Chapter 16

 A sinkhole forms when the roof of a cave weakens from being dissolved by groundwater and suddenly collapses. One recorded sinkhole swallowed a house, several other buildings, five cars, and a swimming pool… 16.1

 Solution Formation ◦ What factors determine the rate at which a substance dissolves?  stirring (agitation)  temperature  the surface area of the dissolving particles 16.1

 A cube of sugar in cold tea dissolves slowly Little surface area, cold temperature and no agitation.

 Granulated sugar dissolves in cold water more quickly than a sugar cube, especially with stirring More surface area and agitation.

 Granulated sugar dissolves very quickly in hot tea More surface area and increased temperature.

 Anything that increases solute to solvent particle contact increases the rate of dissolving. ◦ Stirring, heating, pulverizing

 Solubility ◦ How is solubility usually expressed?  The solubility of a substance is the maximum amount of solute that can dissolve in a given quantity of a solvent at a specified temperature and pressure.  Solubility is often expressed in grams of solute per 100 g of solvent.  Reference table G 16.1

 A saturated solution contains the maximum amount of solute for a given quantity of solvent at a given temperature and pressure.  Points on the curve  An unsaturated solution contains less solute than a saturated solution at a given temperature and pressure.  Points below the curve 16.1

 In a saturated solution, the rate of dissolving equals the rate of crystallization, so the total amount of dissolved solute remains constant.  This is a dynamic equilibrium. 16.1

 The mineral deposits around hot springs result from the cooling of the hot, saturated solution of minerals emerging from the spring. 16.1

◦ What conditions determine the amount of solute that will dissolve in a given solvent?  Temperature affects the solubility of solids, liquids, and gases  Pressure also affects the solubility of gases (not liquids or solids) 16.1

◦ Increasing temperature:  Increases the solubility of most solids  Decreases the solubility of gases 16.1

◦ Changing pressure  Changes in pressure have little effect on the solubility of solids and liquids, but pressure strongly influences the solubility of gases.  Gas solubility increases as the vapor pressure of the gas above the solution increases.  Open bottles of soda go flat… 16.1

 A supersaturated solution is unstable and contains more solute than it can theoretically hold at a given temperature.  The crystallization of a supersaturated solution can be initiated if a very small crystal, called a seed crystal, of the solute is added. 16.1

 A supersaturated solution is clear before a seed crystal is added. 16.1

 Crystals begin to form in the solution immediately after the addition of a seed crystal. 16.1

 Excess solute crystallizes rapidly. 16.1

◦ Water must be tested continually to ensure that the concentrations of contaminants do not exceed established limits. These contaminants include metals, pesticides, bacteria, and even the by- products of water treatment. 16.2

 Molarity ◦ How do you determine the concentration of a solution?  The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent.  A dilute solution is one that contains a small amount of solute.  A concentrated solution contains a large amount of solute.

◦ Molarity (M) is the unit of concentration for solutions.  To calculate the molarity of a solution, divide the moles of solute by the volume (in liters) of the solution.  Reference Table T 16.2

 To make a 0.5-molar (0.5M) solution, first add 0.5 mol of solute to a 1-L volumetric flask half filled with distilled water. 16.2

 Swirl the flask carefully to dissolve the solute. 16.2

 Fill the flask with water exactly to the 1-L mark. 16.2

 Making Dilutions ◦ What effect does dilution have on the total moles of solute in a solution?  Nothing. Diluting a solution reduces the number of moles of solute per unit volume, but the total number of moles of solute in solution does not change.

 Making a Dilute Solution 16.2

 Volume-Measuring Devices 16.2

◦ Since the total number of moles of solute remains unchanged upon dilution, you can use this equation: M 1 V 1 = M 2 V 2  Where M 1 and V 1 are the molarity and volume of the initial solution, and M 2 and V 2 are the molarity and volume of the diluted solution. Write this in ref. table T 16.2

 How do you prepare 100 ml of 0.40M MgSO 4 from a stock solution of 2.0M MgSO 4 ? 16.2

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◦ Another way to express concentration is parts per million (ppm).  This is useful when very little solute is dissolved in a very large volume of water; for example the amount of dissolved oxygen in a lake.  Ref. table T ppm = (mass solute/mass solution) X 1,000,000

◦ A 1.0 L sample of water from a stream contains 0.008g of dissolved oxygen. What is the concentration in parts per million?

 The wood frog is a remarkable creature because it can survive being frozen. Scientists believe that a substance in the cells of this frog acts as a natural antifreeze, which prevents the cells from freezing. A solute can change the freezing point of a solution… 16.3

◦ A property that depends only upon the number of solute particles, and not upon their identity, is called a colligative property. 16.3

◦ What are three colligative properties of solutions?  vapor-pressure lowering  boiling-point elevation  freezing-point depression 16.3

 Dissolving a solute in a solvent will ◦ Lower the vapor pressure ◦ Raise the boiling point and ◦ Lower the freezing point The more you dissolve the greater the change. 16.3

 In a pure solvent, equilibrium is established between the liquid and the vapor. 16.3

 In a solution, solute particles reduce the number of free solvent particles able to escape the liquid. Equilibrium is established at a lower vapor pressure. 16.3

◦ Three moles of glucose dissolved in water produce 3 mol of particles because glucose does not dissociate. 16.3

◦ Three moles of sodium chloride dissolved in water produce 6 mol of particles because each formula unit of NaCl dissociates into two ions. This solution will have a lower vapor pressure than the previous one. 16.3

◦ Three moles of calcium chloride dissolved in water produce 9 mol of particles because each formula unit of CaCl 2 dissociates into three ions. This solution will have the lowest vapor pressure. 16.3

 Which solution has the lowest vapor pressure at STP? ◦ 1 M KCl ◦ 1 M CH 3 OH ◦ 1 M MgBr 2

 The freezing-point depression of aqueous solutions makes walks and driveways safer when people sprinkle salt on icy surfaces to make ice melt. The melted ice forms a solution with a lower freezing point than that of pure water. 16.3

 Dissolving a solute in water will lower its freezing point. ◦ The difference in temperature between the freezing point of a solution and the freezing point of the pure solvent is the freezing-point depression.  The more concentrated the solution the lower the freezing point. 16.3

 Which solution would have the lowest freezing point? ◦ 1 mole of MgBr 2 in 1 kg water ◦ 1 mole of CH 3 OH in 1 kg water ◦ 1 mole NaNO 3 in 1 kg water

 Dissolving a solute in water will raise its boiling point. ◦ The difference in temperature between the boiling point of a solution and the boiling point of the pure solvent is the boiling-point elevation.  The more concentrated the solution the higher the boiling point. 16.3

 Which solution would have the highest boiling point? ◦ 1 mole of NaCl in 1 kg water ◦ 1 mole of NaCl in 2 kg water ◦ 1 mole of NaCl in 3 kg water