2020 05 16 A 13 Lecture

Microsoft PowerPoint - 2020 05 16 A 13_Lecture.pptxHamden, CT .
/ /

Solutions
/
( )

https://sustainabledevelopment.un.org/sdgs
Solutions
© 2015 Pearson Education
Solution Chemistry
The majority of chemical processes are reactions that occur in solution. Important industrial processes often utilize solution chemistry. “Life” is the sum of a series of complex processes occurring in solution. Air, tap water, tincture of iodine, beverages, and household ammoniaare common examples of solutions.
Droplets of a solution of water and oil, exposed to polarized light and magnified.
http://www.chemistryexplained.com/Ru-Sp/Solution-Chemistry.html#ixzz5mqibRuLt
• 13.2 Saturated solutions and solubility
• 13.3 Factors affecting solubility
• 13.4 Expressing solution concentration
• 13.5 Colligative properties
The fundamental difference between states of matter is the strength of the intermolecular forces of attraction.
Stronger forces bring molecules closer together.
Solids and liquids are referred to as the condensed phases.
States of Matter
of two or more pure substances .
• In a solution, the solute (minor, trace, ultra trace) is dispersed uniformly throughout the solvent (major).
• The ability of substances to form solutions depends on
natural tendency toward mixing.
intermolecular forces.
Natural Tendency toward Mixing • Mixing of gases is a spontaneous process./
• Each gas acts as if it is alone to fill the container.
• Mixing causes more randomness in the position of the molecules, increasing a thermodynamic quantitycalled entropy .

• The formation of solutions is favored by the increase in entropy that accompanies mixing.
Solutions
Intermolecular Forces of Attraction
Any intermolecular force of attraction (Chapter 11) can be the attraction between solute and solvent molecules.
- -
Compounds containing these forces are important in biological systems
Solutions
Dispersion Forces
• The figure below shows how a nonpolar particle (in this case a helium atom) can be temporarily polarized to allow dispersion force to form.
• The tendency of an electron cloud to distort is called its polarizability.
Solutions
Factors Which Affect Amount of Dispersion Force in a Molecule
• number of electrons in an atom (more electrons, more dispersion force)
• size of atom or molecule/molecular weight
• shape of molecules with similar masses (more compact, less dispersion force)
Solutions
Polarizability & Boiling Point
• If something is easier to polarize, it has a lower boiling point.
• Remember: This means less intermolecular force (smaller molecule: lower molecular weight, fewer electrons).
Solutions
Dipole–Dipole Interactions
• Polar molecules have a more positive and a more negative end–a dipole (two poles, δ+ and δ−).
• The oppositely charged ends attract each other.
Solutions
Dipole–Dipole Interactions
For molecules of approximately equal mass and size, the more polar the molecule, the higher its boiling point.
Solutions
Dispersion Forces?
• If two molecules are of comparable size and shape, dipole–dipole interactions will likely be the dominating force.
• If one molecule is much larger than another, dispersion forces will likely determine its physical properties.
Solutions
Hydrogen Bonding • The dipole–dipole interactions
experienced when H is bonded to N, O, or F are unusually strong.
• We call these interactions hydrogen bonds.
• A hydrogen bond is an attraction between a hydrogen atom attached to a highly electronegative atom and a nearby small electronegative atom in another molecule or chemical group.
Solutions
• Ion–dipole interactions are found in solutions of ions.
• The strength of these forces is what makes it possible for ionic substances to dissolve in polar solvents.
Solutions
Attractions Involved When Forming a Solution
• Solute–solute interactions must be overcome to disperse these particles when making a solution.
• Solvent–solvent interactions must be overcome to make room for the solute.
• Solvent–solute interactions occur as the particles mix.
Solutions
Solutions
Exothermic or Endothermic
• For a reaction to occur, ΔHmix must be close to the sum of ΔHsolute and ΔHsolvent.
• Remember that the randomness from entropy will affect the process, too.
Solutions
When NaOH dissolves, the container becomes hot. Therefore, we can conclude that the magnitude of ΔHsolute + ΔHsolvent ___ ΔHmixing.
a. is greater than
b. is less than
d. is unrelated to
Aqueous Solution vs. Chemical Reaction
Just because a substance disappears when it comes in contact with a solvent, it does not mean the substance dissolved. It may have reacted, like nickel with hydrochloric acid.
Solutions
•
• When the rate of the opposing processes is equal, additional solute will not dissolve unless some crystallizes from solution. This is a saturated solution.
• If we have not yet reached the amount that will result in crystallization, we have an unsaturatedsolution.
Solutions
Solubility
• Solubility is the maximum amount of solute that can dissolve in a given amount of solvent at a given temperature.
• Saturated solutions have that amount of solute dissolved.
• Unsaturated solutions have any amount of solute less than the maximum amount dissolved in solution.
• Surprisingly, there is one more type of solution.
Solutions
SupersaturatedSolutions
• In supersaturated solutions, the solvent holds more solute than is normally possible at that temperature.
• These solutions are unstable; crystallization can usually be stimulated by adding a “seed crystal”
or scratching the sideof the flask. • These are uncommon solutions.
Solutions
• Solute–solvent Interactions -
• Pressure (for gaseous solutes)
• That does not explain everything!
• The stronger the solute–solvent interaction, the greater the solubility of a solute in that solvent.
• The gases in the table only exhibit dispersion force. The larger the gas, the more soluble it will be in water.
Solutions
Which of the following compounds is miscible with water? a. CH3OH
b. CH4
c. C6H6
d. CH3CH2OCH2CH3
Which of the compounds below is the least miscible with water?
a. CH3OH
b. CH3CH2OH
c. CH3CH2CH2OH
d. CH3CH2CH2CH2OH
dissolve in water better than nonpolar organic molecules.
• Hydrogen bonding increases solubility, since C–C and C–H bonds are not very polar.
Solutions
proportions are miscible.
• Liquids that do not mix in one another are immiscible.
• Because hexane is nonpolar and water is polar, they are immiscible.
Solutions
Solubility and Biological Importance • Fat-soluble vitamins (like vitamin A)are
nonpolar; they are readily stored in fatty tissue in the body.
• Water-soluble vitamins (like vitamin C)need to be included in the daily diet.
Solutions
Solution
Analyze We are given two solvents, one that is nonpolar (CCl4) and the other that is polar (H2O), and asked to determine which will be the better solvent for each solute listed.
Plan By examining the formulas of the solutes, we can predict whether they are ionic or molecular. For those that are molecular, we can predict whether they are polar or nonpolar. We can then apply the idea that the nonpolar solvent will be better for the nonpolar solutes, whereas the polar solvent will be better for the ionic and polar solutes.
Solve C7H16 is a hydrocarbon, so it is molecular and nonpolar. Na2SO4, a compound containing a metal and nonmetals, is ionic. HCl, a diatomic molecule containing two nonmetals that differ in electronegativity, is polar. I2, a diatomic molecule with atoms of equal electronegativity, is nonpolar. We would therefore predict that C7H16 and I2 (the nonpolar solutes) would be more soluble in the nonpolar CCl4 than in polar H2O, whereas water would be the better solvent for Na2SO4 and HCl (the ionic and polar covalent solutes).
Predict whether each of the following substances is more likely to dissolve in the nonpolar solvent carbon tetrachloride (CCl4) or in water: C7H16, Na2SO4, HCl, and I2.
Sample Exercise 13.1 Predicting Solubility Patterns
Solutions
Practice Exercise 1
Which of the following solvents will best dissolve wax, which is a complex mixture of compounds that mostly are CH3–CH2–CH2–CH2–CH2—?
Continued
Solutions
solids and liquids are not appreciably affected by pressure.
• Gas solubility is affected by pressure.
Solutions
Henry’s Law
•
Solution
Analyze We are given the partial pressure of CO2, PCO2 , and the Henry’s law constant, k, and asked to
calculate the concentration of CO2 in the solution.
Plan With the information given, we can use Henry’s law, Equation 13.4, to calculate the solubility, SCO2 .
Solve SCO2 = kPCO2
= (3.4 × 10–2 mol ⁄ L-atm) (4.0 atm)= 0.14 mol ⁄ L = 0.14 M
Check The units are correct for solubility, and the answer has two significant figures consistent with both the partial pressure of CO2 and the value of Henry’s constant.
Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.0 atm over the liquid at 25 . The Henry’s law constant for CO2 in water at this temperature is 3.4 × 10–2 mol ⁄ L-atm.
Sample Exercise 13.2 A Henry’s Law Calculation
Solutions
Temperature Effects
• For most solids, as temperature increases, solubility increases. However, clearly this is not always true—some increase greatly, some remain relatively constant, and others decrease.
• For all gases, as temperature increases, solubility decreases. Cold rivers have higher oxygen content than warm rivers.
Solutions
13.4 Expressing solution concentration Solution Concentration
• We have discussed solubility and solutions qualitatively : saturated (which is quantitative), unsaturated, and supersaturated.
• Now we will give specific amounts to solutions.
Solutions
4) Mole fraction
• Percent means “out of 100.”
• Take the ratio of the mass of the solute to the total solution mass.
• Multiply by 100 to make it a percent.
Solutions
2) Parts per Million (ppm) 3) Parts per Billion (ppb)
• still relating mass of a solute to the total mass of the solution
• Since percent is out of 100, we multiplied by 100.
• ppm is per million, so we multiply by 106.
• ppb is per billion, so we multiply by 109.
Solutions
(4) Mole Fraction (χ)kai
• Mole fraction is the ratio of moles of a substance to the total number of moles in a solution.
• It does not matter if it is for a solute or for a solvent.
Solutions
(5) Molarity (M) (6) Molality (m)
• Be careful of your penmanship !
• Molarity was discussed in Chapter 4 as moles of solute per liter of solution.
• Molality is moles of solute per kilogram of solvent.
Solutions
Solution
Analyze We are asked to calculate the concentration of the solute, HCl, in two related concentration units, given only the percentage by mass of the solute in the solution.
Plan In converting concentration units based on the mass or moles of solute and solvent (mass percentage, mole fraction, and molality), it is useful to assume a certain total mass of solution. Let’s assume that there is exactly 100 g of solution. Because the solution is 36% HCl, it contains 36 g of HCl and (100 − 36) g = 64 g of H2O. We must convert grams of solute (HCl) to moles to calculate either mole fraction or molality. We must convert grams of solvent (H2O) to moles to calculate mole fractions and to kilograms to calculate molality.
Solve (a) To calculate the mole fraction of HCl, we convert the masses of HCl and H2O to moles and then use Equation 13.7:
An aqueous solution of hydrochloric acid contains 36% HCl by mass. (a) Calculate the mole fraction of HCl in the solution. (b) Calculate the molality of HCl in the solution.
Sample Exercise 13.5 Calculation of Mole Fraction and Molality
Solutions
vs. Molality
• When water is the solvent, dilute solutions have similar molarity and molality.
• Molality does not vary with temperature (mass does not change).
• Molarity varies with temperature (volume changes).
Solutions
• Follow dimensional analysis techniques from Chapter 1.
• To convert between molality and molarity, the density of the solution must be used.
Solutions
Solution
Analyze Our goal is to calculate the molarity of a solution, given the masses of solute (5.0 g) and solvent (225 g) and the density of the solution (0.876 g/mL).
Plan The molarity of a solution is the number of moles of solute divided by the number of liters of solution (Equation 13.8). The number of moles of solute (C7H8) is calculated from the number of grams of solute and its molar mass. The volume of the solution is obtained from the mass of the solution (mass of solution = mass of solute + mass of solvent = 5.0 g + 225 g = 230 g) and its density.
Solve The number of moles of solute is:
The density of the solution is used to convert the mass of the solution to its volume:
Molarity is moles of solute per liter of solution:
A solution with a density of 0.876 g/mL contains 5.0 g of toluene (C7H8) and 225 g of benzene. Calculate the molarity of the solution.
Sample Exercise 13.6 Calculation of Molarity Using the Density of the Solution
Solutions
The unit of solution concentration that is independent of temperature is
a. molality.
b. molarity.
13.5 Colligative properties Vapor Pressure
Because of solute–solvent intermolecular attraction, higher concentrations of nonvolatile solutes make it harder for solvent to escape to the vapor phase. Therefore, the vapor pressure of a solution is lower than that of the pure solvent.
Solutions
•
Solutions
Raoult’s Law
• The vapor pressure of a volatile solvent over the solution is the product of the mole fraction of the solvent times the vapor pressure of the pure solvent.
• In ideal solutions, it is assumed that each substance will follow Raoult’s Law.
Solutions
Solution
Analyze Our goal is to calculate the vapor pressure of a solution, given the volumes of solute and solvent and the density of the solute.
Plan We can use Raoult’s law (Equation 13.10) to calculate the vapor pressure of a solution. The mole fraction of the solvent in the solution, Xsolvent, is the ratio of the number of moles of solvent (H2O) to total moles of solution (moles C3H8O3 + moles H2O).
Solve To calculate the mole fraction of water in the solution, we must determine the number of moles of C3H8O3 and H2O:
We now use Raoult’s law to calculate the vapor pressure of water for the solution:
Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25 . Calculate the vapor pressure at 25 of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25 is 23.8 torr (Appendix B), and its density is 1.00 g/mL.
Sample Exercise 13.7 Calculation of Vapor Pressure of a Solution
Solutions
Boiling-Point Elevation
Since vapor pressures are lowered for solutions, it requires a higher temperature to reach atmospheric pressure. Hence, boiling point is raised.
Solutions
Freezing-Point Depression
The construction of the phase diagramfor a solution demonstrates that the freezing point is lowered while the boiling point is raised.
Solutions
Solution
Analyze We are given that a solution contains 25.0% by mass of a nonvolatile, nonelectrolyte solute and asked to calculate the boiling and freezing points of the solution. To do this, we need to calculate the boiling-point elevation and freezing-point depression.
Plan To calculate the boiling-point elevation and the freezing-point depression using Equations 13.12 and 13.13, we must express the concentration of the solution as molality. Let’s assume for convenience that we have 1000 g of solution. Because the solution is 25.0% by mass ethylene glycol, the masses of ethylene glycol and water in the solution are 250 and 750 g, respectively. Using these quantities, we can calculate the molality of the solution, which we use with the molal boiling-point-elevation and freezing-point-depression constants (Table 13.3) to calculate ΔTb and ΔTf. We add ΔTb to the boiling point and ΔTf to the freezing point of the solvent to obtain the boiling point and freezing point of the solution.
Solve The molality of the solution is calculated as follows:
Automotive antifreeze contains ethylene glycol, CH2(OH)CH2(OH), a nonvolatile nonelectrolyte, in water. Calculate the boiling point and freezing point of a 25.0% by mass solution of ethylene glycol in water.
Sample Exercise 13.8 Calculation of Boiling-Point Elevation and Freezing-Point Depression
Solutions
Continued
We can now use Equations 13.12 and 13.13 to calculate the changes in the boiling and freezing points:
Hence, the boiling and freezing points of the solution are readily calculated:
Comment Notice that the solution is a liquid over a larger temperature range than the pure solvent.
Sample Exercise 13.8 Calculation of Boiling-Point Elevation and Freezing-Point Depression
Solutions
Boiling-Point Elevation and Freezing-Point Depression
• The change in temperature is directly proportional to molality (using the van’t Hoff factor).
Solutions
The van’t Hoff Factor (i) • What is the van’t Hoff factor?
• It takes into account dissociation in solution!
• Theoretically, we get 2 particles when NaCl dissociates. So, i = 2.
• In fact, the amount that particles remain together is dependent on the concentration of the solution.
Solutions
Osmosis • Some substances form semipermeable membranes
, allowing some smaller particles to pass through, but blocking larger particles.
• The net movement of solvent molecules from solution of low to high concentration across a semipermeable membrane is osmosis. The applied pressure to stop it is osmotic pressure.
Solutions
• Osmotic pressure is a colligative property.
• If two solutions separated by a semipermeable membranehave the same osmotic pressure, no osmosis will occur.
Solutions
Solution
Analyze We are asked to calculate the concentration of glucose in water that would be isotonic with blood, given that the osmotic pressure of blood at 25 is 7.7 atm.
Plan Because we are given the osmotic pressure and temperature, we can solve for the concentration, using Equation 13.14. Because glucose is a nonelectrolyte, i = 1.
Solve
Comment In clinical situations, the concentrations of solutions are generally expressed as mass percentages. The mass percentage of a 0.31 M solution of glucose is 5.3%. The concentration of NaCl that is isotonic with blood is 0.16 M, because i = 2 for NaCl in water (a 0.155 M solution of NaCl is 0.310 M in particles). A 0.16 M solution of NaCl is 0.9% mass in NaCl. This kind of solution is known as a physiological saline solution.
The average osmotic pressure of blood is 7.7 atm at 25 . What molarity of glucose (C6H12O6) will be isotonic with blood?
Sample Exercise 13.9 Osmotic Pressure Calculations
Solutions
Types of Solutions & Osmosis 1) Isotonic solutions: Same osmotic
pressure; solvent passes the membrane at the same rate both ways.
2) Hypotonic solution: Lower osmotic pressure; solvent will leave this solution at a higher rate than it enters with.
3) Hypertonic solution: Higher osmotic pressure; solvent will enter this solution at a higher rate than it leaves with.
Solutions
Osmosis and Blood Cells • Red blood cells have semipermeable membranes.
• If stored in a hypertonic solution, they will shrivelas water leaves the cell; this is called crenation.
• If stored in a hypotonic solution, they will grow until they burst; this is called hemolysis.
Solutions
13.6 Colloids
Suspensions of particles larger than individual ions or molecules, but too small to be settled out by gravity, are called colloids.
Solutions
a. butter
b. smoke
• Colloidal suspensions can scatter rays of light. (Solutions do not.)
• This phenomenon is known as the Tyndall effect.
Solutions
(water-loving) end and a nonpolar, hydrophobic (water-fearing) end.
Solutions
Stabilizing Colloids by Adsorption
• Ions can adhere to the surface of an otherwise hydrophobic colloid.
• This allows it to interact with aqueous solution.
Solutions
of fats and oils in aqueous solutions.
• An emulsifier
causes something that normally does not dissolve in a solvent to do so.
Solutions
Brownian Motion
Motion of colloids due to numerous collisions with the much smaller solvent.
Video reference Brownian Motion - Definition, Example, Experiment, Applications https://www.youtube.com/watch?v=hy-clLi8gHg
Solutions
20140414

https://news.ltn.com.tw/news/society/breakingnews/987953
20180323

Solutions

Documents

2020 05 16 A 13 Lecture