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Monday January 9, 2023   Day 1
Solubility and Models of Solution Formation

          


Textbook Readings

11.3: Hydration of Ions (LibreText)

13.2: Types of Solutions and Solubility


Course Lectures

                  1.1 pdf   Video*   Solutions
                  1.2 pdf   Video*   Like Dissolves Like
                  1.3 pdf   Video*   Solution Formation Energetics
                  1.4 pdf   Video*   Solution Energetics Examples                  

Objectives

     1. Know and identify the different types
          of intermolecular forces and their
          effects on solubility

     2. Predict relative solubilities based on
         molecular polarity and intermolecular
         forces.

         "like dissolves like"

     3. Understand the "Energetics of solution
         formation" and the factors that make
         solution formation endothermic or
         exothermic"


Chemistry: Enthalpy of Hydration


Homework Problems


1.1 Construct Lewis structures for each of the following and identify which are polar.
     a. H2O                                     b. NH3                             c. CH3OCH3    
     d. CH3CH2OH                       e. HF

1.2 
Construct Lewis structures for each of the following and put them in order from
     least polar to most polar.


     a. CH3CH2CH2OH                     b. CH3CH2OH                                       c. CH3OH  
     d. CH3CH2CH2CH2OH             e.   CH3CH2CH2CH2CH2CH2OH

1.3 Use what you've learned from question #2 to arrange the molecules from least to
     most soluble in water, a very polar solvent.

1.4
Use what you've learned from question #2 to arrange the molecules from least to
     most soluble in liquid hexane (C6H14), a non-polar solvent.


1.5  To better understand solution formation,
     we break the process up into the three
     steps illustrated at right.  Each step is
     associated with a
ΔH value that tells us
     how much heat energy is either required
     or released for that specfic change to
     occur.

     The textbook refers to these heat energy
     terms as
ΔH1,   ΔH2   and ΔH3

     We will instead refer to these values as
    
ΔHsolvent,   ΔHsolute   and ΔHmixing
     as you see  in the figure at right.

     Describe, in your own words,  the process
     that each of these terms refers to.
Enthalpy of solution energy diagram

1.6  As you can see from the figure above, the
ΔHsolution can be determine by adding up the
    ΔH values for each of the three steps.  Calculate a value for
ΔHsolution using the values
    given above. 
    Is the process  endothermic or exothermic?


1.7   Draw the graph corresponding to the following values
ΔH values using the one above
      as a model:
 
                  
ΔHsolute   = 54 kj/mol          ΔHsolvent  = 23 kj/mol           ΔHmixing = -34 kJ/mol

      Determine
ΔHsolution and compare to the situation above.  Explain the differences.

1.8 For ionic solutions we identify   ΔHsolute      =    -ΔHlattice

     We also define    
(ΔHsolvent   +   ΔHmixing )  =    ΔHhydration . 

     Thus,    
ΔHsolution  =  -  ΔHlattice   +   ΔHhydration

     Calculate the
ΔHhydration values for LiCl and NaCl given the following information:

                     LiCl:   
ΔHlattice   =    -834 kJ/mol     ΔHsolution = -37.0 kJ/mol 
                     NaCl:   ΔHlattice   =    -769 kJ/mol     ΔHsolution =  3.88 kJ/mol

      Which of the two cations has stronger ion-dipole attractive forces with water?  Why?


Click and drag the region below for correct answers

1.1 a. polar    b. polar    c. polar    d. polar   e. polar   (That's right, they're all polar!)

1.2 
Least polar ...CH3CH2CH2CH2CH2CH2OH         
                                  
CH3CH2CH2CH2OH
                                       
CH3CH2CH2OH
                                           
CH3CH2OH
                                              
CH3OH        ...most polar

1.3
Least soluble ...CH3CH2CH2CH2CH2CH2OH         
                                  
CH3CH2CH2CH2OH
                                       
CH3CH2CH2OH
                                           
CH3CH2OH
                                              
CH3OH        ...most soluble in water

1.4 
Least soluble ...                     CH3OH
                                          
CH3CH2OH
                                       CH3CH2CH2OH
                                   
CH3CH2CH2CH2OH
                            
CH3CH2CH2CH2CH2CH2OH    ...most soluble in hexane.

1.5    ...looking forward to your explanation!  :)

1.6    
ΔHsolution  =  -35 kJ/mol    Exothermic

1.7    Graph will be different in that the final enthalpy of the solution is above the initial
         level.  C
onsequently, this will be an endothermic process with ΔHsolution = +43 kJ/mol 

1.8 
For LiCl   ΔHhydration = -871 kJ/mol
       For NaCl 
ΔHhydration = - 765 kJ/mol

       Lithium cations experience a more owerful interaction with water molecules because
       the lithium ion, Li+, is small and allows water molecules closer to the center of the
       positive charge.  The closer the water molecules get, the more energy that's released and the
       more negative the
ΔHsolution value



Tuesday January 10, 2023  Day 2
Solubility and Henry's Law

Textbook Reading

11.3 Solubility (OpenStax)



Course Lectures

2.1  pdf  Video*   Solution Formation Spontaneity
2.2  pdf  Video*   Solubility
2.3  pdf  Video*   Gas Solubility and Henry's Law

Solubility Curves


Henry's Law

Objectives

1. Define the terms unsaturatedsaturated  &  super saturated

2. Identify a solution as unsaturated,  saturated or supersaturated using solute and solvent
    information and a solubility curve.

3.  Describe how changes in pressure and temperature affect the solubility of solutes in solvents

4. Use solubility curves to predict the maximum amount of solute that can be dissolved
    in a prescribed amount of solute.

5. Use Henry's law to determine the solubility of  a gas using the  Henry's law gas constant
     and known gas pressures.

Homework Problems

2.1 Referring to the solubility graph at right,
    how many grams of KCl  will dissolve in
    100. grams of water at 40oC?

2.2   70. grams of KCl are added to 100. grams
    of water at 10oC. How many grams of KCl
    will dissolve and how many grams will remain
    undissolved at the bottom of the beaker at
    this temperature?

2.3 80.0 grams of KCl are added to 200. grams
    of water at 50oC.   The resulting solution
    is unsaturated.  Why?

2.4  Does the presense of solid KCl at the
     bottom of a KCl solution guarentee the  
     solution is saturated?  Why?  Why not?
KCl Solubility curveKCl Solubility curve

2.5 A solution formed by mixing 40.0 grams of KCl in 100. grams of water at 20 oC should be
    supersaturated according to the graph above.  However, simply mixing the KCl and water will
    not produce a supersaturated solution.  What additional steps are required to prepare the
    supersaturated solution?

2.6 Because it's non-polar, oxygen isn't very
    soluble in water.  However, enough
    oxygen does dissolve to support
    aquatic life. 

    Use the graph at right to determine the
    mass of oxygen gas dissolved in 1 liter of
    water at 25oC and 35oC.
   

2.7.  Will a fish that requires 12 mg/L of
     dissolved oxygen to survive be able to live
     in water at 10 oC?
Oxygen Solubility vs temperature
https://www.ck12.org/book/ck-12-chemistry-basic/section/16.1/
2.8 Just like O2, CO2 is a non-polar molecule that shouldn't dissolve very much in liquid
        water, a polar solvent.  However,  by increasing the pressure of CO2 at the surface of the liquid
        more CO2 can be dissolved.  This idea is the basis of beverage "carbonation"

         Why does increasing the pressure of the carbon dioxide (or any other) gas, increase its solubility?

2.Henry's law allows us to determine how much gas will dissolve in a specific amount
           of water.
  
           How many grams of carbon dioxide gas is dissolved in a 1.0 L bottle of carbonated water if the
           manufacturer uses a pressure of 2.4 atm in the bottling process at 25 °C?
   
            Useful information: k
H of CO2 in water = 3.4 x 10-2 M/atm at 25 °C

2.10 Oxygen is much less soluble in water than carbon dioxide,  0.00412 g/100 mL at 20 °C
          and 760. mm Hg.

           Calculate the MOLAR solubility of oxygen gas in water at 20 °C
           and a pressure of 1150 mm Hg.

 Click and drag the region below for correct answer

2.1  Approximately 40 grams of KCl will dissolve in 100. g of water at 
40oC

2.2  30 grams of KCl should dissolve at this temperature leaving 40 grams as undissolved ppt
       found at the bottom of the container.

2.3  
  80 grams of KCl in 200 g of water is equivalent to 40gKCl/100gH2O.  At 50oC, the solubility
         of KCl is approximately 43
gKCl/100gH2O.  Thus, the solution still has room for additional
         solute and is considered unsaturated.

2.4   In most cases, the presense of a solid solute at the bottom of the container suggests the
        solution is saturated.  However, to be sure, the solution should be thoroughly stirred
        at constant temperature.

2.5  A supersaturated solution is prepared by dissolving a lot of solute in solvent by increasing
       the temperature.  Then the solution is slowly cooled in hopes that as the saturation temp.
       is reached, the dissolved solute STAYS dissolved.  Sometimes it works and the resulting
       solution is supersaturated.  A slight bump, seed crystal, or dust speck is all that's required
       to "trigger" the supersaturated solution causing it to release the solute that isn't supposed
       to be dissolved.

2.6   Here's the graph.   At 25oC, S = 8.1 mg/L     ...   at  35 oC,  S = 7.0 mg/L
        This demonstrates that the solubility of a gas actually decreases as temperature increases.

2.7    The O2 solubility at this temperature is approximately 10.8 mg/L  (See graph).
         This level is below the required 12 mg/L for the fish and will be too low for survival.

2.8    The amount of CO2 dissolved in the liquid solvent depends on the number of collisions
         gas phase
CO2 experiences with the liquid's surface.  Higher gas pressures mean more
         frequent
CO2/liquid-surface collisions and an increased likelyhood of a CO2 molecule
         dissolving in the solvent.

2.9  3.59 grams of
CO2

2.10   1. 95 x 10-3 mol/L.

Wednesday January 11, 2023
Molarity Manipulations and Dilution

Textbook Readings

6.3 Molarity



Course Lectures

3.4  pdf  Video*   Solutions: Molarity and Titrations
3.1  pdf  Video*   Concentrations and Conversions


Molarity (Chemistry Solution)
Dilution (Chemistry Solution)

Objectives

1. Calculate the molarity of a solution  given moles or grams of solute and total solution volume.
                                        Molarity = molessolute/Literssolution
2. Determine moles, volume or Molarity given any two parameters.
3. Determine the new concentration after dilution (addition of pure solvent)
4. Determine the volume of solvent required to dilute a solution to obtain a new concentration
5. Determine the concentration of a solution made by mixing two different solutions of
    different concentrataions.
6. Determine the individual ion concentrations of species present in solutions.


Homework Problems

3.1  What is the molar concentration of the solution formed when 2.54 grams of KNO3 are
       dissolved in 325 mL of water?

3.2  145 mL of distilled water is added to 500. mL of 4.50 M HCl. 
       What is the new concentration of the solution?

3.3  How many mL of distilled water must be added to 335 mL of 0.500 M NaOH to lower the
       solution's concentration to 0.100 M?

3.4  What are the individual ion concentrations present in 4.50 M K3(PO4)2?

3.5   Why must you never use the dilution equation, M1V1 = M2V2 in a titration calculation?
 

Click and drag the region below for correct answers

3.1    0.0773 M                                      
3.2     3.49 M         
3.3     1340 mL additional water        
3.4      [ K+] = 13.5 M            [PO4-3] = 9.00 M
3.5   The dilution equation assumes a 1:1 mole ratio.  However, in many titration calculations,
         the mole ratio is NOT 1:1.

Thursday January 12, 2023   Day 4
Concentration:   
Molality ... Mole Fraction  ... Mass Percent

Textbook Readings

11.4 Colligative Properties




Course Lectures

3.1  pdf  Video*   Concentrations and Conversions

i. Molality

ii. Mole Fraction


Objectives

1.  Use the Molarity definition to determine
     molessolute, volumesolution or Molarity
     of the solution.

            Molarity = molessolute/Literssolution

2. Use the Molality
definition to determine
     molessolute, volumesolution or
Molality
     of the solution.

            Molality = molessolute/kgsolvent
iii. Mass Percent



3. 
Use the Mass Percent definition to determine  masssolute, masssolvent or Mass Percent
     of the solute and solvent

             Mass %  =
masssolute / masssolution  x 102

4.   Use the mole fraction (X) definition to determine molessolute, moles solvent or mole fraction
      of the solute and solvent.

                     Xsolute = molessolute/molesTotal             
Xsolvent = molessolvent/molesTotal

5.  Use the ppm or ppb definitions to determine
masssolute, masssolvent or ppm/ppb of solute.

                  ppm = masssolute / masssolution  x 106           ppb = masssolute / masssolution  x 109


Homework Problems

4.1  
ppm is approximately the same as mgsolute/LiterH2O.  Why is it safe to assume the mass
      of the solution is the same as the mass of the water??

4.2
What are the NaCl mass percent, ppm and ppb levels in a solution that is prepared by
     dissolving 0.30 mg of NaCl in 1.00 Liter of water? (assume Dsolution = DH2O)

4.3  How are molarity and molality concentration units the same?  How are they different?

4.4 Consider 350. mL  of a 0.100 M Ca(NO3)2 solution
     a.  How many moles of
Ca(NO3)2  are there in this solution?
     b. 
How many grams of Ca(NO3)2  are there in this solution?
     c.  What is the molar concentration of the solution after the addition of
           500. mL of distilled water?

4.5  Your goal is to prepare a 0.500 m KCl solution in 350. mL of water.
     a.  How many kg of solvent will there be?
     b.  How many moles of KCl must you add to the water?
     c.   How many grams of KCl
must you add to the water?
     d.   What is the mass of the solution once you've made it?


4.6  Use your results from question 5 to determine the mole fraction of KCl (XKCl) and
       mole fraction of H2O (XH2O).  What do the two mole fraction values add up to?

4.7  Describe the actual steps taken to prepare 500. mL of a 1.50 M CuSO4 solution?

4.8  Describe the actual steps taken to prepare approximately 500. mL of a  1.50 m CuSO4 solution?
     


Answers:  Click and drag in the space below.

4.1  For dilute solutions, such as those appropriately described by ppm or ppb concentration
       units, the mass of dissolved solute is so low that it is reasonable to assume the mass of the
       entire solution is approximately equal to the mass of the solvent.

4.2    3.00 x 10-5 %,        0.300 ppm,         300. ppb

4.3   Molarity and molality both utilize "moles" of solute in their determinations.  However,
        molarity is calculated using the solution's volume in liters whereas molality is determined
        using the solvent's mass in kg.

4.4.   a.  0.0350 moles     b. 5.743 grams     c.  0.04117 M

4.5   a.  0.350 kg H2O (...using DH2O = 1.00 g/mL @ 25oC) 
        b.  0.175 moles of KCl
        c.  13.04

        d.   363 grams

4.6  XKCl = 0.00892    XH2O = 99.1   Mole fractions add up to 1.00

4.8  In a large beaker, weigh out 500. grams of distilled water.

       Add 119.71 g (0.750) moles of
CuSO4  and stir until all solid is dissolved.
    



Friday January 13, 2023   Day 5
Concentration Conversions

Textbook Readings

11.4 Colligative Properties

Course Lectures

3.1  pdf  Video*   Concentrations and Conversions

Molarity-Molality-Mass percent

Mole Fraction & Solution Concentration
Practice Problems

Objectives

1. Convert between concentration  systems using...
       a.  solution density
       b.   the 100 g assumption
       c.   the 1.00 Liter assumption

Homework Problems

5.1  Concentrated hydrochloric acid is  received as an 11.65 M solution with a density of  1.180 g/mL.
        a. Assume V = 1.00 L and determine the mass of the solution
        b. Assume V = 1.00 L and determine the moles & grams of HCl present in the solution.
        c.  Use your results from a & b to determine the mass & moles of H2O present in the solution
        d.  Use your results above to calculate the molality and mole fraction for HCl.
        e.  Why was it both acceptable and important to assume V= 1.00 L in the above calculations?

5.2  An aqueous NaCl solution  is 2.55 %NaCl by mass.
        a. Assuming there are 100 grams of solution, determine the mass of NaCl and
H2O
             present in the solution
        b.  Use your results from a & b above to determine the moles of NaCl and H2O present in
              the 100 gram solution sample.
        c.  Use your results from "b"  to determine the molality of the solution , XNaCl
& XH2O
        d.  This is a very dilute solution.  Make an appropriate assumption and
             determine the molarity of the solution.


5.3  Concentrated acetic acid solution is 99.5% by mass CH3COOH and has a density of 1.05 g/mL
       What is the molarity  and X
CH3COOH of the solution?  (Assume water is the solvent)

5.4   Consider an aqueous solution of Ca(NO3)2 where the mole fraction,  XCa(NO3)2 = 0.25
         What is the molality of the solution?

Answers:  Click and drag in the space below.

5.1    a. 1180. gramssolution                    
         b. 11.65 molesHCl  =   424.759 gramsHCl
         c. 755.241 gramsH2O   =   41.9222 molesH2O    
         d. 15.425 m        XHCl = 0.2175
         e.  The solution's concentration is independent of the volume we consider. 
              Assuming 1L as the solution's volume, we can easily calculate the solution's mass.
              Additionally, the assumption let's us know there are 11.65 moles of solute present which
              is necessary for later calculations.

5.2  a.   2.55 grams NaCl              97.45 grams H2O
       b.   molNaCl = 0.043634         molH2O = 5.409297
       c.   molality = 0.4477 m      
XNaCl  =   8.002 x 10-3   &    XH2O  = 0.9919
       d.   Assume solution has same density as water (i.e. the solution is very dilute)
              Molarity = 0.4363 M

5.3    [
CH3COOH] =  17.399M   (Note the brackets are used to designate Molar Concentration)
          X
CH3COOH = 0.9835

5.4     18.50 m


















































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