MSJ2e_Ch15_ISM

MSJ2e_Ch15_ISM - Chapter 15: The Chemistry of Solutes and...

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Chapter 15: The Chemistry of Solutes and Solutions 641 Chapter 15: The Chemistry of Solutes and Solutions Questions for Review and Thought Review Questions 1. Solubility in water, a highly polar covalent molecule, depends on the solute's ability to interact with water’s large dipole. Solutes that are composed of ions or ones that are capable of hydrogen bonding will generally have high solubility, and solutes without that capability will not. Of the choices provided, (f) salts and (a) alcohols will have the best solubility, and (e) polar molecules should be able to dissolve reasonably well. The (b) hydrocarbons and (d) nonpolar molecules will not be soluble in water, and (c) metals will not dissolve in water unless they react with water. 2. The components of blended motor oils are all long-chain hydrocarbons, and as such are largely nonpolar molecules that interact with London dispersion forces. Since the interactive attractions between different molecules are as good as the interactive attractions between like molecules, the blended mixture will not separate. 3. An unsaturated solution has some solute, but not the maximum possible. A saturated solution has the most solute dissolved as is possible at that temperature. A supersaturated solution is a solution with more than the maximum amount of solute dissolved. This is possible only if a solution is formed at a different temperature where the solute is more soluble and then allowed to sit calmly as it achieves the new temperature. Small disturbances of a supersaturated solution will result in crystallization of the solute back to the saturated solution’s concentration. 4. Look at the solubility equation: original-phase-solute solution-phase-solute. The equilibrium expression is given in this equation: K = [solution-phase-solute] [original- phase-solute] . The expression for the reaction quotient is given in this equation: Q = (conc. solution-phase-solute) (conc. original- phase-solute) . We will assume that the concentration of the original- phase-solute is constant. This is always true of solids, as long as some solid is there in its pure form. This assumption must be made for gas-phase-solutes in order to compare these three solutions. With this assumption, the only thing that affects the size of Q is the concentration of the solution-phase-solute. If the concentration of the solution-phase-solute is smaller than the equilibrium concentration, then Q < K and the solution is unsaturated. If an unsaturated solution is in the presence of the original-phase-solute, more solute will dissolve. If the concentration of the solution-phase-solute is the same as the equilibrium concentration, then Q = K and the solution is saturated. If a saturated solution is in the presence of the original-phase-solute at a constant temperature, no more solute will dissolve.
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MSJ2e_Ch15_ISM - Chapter 15: The Chemistry of Solutes and...

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