Chapter story: Use of mannitol to increase the osmotic pressure of the filtrate in the kidneys’ tubules for reducing excess fluid

Solution: A homogenious mixture of one matter (solute) in another matter (solvent). Solute and solvent can be either liquid, solid, or gas., e.g., air (O2/N2), dental amalgam (Ag/Hg), alloy (e.g., Zn/Cu)

Solvation: (Secondary) interaction between solute and solvent Aqueous solutions: Forming solutions with water as the solvent

            e.g., Grease in gasoline;detergent and grease e.g., CaCO3 and BaSO4 (What is this compound for?)             e.g., sugar

Hydration: solvation by water
    Ionic compounds in water
    Water as an integral part of the crystal structure;
    e.g., water in CuSO4•5H2O (Six H2O’s are surrounding the Cu2+.)
    Water molecules in protein crystal structures (What's the significance of this?)

The water can be removed to give dehydrated form
    e.g., heating CuSO4•5H2O

Solubility—the maximum amount of solute dissolves in a unit volume of a solvent at a given temperature (to produce a saturated solution).

Soluble salts                                 Insoluble salts

BaCl2 35.7 g/100mL                 BaSO4 2.4 ´ 10–4 g/100mL
FeSO4 26.5                             CaF2 1.6 ´ 10–4
BaI2 203.1                               FeS 6.2 ´ 10–4
HgCl2 6.1                                Hg2Cl2 2.0 ´ 10–4

Solubility rules: (Table 7.3)
Na+, K+, NH4+, NO3, acetate, salts are soluble
PO43–, CO32–, and S2– salts are insoluble, except with above cations.

Factors affecting solubility
• Temperature—In general, solubility becomes higher (to different extents) for solid compounds at higher temperature.
    —Gases become less soluble at higher temperature

• Pressure—Henry's law (Chapter 5): At a given temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid (e.g., Soda can and the bends), i.e., S1/P1 = S2/P2

• Supersaturation—of solids? of gases? What’s going to happen when disturbed?

Concentration—a measure of the amount of solute dissolved in a given quantity of the solution or solvent

i.e., concentration = amount of solute (mass, volume, or moles)/unit of solution or solvent (mass or volume)
        Name/define different units of concentration you know.

Percent concentration (composition)
(using mass units or/and volume units)

(A) mass percent concentration
(w/w)% = (mass of solute/mass of solution) ´ 100%

(B) Volume percent concentration
(v/v)% = (volume of solute/volume of solution) ´ 100% Note: The "pure" ethanol you use is normally 95% because the 5% water can not be removed by distillation of ethanol. 100% ethanol is called "absolute ethanol". What is the real percentage of the above ethanol solution?

          [(75 ´ 0.95)/250] ´ 100% = 28.5%

(C) mass/volume percent
    = (mass of solute in g/solution volume in mL) ´ 100% Molarity: number of moles of a solute dissolved in 1 L of solution
    molarity (molar; M) = moles solute/ liters solution

Molarity is the most important concentration unit in chemistry as it is based on mole!

e.g., 9.0 g glucose in 500 mL (0.5 L) water gives a solution of
    (9.0/180)/0.5 = 0.1 M

ppm and ppb: parts per million and parts per billion
    —useful for trace pollutants and toxic substances
1 ppm = 1g solute/106 g solution, i.e., 1 mg per kg
          = 1 mg solute/1 kg solution (~1 mg/1 L water)

Thus, 0.50 mg of Pb(II) in 2.0 L of polluted water is
        0.50 mg/2.0 L = 0.25 ppm

1 ppb = 1 g solute/109 g solution
         = 1 m g solute/1 kg solution (1 L aqueous solution)

Thus the above solution is
        500 m g/2.0 L = 250 ppb

1) What is the mass-mass percent concentration of a 5.0% (m/v) NaCl solution? Assuming the addition of small amount of salt into water does not change the volume of the water.

5.0% (m/v) = 5.0 g/100 mL
100 mL of above solution contains 100 mL water and 5.0 g salt to give a total mass of 105.0 g.
Thus, (m/m)% concentration of the solution is:
5.0 g solute/105.0 g solution = 4.8%
2) What is the (m/v)% concentration of a 5.0% (m/m) NaCl solution?

                    5.0% (m/m) = 5.0 g solute/100 g solution
                    In this 100 g solution, there is 5.0 g salt and 95 g (= 95 mL) water.
                    Thus, (m/v)% = 5.0 g salt/95 mL = 5.3%

3) What is the molarity of the 5.0% (m/v) NaCl solution?

M = # mol solute/1 L solution
5.0% (m/v) NaCl = 5.0 g NaCl/100 mL solution
       = 50 g NaCl/1 L solution
50 g NaCl = 50/58.5 mol = 0.85 mol
Thus, the solution is 0.85 M

Alternative thought:
5.0% (m/v) NaCl = 5.0 g NaCl/0.1 L solution
      = (5.0/58.5)/0.1 = 0.85 M

(What is the molarity of the physiological saline solution?)

4) What is the 5.0% (m/m) NaCl solution in molarity?

From question 2 above, 5.0% (m/m) = 5.3% (m/v)
5.3% (m/v) NaCl = 5.3 g NaCl/0.1 L solution
       = (5.3/58.5)/0.1 = 0.91 M
5) What is the 5.0% (m/m) NaCl solution in ppm? 5.0% (m/m) = 5.3% (m/v) = 5.3 g NaCl/100 mL solution
       = 5.3 ´ 103 mg/0.1 L = 5.3 ´ 104 ppm (i.e., mg/L)
6) What is the molarity of 2.0 ppm lead(II) aqueous solution? 2.0 ppm Pb2+= 2.0 mg Pb2+/1 L solution
       = (0.002/207.19)mol/1 L = 9.7 ´ 10–6 M = 9.7 mM
7) How about changing 2.0 ppb mercury(II) solution into M? 2.0 ppb Hg2+ = 2.0 m g Hg2+/1 L solution
       = [(2.0 ´ 10–6)/200.59]/1 L solution
       = 10 ´ 10–9 M = 10 nM
(How about changing the NaCl in questions 1–5 into glucose and answer the questions?)

Dilution of solutions

1) Number of moles before dilution = number of moles after dilution
            M1 (mol/L) ´ V1 (L) = M2 (mol/L) ´ V2 (L)

2) Amount in mass before dilution = amount in mass after dilution

1) How much solvent is needed to dilute 100 mL of a 0.090 M solution into 0.0015 M?

You need to know: the number of moles and the final volume.
final volume = (100 ´ 0.090)/0.0015 = 6.0 ´ 103 mL = 6.0 L
(i.e., V2 = (V1 ´ M1)/M2)
6.0 L – 100 mL = 5.9 L (solvent needed)
2) What is the final concentration of a 0.10 M solution when it is diluted with 4 times the amount of solvent? The initial volume is V, then the final volume is V + 4V = 5V
MF = (0.10 ´ V)/5V = 0.020 M
3) What volume of a 0.730 M solution must be obtained to prepare 1.36 L of a 0.27 M solution? VI = (1.36 ´ 0.27)/0.730 = 0.503 L 4) A bottle of 750 mL 110 proof (i.e., v/v 55.0%) Vodka is added into 5 L of a cocktail containing 6.00% alcohol. What is the final alcohol concentration of the cocktail in v/v%? You need to know the final amount of alcohol and the final volume.
The total amount of alcohol does not change, which is
    (750 ´ 0.550) + 5000 ´ 0.0600 = 713 mL
The final volume is 750 mL + 5000 mL = 5750 mL
Thus, the final concentration is 713/5750 ´ 100% = 12.4%
5) An amount of 25.0 g glucose (C6H12O6) is added to 200 mL 5.00% (m/v) glucose solution, what is the final concentration in (m/v)% and in M? Assuming there is no change in volume upon glucose addition. Total amount of glucose in the final solution is
    25.0 g + 0.0500 g/mL ´ 200 mL = 35.0 g
Final concentration = 35.0/200 ´ 100% = 17.5%
(What about if there is a 5% increase in volume? 16.7%)

The following are important properties associated with concentration that are not mentioned in the textbook!!
Collegative properties: Solution properties that depend on the number of particles dissolved in a given mass of solvent
Vapor pressure lowering—solution with nonvolatile solute has lower vapor pressure than pure solvent (thus, higher temperature is needed to raise the vapor pressure the same as pure solvent, e.g., adding salt into water)
Boiling point elevation—Because of the lowering of the vapor pressure of solvent with nonvolatile solute.
Freezing point depression—Such as the addition of salt to water can prevent water from freezing at 0 °C (also, spreading salt on highway to "melt" ice).

Solubility and Chemical reactions
Insolubility can result in chemical reactions in solutions:

KCl + AgNO3 (in water) ® AgCl¯ + K+ + NO3
BaCl2 + K2SO4 (in water) ® BaSO4¯ + 2K+ + SO42–
3CaCl2 + 2K3PO4 (in water) ® Ca3(PO4)2¯ + 6 K+ + 6 Cl
CaCl2 + Na2CO3 (in water) ® CaCO3¯ + 2Na+ + 2Cl
    cf. Tables 7.2 and 7.3 for insoluble compounds.

Osmosis and dialysis: The flow of water through a semipermeable membrane (from lower to higher concentrations).
examples: Swimming in the sea for a while, your skin may become wrinkled because water flows "out of your body" into the salt water.
              A red blood cell will burst open when it is put in pure water, because water flows into the cell.

"Osmolarity": The "concentration" equivalent of molarity in osmosis
        P = i(nRT/V)
Do you notice that the equation is familiar to you?  Sure, the ideal gas law P = nRT/V.
P is the osmotic pressure in atm
R is the gas constant
T is the temperature in K
V is the volume in L
n is the number of moles (thus, n/V is the molar concentration)
i is a factor that correct the number of particles per mole of solute. Thus, it is 1 for glucose, 2 for NaCl (dissociates into Na+ and Cl), 3 for CaCl2, and 4 for soluble MCl3, and so on.

Ultrafiltration: water and small molecules are forced to go through a semipermeable membrane by hydrostatic pressure, such as "squeezing" water out of protein solution in the kidneys.

isotonic: outer solution has the same concentration as the internal solution of a cell; thus there is no flow of water across the cell membrane

hypotonic: outer solution has lower concentration than the internal solution of a cell; thus there is a net flow of water into the cell, e.g., hemolysis of red blood cells

hypertonic: outer solution has higher concentration than the internal solution of a cell; thus there is a net flow of water out of the cell, causing contraction of the cell.

Other terms: colloids, micelles, CMC, and diffusion