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CHAPTER 15 SOLUTIONS Powered By Docstoc
					                    CHAPTER 15: SOLUTIONS

Qualitative description of solution
Quantitative description of solution
Making dilutions
Solution stoichiometry
Qualitative description of solution

     Homogeneous mixture

     Solvent = substance present in largest amount

     Solute = the other substance(s)

     All the same phase (solid, liquid, gas)


           AIR: composed of _____ & ____

           LATTE: composed of _____ & _____ & ______

           BRASS: composed of _____ & _____

   In each of the above cases, which is the solvent and which is the

   What is the solvent in an aqueous solution?

How are mixtures formed?

RECALL: Like dissolves like

  NaCl in water:   anions and cations more attracted to dipole of water
  than to each other

  Ethanol in water: both molecules have OH groups so ethanol can

  form hydrogen bonds with water

  Grease in acetone: both molecules are nonpolar
  Oil in water:   What happens? Why?

  Is paint a solution? Why or why not?

How do you indicate how much solute is dissolved in a

  Dilute              =       “not much”                           little

  Concentrated        =       a relatively large amount

  Saturated           =       as much as can dissolve at given T

  Supersaturated: How does that happen?                            lots

Quantitative description of solution

      Mass percent (or weight percent):      units = %

      Molarity: units = moles/liter = M

Mass percent         = mass solute x 100%
                   mass solution

                  =        grams solute                   x 100%
                      grams solute + grams solvent

How do I make one of these solutions? Weigh out the amounts of solute
and of solvent, and mix together.

Example: dissolve 1.0 grams of sodium chloride in 48. grams of water.
          Grams solute = 1.0
          Grams solvent = 48.
          Grams solute + solvent = 1.0 + 48. = 49.

                  1.0   x 100% = 0.020 x 100% = 2.0% NaCl

Look at it from another angle: a 135. g sample of seawater is evaporated
to dryness, leaving 4.73 g of solid residue (the salts that were dissolved in
the seawater). Calculate the mass percent of solute present in the original

To solve the problem, answer these questions:

      What is the solute and what is its weight?
      What is the solvent and what is its weight?
      What is the solution and what is its weight?

And the answer is……

What if you want to find the mass of solute necessary to prepare a solution
with a given mass % ? For example, what mass of water must be added
to 425. g of formaldehyde to prepare a 40.0% (by mass) solution of

What do you know?
     Mass of solute
     Desired mass %

Substitute into the expression for mass % and solve for the unknown.

And the answer is….

Molarity = M           = moles of solute = mol
                          Liters of solution L

1.0 M = 1.0 molar = 1.0 mol of solute per liter of solution

How do I make one of these solutions?
  1. Pick a volumetric flask of the desired final volume (i.e., 1 liter)
  2. Put some solvent in the flask.
  3. Weigh out the desired amount of solute and add to flask.
  4. Add solvent to the mark on the flask.

Four cases:
1. Given grams of solute and final desired volume, calculate molarity
2. Given molarity of salt dissolved in solvent, calculate the molarity of
   each ion
3. Given the molarity of a solution, calculate the number of moles of
   solute in a given volume of solution
4. Given the desired molarity of a solution, calculate the mass of solute
   required to obtain that molarity

What is the molarity of a solution made by dissolving 1.00 g of ethanol,
C2H5OH, in enough water to give a final volume of 101 mL?

  1. What do we know?
     We have 1.00 g of ethanol
     Final volume is 101 mL

  2. What do we need to find out?
     The moles of solute per liter of solvent

  3. Determine which is the solute and which is the solvent:
     Ethanol is the solute, water is the solvent

  4. Convert grams of solute to moles:
     MW of ethanol = (2 x 12)+ (6 x 1) + (1 x 16) = 46 g mole-1
     Moles of ethanol = (1.00 g/46 g mole-1) = 0.0217 moles

  5. Convert volume of solution from mL to L, if necessary:
     101 mL x 1 L = 0.101 L
              1000 mL
  6. Divide moles of solute by liters of solution:
     0.0217 moles ethanol = 0.215 M

What are the concentrations of ions in a 0.010 M solution of Al2(SO4)3?

  1. What do we know?
     The moles of aluminum sulfate per liter of solution

  2. What do we need to find?
     The moles of each ion per liter of solution

  3. Determine what ions are produced when the salt dissolves. Don’t
     forget stoichiometry!:
     Al2(SO4)3 → 2Al+3 + 3SO4-2

  4. Write mole ratios for each ion:
        2 mole Al+3                   3 mole SO4-2
     1 mole Al2(SO4)3                1 mole Al2(SO4)3

  5. Use the ratios to determine concentrations:
     2 mole Al+3    so     2 M Al+3   x 0.010 M Al2(SO4)3 = 0.020 M Al+3
     1 mole Al2(SO4)3 1 M Al2(SO4)3

      What is the concentration of SO4-2?

How many moles of Cl- ions are present in 1.75 L of 1.0 x 10-3 M AlCl3?

  1. What do we know?
     The molarity of the solution
     The volume of the solution

  2. What do we need to find?
     The moles of anions in a given volume of solution

  3. Determine the moles of AlCl3 present by multiplying the total volume
     by the molarity:
     1.75 L x 1.0 x 10-3 mol AlCl3 = 1.75 x 10-3 mol AlCl3
  4. Determine what ions are produced when the salt dissolves. Don’t
     forget stoichiometry!:
     AlCl3 → Al+3 + 3Cl-1

  5. Write mole ratio for the ion of interest:
      3 mole Cl-1
     1 mole AlCl3

  6. Use the ratio to determine moles:
     3 mole Cl-1    x 1.75 x 10-3 mol AlCl3 = 5.25 x 10-3 mol Cl-1
     1 mole AlCl3

How many grams of formaldehyde, HCHO, must be used to prepare 2.50 L
of 12.3 M formalin (aqueous solution of formaldehyde)?

  1. What do we know?
     The desired moles formaldehyde per liter of solution
     The desired final volume

  2. What do we need to find?
     Grams of formaldehyde (solute) needed

  3. Determine the moles of formaldehyde required by multiplying the final
     volume times the concentration:
     2.50 L x 12.3 mol formaldehyde = 30.75 mol formaldehyde
                L solution

  4. Convert moles of solute to grams:
     MW of HCHO = (1 x 12)+ (2 x 1) + (1 x 16) = 30 g mole-1
     grams of HCHO = 30 g mole-1 x 30.75 mol = 923 grams

Making Dilutions

Concentration indicates a certain number of moles of solute per liter of

                        Dilute by adding solvent:

Initial solution              Add water              Diluted solution
(10 molecules in 1 L)         (+ 1 L)                (10 molecules in 2 L)

  MMMM                         MMMM
   M                            M                           M  M
  MMM                          MMM                            M   M
    M M                          M                              M
                                                            M M
                          +                                   M
                                                            M    M

  Dilute by taking out a volume of solution and adding solvent to
                          desired volume:

Initial solution              Remove ½ L                   Add ½ L solvent
(10 molecules in 1 L)

  MMMM                          M                           M
   M                                                            M
  MMM                                                           M
    M M                   -                                 M M
                               MMM                            X
                                 M M

A short cut:
               M1 x V1 = M2 x V2

   Where M1 = initial (before dilution) concentration
         V1 = amount of concentrated solution used
         M2 = final (diluted) concentration
         V2 = final (desired) volume

   Check yourself:

      M     > M2 because you are diluting the solution (reducing

      V   1< V2 because you are using a small amount of the
         concentrated solution to make a larger amount of dilute solution

      V   1       and V2 must have the same units ( either both mL or both L)!

Try this: What volume of 12 M HCl must be used to prepare 0.75 L of 0.25
M HCl?

   1. What do we know?
      M1 = 12 M
      M2 = 0.25 M
      V2 = 0.75 L

   2. What do we need to find?

   3. Rearrange the equation:
      V1 = M2 x V2/ M1

   4. Substitute in the known values to solve. What are the units of V1?

Solution Stoichiometry: How to do stoichiometry when you
are using solutions instead of pure solids, liquids, or gases

Follow the same basic steps as you did before (chapter 9)!
  1. Write the balanced equation for the reaction.

  2. Calculate the moles of the known ingredients.

  3. Determine which, if any, reactant is limiting.

  4. Calculate the moles of the desired unknown.

  5. Convert moles to grams or concentration, as required.

Example: When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed,
PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.25 L of
0.500 M Pb(NO3)2 and 2.00 L of 0.025 M Na2SO4 are mixed.

  1. What do we know?

     Pb(NO3)2 + Na2SO4 → PbSO4 + ?
     0.500 M    0.025 M
     1.25 L     2.00 L

  2. What do we need to find?
     Grams of PbSO4
      you aren’t asked about the other product
  3. Balance the equation, if necessary (this equation is already

  4. Determine the number of moles of each reactant.
     (0.500 mole Pb(NO3)2/L) x 1.25 L = 0.625 mol Pb(NO3)2
     (0.025 mole Na2SO4/L) x 2.00 L = 0.0500 mol Na2SO4

5. Use mole ratios to determine the limiting reagent.
   1 mol Pb(NO3)2 = 0.625 mol Pb(NO3)2
   1 mol Na2SO4           x mol Na2SO4

  You need 0.625 mol Na2SO4, but you only have 0.0500 mol.
  Therefore, Na2SO4 is the limiting reagent.

   you can use either reagent to determine which is limiting
6. Use the limiting reagent to determine moles of product.
   1 mol PbSO4 x 0.0500 mol Na2SO4 = 0.0500 mol PbSO4
   1 mol Na2SO4

7. Convert mol of PbSO4 to grams of PbSO4.
   0.0500 mol PbSO4 x 303.3 g PbSO4 = 15.2 g PbSO4

Neutralization is when just enough acid and base react to
form a solution that is neither acidic nor basic.

 Review how to write net ionic equations, chapter 7.3 and acid-base
reactions, chapter 7.4

                 HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)

How do I know that I have added just enough?

    By using an indicator, a chemical that changes color when the amount
    of acid (or base) in a solution changes.

Say I have 25.0 mL of HCl(aq) and I want to find its concentration (moles
HCl per liter of solution). What do I do?

  1. Add a drop or two of indicator to the acid solution.
  2. (can be optional) Add some water to the HCl to make a larger
     volume, so the color change will be easier to see. Say we add 25.0
     mL, for a total final volume of 50.0 mL.

  Does this change the number of moles of HCl in the flask?

  3. Pour NaOH(aq) of known concentration (say, 0.10 M) into a burette.
     A burette allows you to add NaOH dropwise to the HCl, so you can
     stop adding NaOH as close as is possible to when the acid is
     neutralized (indicator changes color), and measure the amount of
     NaOH added to 0.1 mL accuracy.
  4. Add the NaOH to the HCl and watch for a color change. Swirl the
     flask to make sure the two chemicals are completely mixed. When
     you observe a permanent color change (one that doesn’t disappear
     when you swirl the flask), the neutralization is complete.
  5. Record the mL of NaOH(aq) used (say it was 50.0 mL).
  6. Now it’s time to calculate!

Go back to the original equation:

                 HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)

What do you know?

                 HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
                 25.0 mL    50.0 mL
                 xM         0.10 M

Do you need to know anything about the products?

Determine the moles of NaOH used.

From the stoichiometry, you know that HCl(aq) reacts with NaOH(aq) in a
one to one ratio, so moles HCl= moles NaOH.

The concentration of HCl, then, is:

     Moles HCl x 1000 mL = ?
      25.0 mL      1L

Why do you use the initial volume of HCl(aq), not the diluted volume used
in the titration?


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