociation of is What is the expression for the solubility product? ANSWER: Part B.2 Express K sp in terms of the molar solubility of Ag 2 SO 4 If represents the number of moles per liter of that dissolves to form a saturated solution, then we can construct the following table: excess 0 0 What is in terms of ? Express in terms of . ANSWER: = Knowing that , you can now solve this expression for and find the molar solubility of . Part B.3 Calculate the molar solubility of Ag 2 SO 4 The solubility product is . Calculate , the molar solubility of . Express the molar solubility numerically. ANSWER: = 1.44×10 2 Now you can convert the solubility from moles per liter to grams per liter using the molar mass of .
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Precipitation Learning Goal: To understand the relationship between precipiatation and the solubility product. To be able to predict whether a substance will precipitate or not. Precipitation is the formation of an insoluble substance. For the equation , precipitation represents a shift to the left, and the production of a solid. From Le Chatelier's principle, we know that when the product of the concentrations of and get above a certain level, the reaction will respond by shifting left to decrease the concentrations of and . This critical level, , is a contstant at a certain temperature. In this case, . When
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This note was uploaded on 04/03/2008 for the course CHEM 162 taught by Professor Siegal during the Spring '08 term at Rutgers.

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Masteringchem - Print View CHEM162S08DC Assignment 5 Due at 3:00am on Friday View Grading Details Introduction to Solubility and the Solubility

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