Chapter 9

Solutions

Chapter 9 suggested problems: 43, 52, 62, 68, 70, 94, 96

Class Notes

  1. Definitions

    1. Solution: homogeneous mixture, equally dispersed at the molecular level, uniform throughout in its physical and chemical properties

    2. Solvent: does the dissolving, present in greater abundance

    3. Solute: gets dissolved, present in lesser abundance

    4. Solvation: the molecular process by which solutes are dissolved by solvents

      1. Hydration: the molecular process by which solutes are dissolved by water

    5. "Like attracts like" governs solvent-solute interactions and dissolution, or the lack of reaction between a solvent and solute

    6. Saturated solution: a solution in which as much solute as is physically possible has been dissolved in the solvent, with some solute remaining undissolved

      1. Solubility: the maximum amount of a substance that will dissolve in a specific amount of solvent to give a thermodynamically stable solution, usually expressed in grams and at specific temperature and pressure conditions

        1. NaCl: 35.7 g per 100 cc of water at 0°C

        2. CO2: 171.3 g per 100 cc of water at 0°C

        3. Barium sulfate: 0.000222 g per 100 cc of water at 18°C

      2. Molar solubility: solubility expressed in moles rather than in grams

    7. Unsaturated solutions and supersaturated solutions

    8. Suspension: molecular-level aggregations of particles physically suspended in liquid, sometimes so small that gravity has no effect on the particles

    9. Miscible: two (or more) substances that can form solutions in all proportions

    10. Concentration: a measure of the amount of solute dissolved in either a specific amount of solvent or solution

  2. Calculating concentration

    1. Percent composition: in general, the ratio of amount of solute to amount of solution multiplied by 100

      1. Weight/weight (w/w): (grams solute / grams solution) x 100

      2. Weight/volume (w/v): (grams solute / volume solution) x 100

      3. Volume/volume (v/v): (volume solute / volume solution) x 100

      4. Examples

        1. What is the (w/w) concentration of 100 g of solution containing 1.22 g of sodium chloride? 1.22% w/w

        2. What is the (w/v) concentration of 50.0 mL of solution containing 1.22 g of sodium chloride? 2.44% w/v

        3. What is the (v/v) concentration of a solution containing 100 mL of isopropyl alcohol that is diluted with water to 150 mL? 66.7% v/v

    2. Concentrations of dilute solutions

      1. mg %: (mg solute / 100 mL solution) x 100

      2. ppm: parts per million, (mg solute / L solution)

      3. ppb: parts per billion, (ug solute / L solution)

    3. Molality: (m), molal concentration, moles solute / kg solvent

      1. What is the molal concentration of 125 grams of glucose that is dissolved in 750 grams of water?

      2. (125 g glucose) x (mole glucose / 180.2 g glucose) x (1 / 750 g water) x (1000 g water / 1 kg water) = 0.925 m

    4. Molarity: (M), molar concentration, moles solute / L solution

      1. What is the molar concentration of 125 grams of glucose that is dissolved in 750 mLs of water?

      2. (125 g glucose) x (mole glucose / 180.2 g glucose) x (1 / 0.750 L solution) = 0.925 M

      3. Molar and molal concentrations will not always be the same

      4. Calculating molarity

        1. If 0.100 moles of H2SO4 is dissolved in 450 mL of water, what is the molarity of the resulting solution? (0.222 M)

        2. If 37.6 grams of copper (II) nitrate is dissolved in 500 mL of water, what is the molarity of the resulting concentration? (MM = 187.56 g/mol; 0.400 M)

        3. How many moles of copper (II) nitrate are contained in 25.0 mL of 2.5 M solution? (0.0625 moles copper (II) nitrate)

        4. How many grams of ammonium hydroxide are contained in 75 mL of concentrated (15M) solution? (MM = 35.05 g/mol; 39.43 g)

  3. Dilutions

    1. The result of diluting solutions can be calculated using the equation M1V1 = M2V2

    2. Examples

      1. How many moles of concentrated NH4OH are required to form 100 mL of 1.0 M solution? (6.67 mL)

      2. If 10.0 mL of 12 M HCl is diluted to 600 mL, what is the new concentration of the acid? (0.200 M)

      3. The dilution of 50.0 mL of a solution of phosphoric acid results in 1.00 L of a 0.75 M solution. What was the initial concentration of the acid? (15 M)

  4. Properties of solutions

    1. Surface tension - the molecules at a phase boundary (e.g., l - g) behave differently than the molecules far away (5 - 10 molecular diameters) from the phase boundary

      1. In bulk liquid molecules attract equally and are equally attracted in all directions

      2. At a phase boundary the attraction is one-sided; it is only inward toward other molecules of the same type

      3. This makes the liquid resist expansion and attempt to minimize its surface area

      4. The molecules closest to the interface act like a kind of "skin"

      5. This also explains the behavior of water on a waxed surface, rain drops falling through the air, etc.

      6. There are exceptions to this behavior: if the intermolecular forces in the substances on both sides of interface have similar bonding interactions, the meniscus will shrink to the point of being barely observable

    2. Vaporization and vapor pressure

      1. An important difference between molecules in the liquid and gas phases is the KE of the particles

      2. If we examine the KE of all of the particles in any phase we find not a single KE but a Gaussian distribution of energies - i.e., there are many molecules with very high energies and many with very low energies as well

      3. Molecules in the liquid phase with high KE and near the l-g phase boundary can actually have enough KE to "escape" from the liquid and join the gas phase

      4. This is why heating a liquid facilitates the l-g transition

      5. Molecules in the vapor phase collide with the liquid surface during their random travels

        1. High KE - bounce off surface

        2. Low KE - "stick" and become part of liquid phase

        3. By cooling a vapor we lower the average KE of the molecules in the gas phase and make it easier for them to " stick" to the liquid surface if they collide with it

      6. At any given temperature an equilibrium exists between the rates of l —> g and g—> l

      7. The pressure exerted by the vapor when this equilibrium exists is called the vapor pressure

      8. Boiling point: when the vapor pressure of a liquid is equal to atmospheric pressure

      9. Normal boiling point: when the vapor pressure of a liquid is equal to exactly 1 atm

      10. Vapor pressure is like BP in that it gives a general relative sense of the strengths of intermolecular forces

        1. Propane - 8.6 atm, acetone - 266 T, i-PrOH - 33 T

        2. Hexane - 124 T, 2-hexanone - 3 T

    3. Diffusion: molecules tend to move from areas of high to low concentration until uniformly distributed

      1. This randomizing effect will occur independent of any external forces (e.g. mixing) and is due to random collisions between molecules (Brownian motion; see Einstein's Explanation of Brownian Motion)

      2. The rate of diffusion depends on several factors including temperature, viscosity and pressure, the molecular weight of the solute, and solute concentration

      3. This is how a solute achieves uniform distribution throughout a solvent in the process of forming a solution - g/g, g/l, l/l, l/s

    4. Osmosis

      1. Definitions

        1. Membrane: a sheet-like structure, often porous, that can regulate the passage of substances from one side of the membrane to the other based on molecular size or charge

        2. Membranes that allow some substances to pass but not others are called semi-permeable

        3. Osmosis: the flow of water through a semi-permeable membrane from areas of low to high solute concentration
          the process of osmosis

        4. Osmotic pressure: a force is generated by osmosis. The pressure that would be required to stop the net flow of water (i.e., the force required to stop osmosis) is the osmotic pressure

          1. Reverse osmosis: forcing water across a semi-permeable membrane from areas of high to low solute concentration

          2. Used to desalinize sea water for drinking

        5. Osmolarity: the behavior of dilute solutions is similar to the behavior of gases in some respects, to the extent that the behavior of dilute solutions can be approximated using the Ideal Gas Law

          1. Π= inRT/V = iMRT

          2. i = # of particles per solute molecule, i = 1 for nonelectrolytes, i = 2 or greater for all electrolytes

          3. n = # moles of solute

          4. M = solute molarity

      2. Examples

        1. What is the osmotic pressure (pi) of a 0.0020 M solution of sucrose (C12H22O11) at 20°C? (0.048 atm = 37 T)

        2. The average osmotic pressure of blood is 7.7 atm at 25°C. What concentration of glucose will be isotonic with blood? (0.315 M)

          1. Isotonic: two solutions with equal osmotic pressures

          2. Hypotonic: one solution has a lower osmotic pressure than the other

          3. Hypertonic: one solution has a higher osmotic pressure than the other

        3. Finding the MM of an unknown substance via osmotic pressure

          1. since n = g/ MM then pi = inRT/V = igRT/( MM V) => MM = igRT/(pi x v)

          2. The osmotic pressure of an aqueous solution of an unknown protein was found to be 1.54 T at 20°C. The solution contained 3.50 mg of the protein, which had been dissolved in 5.00 mL of water. Find the molar mass of the protein (assume i = 1 for proteins). (MM = 8451 g/mol)

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