SCH 3U - Predicting Solubility & Factors Affecting Solubility

Predicting Solubility

 

Ionic Compounds 
Calculate the ΔEneg between the two atoms in the compound.  If the difference is small, the compound will be insoluble in water.

ex.  Between silver sulfide and sodium chloride, which compound is more likely water soluble?

AgS     ΔEneg (S, Ag) = 2.58 – 1.93 = 0.65 

NaCl    ΔEneg (Cl, Na) = 3.16 – 0.93 = 2.23  - soluble

Covalent Compounds

Covalent compounds do not have full positive & negative charges like ions, but some are still water soluble.  If the covalent compound has polar regions, it is more likely to be water soluble.

ex. Between ethanol (C₂H₅OH) and butane (C₄H₁₀), which compound is more likely water soluble?

Ethanol has an H directly bonded to an O, which creates a polar region in this compound.  Ethanol will be able to hydrogen bond to water and will be soluble.

Butane has only C-C and C-H bonds, which are non-polar.  Butane will not be attracted to water and will be insoluble.

Remember “like dissolves like.” 

Factors Affecting Solubility

1. Molecule Size

Small molecules are usually more soluble than large ones.

ex. Will water solubility increase, decrease or remain the same in going from methanol ↝ ethanol ↝ propanol ↝ butanol ↝ pentanol?

methanol ↝ ethanol         ↝ propanol             ↝ butanol                      ↝ pentanol

   CH₃OH ↝ CH₃CH₂OH ↝ CH₃CH₂CH₂OH ↝ CH₃CH₂CH₂CH₂OH ↝ CH₃CH₂CH₂CH₂CH₂OH

Notice that the amount of polar region (OH) remains the same going from molecule to molecule.  Conversely, the amount of non-polar region (CH_x) is getting larger as we proceed to the right.  Thus, as the length of the carbon chain increases, the molecule struggles to maintain its attraction to water.  So, as we move from left to right, water solubility decreases.

 

Check out this video to see this in action.

2. Temperature

Temperature affects both the rate of dissolution, as well as solubility.  Quantitatively, solubility is usually given as the mass of solute dissolved in 100 mL of water at a specific temp.

Recall that for dissolution to occur, energy is required to break apart the solid solute particles.  So, an increase in temperature can expedite this process.  Thus, an increase in temperature usually increases the solubility of a solid. 

However, since the forces of attraction between liquid molecules are weaker, temperature does not have a large affect on its solubility. 

Gases already have a lot of energy and when dissolved in liquid, this energy is lost – at higher temperatures, the gas regains some energy and comes out of solution, so gases are less soluble at higher temperatures

3. Pressure

Changes in pressure do not really affect solid/liquid solutions.   

However gas/liquid solutions are affected.  The solubility of the gas is directly proportional to the pressure of the gas above the solution - that is why carbonated drinks are homogeneous when closed and bubble when opened.

Homework: Learning Check p. 368 #7-10, 12

 

 

Student Questions:
 1.  Can you do p.368 #8?
Fluoride ions have a very different electronegativity compared to the other halogens, so fluoride compounds will have different solubility than other halogen compounds.

 

2.  Hey, lady!  Please do the rest of the questions.  I wish to become as smart as possible.

 

p. 368 # 7   Benzene is soluble in fats/oils, which are non-polar.  Thus, benzene is non-polar, so it should be water-insoluble.

 

p. 368 # 9   Liquid solutes dissolving in liquid water are least affected by temperature.   Thus, liquid methanol would be more water-soluble than solid potassium chlorate or gaseous nitrogen.

 

p. 368 # 10   Gases absorb kinetic energy at the warmer temperature.  This makes the gas leave the soda.  So, the flat soda tastes different than fizzy soda.

 

p. 368 # 12   Warm surface water would not contain as much dissolved gaseous oxygen, so fish are less likely to want to hang out there.

SCH 4U - Alkyl Halides (R-X)

Commercial uses of alkyl halides include:

• CFCs (chlorofluorocarbons, like Freon) in refrigerators/air conditioners
• Teflon in non-stick surfaces
• Some are toxic and are banned – the insecticide DDT (dichlorodiphenyltrichloroethane) and the PCBs (polychlorinated biphenyls) used in electrical transformers (interesting/horrifying aside:  DDT used to be sprayed in neighbourhoods - right on the people - yikes! 😱)
  

Properties of the Alkyl Halides 
The halogen substituent on the carbon chain renders the molecule more polar, since halogens are more electronegative than C or H atoms.  The increased polarity of the C-halogen bond creates stronger intermolecular forces, resulting in higher bp than the corresponding hydrocarbons.  Also, the increased polarity makes them more soluble in polar solvents.

When organic halides are synthesized, a mix of products forms, with one to several halogens per molecule.  As the number of halogens per molecule increases, so does the molecule’s polarity, and thus boiling point.  The different molecules can be separated by fractional distillation. 

Preparing Organic Halides 
Recall from the last lesson that alkyl halides are produced in a substitution (halogenation) reaction with an alkane.  Repeated substitution may occur until all the hydrogens are replaced by halogen atoms.

Alkenes and alkynes undergo addition (halogenation, hydrohalogenation) reactions and can easily add halogens or hydrogen halides across their multiple bonds.

We can also produce a halide of a benzene ring through a substitution reaction with a halogen.

Preparing Alkenes from Alkyl Halides:  Elimination Reactions 
Alkyl halides can eliminate a hydrogen and a halide ion from neighbouring C atoms, producing a double bond in their place, becoming an alkene in the process.  A hydroxide ion is necessary to facilitate the reaction.

An elimination reaction is the reverse reaction to the addition (halogenation) of a alkene.

**Note that typically we would not show the H atom on the 2-bromopropane; it is shown to make the process of the reaction more clear.**  The Br and the H are removed by the action of the hydroxide ion, OH-.  The two carbons (that were previously bonded to Br and H respectively) form another bond to each other, resulting in a double bond in that position.  The H bonds to the OH, producing water.  The bromide ion, Br-, is also a product.

Homework #16-20

SCH 3U - Solution Prep Lab

Today, you will create a solution, using the calculations and procedure discussed last day.

The solution you could be asked to create is in the list below.  Using this data, calculate the mass of solid solute or volume of stock solute solution.  

You must get a 😊 from the teacher before you actually create your solution.  Do not lose this calculation page as you will be handing it in.

Number  Substance   Formula Conc  (M) Vol  (mL) Stk  Conc (M) 1 AgNO3 0.100 100.0   2 AgNO3 0.300 25.0   3 AgNO3 0.150 25.0   4 CaCl2 0.150 25.0   5 Ca(NO3)2￮4H2O 0.100 50.0   6 CoCl2￮6H2O 0.100 250.0   7 CrCl3￮6H2O 0.100 25.0   8 CuCl2￮2H2O 0.100 50.0   9 CuSO4 0.100 100.0   10 CuSO4 0.100 250.0   11 CuSO4 0.100 500.0   12 Fe(NO3)3 0.200 500.0   13 FeSO4￮7H2O 0.200 100.0   14 H2SO4 1.00 2000.0 18.0 15 H3PO4 1.00 100.0 16.0 16 H3PO4 1.00 250.0 16.0 17 HC2H3O2 1.00 2000.0 17.0 18 HCl 0.100 1000.0 12.0 19 HCl 1.00 1000.0 12.0 20 HCl 1.00 2000.0 12.0 21 HCl 1.00 250.0 12.0 22 HCl 3.00 1000 12.0 23 HCl 6.00 250.0 12.0 24 HCl  1.00 500.0 12.0 25 HCl  3.00 250.0 12.0 26 HNO3 1.00 100.0 16.0 27 HNO3 1.00 1000.0 16.0 28 KCl 0.250 500.0   29 KI 0.0550 500.0   30 KOH - hygroscopic 1.00 25.0   31 KOH - hygroscopic 1.00 500.0   32 KSCN  0.00200 250.0   33 Li2CO3  0.0500 250.0   34 Na2CO3￮10H2O 0.100 250.0   35 Na2SO4￮10H2O  0.100 100.0   36 Na3PO4￮12H2O  0.0250 250.0   37 NaC2H3O3 0.250 500.0   38 NaCl  1.00 50.0   39 NaI 0.100 50.0   40 NaOH - hygroscopic 1.00 1000.0   41 NaOH - hygroscopic 1.00 500.0   42 NaOH - hygroscopic 1.00 2000.0   43 NH4Cl  0.250 250.0   44 NH4OH  0.100 500.0 15.0 45 Ni(NO3)2￮6H2O 0.350 10.0   46 Pb(NO3)2 0.100 100.0   47 Pb(NO3)2 0.100 25.0   48 Pb(NO3)2 0.250 50.0   49 SnCl2￮2H2O 0.150 25.0