Solutions
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Solutions. Solution : homogeneous mixture components are uniformly intermingled on a molecular level Solutions can be solid: brass (zinc in copper) liquid : salt water, sugar water, etc gas : Air (oxygen & others in nitrogen). Solutions. Unsaturated solution:

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Solutions

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Solutions

Solutions

  • Solution:

    • homogeneous mixture

    • components are uniformly intermingled on a molecular level

  • Solutions can be

    • solid:

      • brass (zinc in copper)

    • liquid:

      • salt water, sugar water, etc

    • gas:

      • Air (oxygen & others in nitrogen)


Solutions1

Solutions

  • Unsaturated solution:

    • a solution that is capable of dissolving more solute

  • Saturated solution:

    • a solution that is in equilibrium with undissolved solid

  • Supersaturated solution:

    • a solution that contains more dissolved solute than is needed to form a saturated solution


Solubility

Solubility

Example: Which of the following solutes would you expect to be soluble in water:

CH3CH2CH3:

CH3CH2OH:

HCl:

Vitamin AVitamin C

Remember: Substances with similar intermolecular forces tend to dissolve in each other


Solubility of gases

Solubility of Gases

  • The solubility of a gas in a solvent depends on the nature of the solute and solvent, the temperature, and the pressure.

  • In general, the solubility of gases in water increases with increasing molar mass.

    • Larger LDF


Solubility of gases1

Solubility of Gases

  • The solubility of a gas in a solvent increases as the pressure of the gas over the solvent increases.

  • Henry’s Law:The solubility of a gas in a solvent is directly proportional to its partial pressure above the solution.

    Cg = kPg

    where Cg = solubility of the gas in the solution phase

    Pg = partial pressure of the gas

    k = proportionality constant (value depends on solute, solvent, and temperature


Solubility1

Solubility

Example: Calculate the concentration of CO2 in a soft drink that was bottled with a partial pressure of carbon dioxide of 3.5 atm over the liquid at 25oC. (k = 3.1 x 10-2 mol/L.atm)


Solubility of gases2

Solubility of Gases

Example: Why does a bottle of soda bubble when the cap is first removed?

  • Carbonated beverages like soda are bottled under a carbon dioxide pressure slightly greater than 1 atm.

  • Opening the bottles, reduces the partial pressure of CO2 above the soda.

    • Solubility of CO2 decreases so CO2 bubbles out of the solution.


Solubility of gases3

Solubility of Gases

  • The solubility of solid solutes generally increases with increasing temperature.


Solubility of gases4

Solubility of Gases

  • The solubility of a gas in a solution decreases with increasing temperature.

    • Gas molecules have greater KE and can escape from the solution more easily.


Solubility of gases5

Solubility of Gases

Example: A warm bottle of soda tends to taste “flat” compared to a cold bottle of soda. Explain why.


Concentration

Concentration

  • Several different units can be used to express the concentration of a solute in a solution:

    • mass (weight) percent

    • parts per million (ppm)

    • parts per billion (ppb)

    • mole fraction

    • Molality

    • Molarity

      • Varies with temperature

Independent of temperature


Concentration1

Concentration

  • Mass Percent = mass of component x 100

    total mass of sol’n

Example: A solution is prepared by dissolving 6.8 g of NaCl in 750.0 g of water. What is the mass percent of the solute?

Mass % NaCl = ___g NaCl____ x 100

g NaCl + g H2O

= ___6.8 g____ x 100 = 0.90 %

6.8 g + 750.0 g


Concentration2

Concentration

  • ppm = mass of component x 106

    total mass of sol’n

Example: A 10.25 g sample of lake water contains 1.28 x 10-2 mg of arsenic. What is the concentration of arsenic in ppm?

ppm As =0.0128 mg As x __1 g As_ x 106

10.25 g water 103 mg As

ppm As = 1.25 ppm


Concentration3

Concentration

  • ppb = mass of component x 109

    total mass of sol’n

Example: A 225 g sample of lake water contains 1.2 mg of pesticide. What is the concentration of pestcide in ppb?

ppb =1.2 mg pest. x _1 g _ x 109

225 g water 106mg

ppb = 5.3 ppb


Concentration4

Concentration

  • Mole fraction = moles of component total moles of all components

Example: Calculate the mole fraction HCl present in a solution prepared by dissolving 0.25 mol HCl in 9.50 mol H2O.

XHCl = ___moles HCl____

mol HCl + mol H2O

XHCl = ___0.25 mol ______ = 0.026

0.25 mol + 9.50 mol


Concentration5

Concentration

  • Molality = m =moles of solute

    kg solvent

Example: Calculate the molality of a solution prepared by dissolving 1.25 g of sodium hydroxide in 250 g of water.

m = _moles NaCl_

kg H2O

m = 1.25 g NaCl x 1 mol NaCl x 103 g H2O

250 g H2O58.5 g NaCl1 kg H2O

m = 0.085 m


Concentration6

Concentration

  • Molarity = M =moles of solute

    L solution

Example: Calculate the molarity of a solution that contains 73.0 g of HCl per 250 mL of solution.

M = _moles HCl_

L soln

M = 73.0 g HCl x 1 mol HCl x 103 mL

250 mL soln36.5 g HCl1 L

M = 8.0 M


Concentration7

Concentration

  • Why does molarity vary with temperature??

  • You must be able to interconvert between the different concentration units.


Concentration8

Concentration

Example: An aqueous solution of sodium hydroxide contains 4.4% NaOH by mass. Calculate the mole fraction and molality of the solution.

To find mole fraction:

Given: 4.4 g NaOH per

100.0 g solution

Find: XNaOH

XNaOH = mol NaOH

total mol


Concentration9

Concentration

mol NaOH = 4.4 g NaOH x 1 mol = 0.11 mol

40.0 g

grams H2O = 100.0 g – 4.4 g = 95.6 g

mol H2O = 95.6 g H2O x 1 mol = 5.31 mol

18.0 g

XNaOH = ___0.11 mol______ = 0.020

0.11 mol + 5.31 mol


Concentration10

Concentration

To find molality:

Given: 4.4 g NaOH per

100.0 g solution

Find: m

m = mol NaOH

kg solvent

m = 0.11 mol NaOH x 1000 g H2O = 1.2 m

95.6 g H2O1 kg H2O


Concentration11

Concentration

Example: Calculate the molarity of a 1.50 m solution of toluene (C7H8) in benzene if the solution has a density of 0.876 g/mL.

Given:1.50 mol toluene

per 1 (exact) kg benzene

d = 0.876 g/mL

Find:M = mol toluene

L solution


Concentration12

Concentration

Solution:

Answer: 1.15 M


Concentration13

Concentration

  • The dilution equation is used to calculate either

    • the new concentration of a solution prepared by diluting a stock solution

      Or

    • The volume of a stock solution needed to prepare a known volume of a more dilute solution

      C1V1 = C2V2


Concentration14

Concentration

Example: Calculate the molarity of a solution prepared by diluting 225 mL of 1.5 M KMnO4 to a total volume of 850.0 mL.

Given:

Find:

Solution:

Answer: 0.40 M


Concentration15

Concentration

Example: Describe in detail how you would prepare 500.0 mL of 0.60 M HCl from a 12.0 M HCl stock solution?


Concentration16

Concentration

Example: A solution is prepared by dissolving 1.50 g of sodium chloride in enough water to give 250.0 g of solution. A 25.0 g aliquot of this solution was then diluted with water to a total mass of 100.0 g. Calculate the weight percent sodium chloride present in the final solution.


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