Lab #6: KSP OF SILVER ACETATE KEY CONCEPTS . Solubility product Saturated soluti
ID: 1015631 • Letter: L
Question
Lab #6: KSP OF SILVER ACETATE KEY CONCEPTS . Solubility product Saturated solution Heterogeneous equilibrium INTRODUCTION HETEROGENEOUS EQUILIBRIUM Equilibriums are said to be homogeneous if they take place within one phase by contrast with heterogeneous equilibrium, which exist between two phases, i.e. between a gas and a solution, or between a solution and a solid or between two immiscible solutions in contact with one another. The theory of heterogeneous equilibrium is essentially the same as for homogeneous equilibriunm and the law of equilibrium applies just as well to multiphase systems. For example, if we heat calcium carbonate in a closed reaction vessel the following equilibrium will be reached; The essential condition of all equilibriums is that the rates of the forward and of the reverse reactions are equal. Here, the reaction is for one chemical species to pass from one phase (state to the other. The equilibrium constant is given below; We notice that neither [Ca0or (CaCO, ol appear in the expression of the equilibrium constant. Since solids and pure liquids are almost incompressible, the concentrations of these at a given temperature will not vary significantly EQUILIBRIUM BETWEEN IONIC SOLIDS AND ITS IONIC CONSTITUENTS IN SOLUTION Equilibriums are homogeneous if they are in a single phase whereas they are heterogeneous if they exist between two phases, for example, between the gas phase and the liquid phase. The equilibrium law still applies between multiphase systems Consider the equilibrium in water between the solid silver chloride and its ions: The equilibrium constant for the dissociation of silver chloride is: The indication sp signifies a solubility product and the K, constant is called the solubility product The general formula for the dissolution of a salt in water is:Explanation / Answer
Number of moles of AgNO3 in 0.2 M of 100 ml
= molarity * volume in L
= 0.2 M moles / L * 0.100 L
= 0.02 moles AgNO3
Amount of AgNO3:
Number of moles * molar mass
= 0.02 moles AgNO3* 169.87 g/ moles
= 3.3974 g
Molar mass AgNO3 = 169.87 g/mol
Thus to prepare the 0.2 M solution of AgNO3 in 100 ml dissolve 3.3974 g AgNO3 in 100 ml
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