Consider a solution that is made from mixing 35ml of 0.22M AgNO 3 and 83ml of 0.22M Na 2 SO 4 ? a. How much silver sulfate forms during the reaction step? b. What is the excess reactant and its concentration? c. What is the concentration of silver ions and sulfate ions at equilibrium? d. How much silver sulfate re-dissolves during the equilibrium step in part (c) e. What is the ânetâ amount of silver sulfate produced? f. What would be the percent error if you neglected the amount of solid re-dissolving?

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Consider a solution that is made from mixing 35ml of .22 AgN03 and 83 ml of .22M Na2SO4. a. how much silver sulfate forms during the reaction step b.what is the excess reactant and its concentration c.what is the concentration of silver ions in solution d.how much silver sulfate re-dissolves during the equilibrium step in part (c) e. what is the net amount of silver prodcued f.what would be the percent error if you neglected the amount of solid re-dissolving

D. 1. Silver chloride is not soluble in water. What are the relative strengths of the intermolecular forces between Ag and Ciions and water molecules as compared to the intermolecular forces between water molecules and the forces that hold the silver chloride as a solid? a. The ionic attractions are stronger than the polar covalent bonds between the H' and O ions of the b. The ionic attractions are weaker than the intermolecular forces between the Ag' and Cr ions and water c. The ionic attractions are stronger than the intermolecular forces between the Ag' and Cr ions and water d. The ionic attractions are the same as the intermolecular forces between the Ag and CI ions and water water molecules. molecules. molecules molecules 2. The value of AHsotn for a water soluble compound is +34 kJ moÄ¾1. Which value for this compound would be numerically larger, its lattice energy or its hydration energy? Select the single best answer. step. b. The lattice energy, because it is the endothermic step c. The hydration energy, because it is the exothermic step. d. The lattice energy, because it is the exothermic step. 3, which of the following are reasons why the value of ì¨,on for a gas such as CO2, dissolving in water, is negative? a. When a gas dissolves in a liquid, there is no endothermic step as with the lattice energy of a solid. b. When a gas dissolves in a liquid, there is only an exothermic step as with the lattice energy of a solid. c. The only enthalpy change is the one associated with hydration, and this is always positive. d. The only enthalpy change is the one associated with hydration, and this is always negative. e. When a gas dissolves in a liquid the lattice energy is always higher than the hydration energy, making the net enthalpy always negative. 4. Why is ammonia so much more soluble in water than is nitrogen? a. The triple bond in nitrogen prevents it from dissolving in water. b. Nitrogen has only London forces, while both water and ammonia form hydrogen bonds. c. The molar mass of nitrogen is much larger than the molar mass of ammonia, which reduces its solubility. d. Diatomic molecules are never soluble in water s which agueous solution, if either, is likely to have the highÏr boiling point 50 m Nal or en-Na e au

Silver bromide, AgBr (s), is an essential reagent in black and white film developing. It is,

however, only sparingly soluble in water. AgBr (s) has K = 5.0 Ã 10-13, making it difficult to

rinse AgBr from the film negative with water.

Instead, excess AgBr is removed by an aqueous solution of sodium thiosulfate (Na2S2O3), which forms the complex ion Ag(S2O3)23-:

Ag+ (aq) + 2 S2O32- (aq) Ag(S2O3)23- (aq) Kf = 4.7 Ã 10+13 a) To see how this helps, first determine the molar solubility of AgBr (s) in water.

b) The large formation constant of Ag(S2O3)23- (aq) means that almost all of the silver is complexed with the thiosulfate. To determine how much, calculate the [Ag+] at equilibrium in a 1.0 L solution that initially contains 0.001 M silver ions and 0.200 M sodium thiosulfate. Do so by setting up an ICE table and assuming that x is small compared to 0.200 (but not 0.001). Show that this results in a value of x = [Ag(S2O3)23-] = 0.001. In other words, all of the Ag+ complexes.

c) Your result from part b suggests that no Ag+ remains in solution. This obviously canât be correct, since it would result in an infinitely large reaction quotient that wouldnât be equal to an admittedly large equilibrium constant. To determine the correct [Ag+], use the equilibrium concentrations determined in part b as your initial concentrations in a new ICE table. Then use Kf to determine [Ag+]. [This approach is called a stoichiometric shift and is useful when a reaction starts with only reactants but goes almost to completion. Is essence, we are approaching the equilibrium from the direction of all Ag(S2O3)23- and no Ag+ as opposed to the direction of all Ag+ and no Ag(S2O3)23-.]

d) Before we determine the solubility of AgBr (s) in a thiosulfate solution, we need to know the appropriate equilibrium constant. To find it, determine the value of the equilibrium constant for the reaction:

AgBr (s) + 2 S2O32- (aq) <-->Ag(S2O3)23- (aq) + Br - (aq) Kc = ??

e) Finally, calculate the molar solubility of AgBr (s) in 1.0 M sodium thiosulfate. In other words, what is [Ag(S2O3)23-] for the reaction given in part d when the initial concentration of thiosulfate is 1.0 M?

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