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Solutions. Solutions. A solution is a homogeneous mixture. A solution is composed of a solute dissolved in a solvent . Solutions exist in all three physical states:. Gases in Solution. Temperature effects the solubility of gases.

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solutions2
Solutions
  • A solution is a homogeneous mixture.
  • A solution is composed of a solute dissolved in a solvent.
  • Solutions exist in all three physical states:
gases in solution
Gases in Solution
  • Temperature effects the solubility of gases.
  • The higher the temperature, the lower the solubility of a gas in solution.
  • An example is carbon dioxide in soda:
    • Less CO2 escapes when you open a cold soda than when you open the soda warm
polar molecules
Polar Molecules
  • When two liquids make a solution, the solute is the lesser quantity, and the solvent is the greater quantity.
  • Recall, that a net dipole is present in a polar molecule.
  • Water is a polar molecule.
polar nonpolar solvents
Polar & Nonpolar Solvents
  • A liquid composed of polar molecules is a polar solvent. Water and ethanol are polar solvents.
  • A liquid composed of nonpolar molecules is a nonpolar solvent. Hexane is a nonpolar solvent.
like dissolves like
Like Dissolves Like
  • Polar solvents dissolve in one another.
  • Nonpolar solvents dissolve in one another.
  • This it the like dissolves like rule.
  • Methanol dissolves in water but hexane does not dissolve in water.
  • Hexane dissolves in toluene, but water does not dissolve in toluene.
miscible immiscible
Miscible & Immiscible
  • Two liquids that completely dissolve in each other are miscible liquids.
  • Two liquids that are not miscible in each other are immiscible liquids.
  • Polar water and nonpolar oil are immiscible liquids and do not mix to form a solution.
solids in solution
Solids in Solution
  • When a solid substance dissolves in a liquid, the solute particles are attracted to the solvent particles.
  • When a solution forms, the solute particles are more strongly attracted to the solvent particles than other solute particles.
  • We can also predict whether a solid will dissolve in a liquid by applying the like dissolves like rule.
like dissolves like for solids
Like Dissolves Like for Solids
  • Ionic compounds, like sodium chloride, are soluble in polar solvents and insoluble in nonpolar solvents.
  • Polar compounds, like table sugar (C12H22O11), are soluble in polar solvents and insoluble in nonpolar solvents.
  • Nonpolar compounds, like naphthalene (C10H8), are soluble in nonpolar solvents and insoluble in polar solvents.
the dissolving process
The Dissolving Process
  • When a soluble crystal is placed into a solvent, it begins to dissolve.
  • When a sugar crystal is placed in water, the water molecules attack the crystal and begin pulling part of it away and into solution.
  • The sugar molecules are held within a cluster of water molecules called a solvent cage.
dissolving of ionic compounds
Dissolving of Ionic Compounds
  • When a sodium chloride crystal is place in water, the water molecules attack the edge of the crystal.
  • In an ionic compound, the water molecules pull individual ions off of the crystal.
  • The anions are surrounded by the positively charged hydrogens on water.
  • The cations are surrounded by the negatively charged oxygen on water.
rate of dissolving
Rate of Dissolving
  • There are three ways we can speed up the rate of dissolving for a solid compound.
  • Heating the solution:
    • This increases the kinetic energy of the solvent and the solute is attacked faster by the solvent molecules.
  • Stirring the solution:
    • This increases the interaction between solvent and solute molecules.
  • Grinding the solid solute:
    • There is more surface area for the solvent to attack.
solubility and temperature
Solubility and Temperature
  • The solubility of a compound is the maximum amount of solute that can dissolve in 100 g of water at a given temperature.
  • In general, a compound becomes more soluble as the temperature increases.
saturated solutions
Saturated Solutions
  • A solution containing exactly the maximum amount of solute at a given temperature is a saturated solution.
  • A solution that contains less than the maximum amount of solute is an unsaturated solution.
  • Under certain conditions, it is possible to exceed the maximum solubility of a compound. A solution with greater than the maximum amount of solute is a supersaturated solution.
supersaturated solutions
Supersaturated Solutions
  • At 55C, the solubility of NaC2H3O2 is 100 g per 100 g water.
  • If a saturated solution at 55C is cooled to 20C, the solution is supersaturated.
  • Supersaturated solutions are unstable. The excess solute can readily be precipitated.
supersaturation
Supersaturation
  • A single crystal of sodium acetate added to a supersaturated solution of sodium acetate in water causes the excess solute to rapidly crystallize from the solution.
concentration of solutions
Concentration of Solutions
  • The concentration of a solution tells us how much solute is dissolved in a given quantity of solution.
  • We often hear imprecise terms such as a “dilute solution” or a “concentrated solution”.
  • There are two precise ways to express the concentration of a solution:
    • mass/mass percent
    • molarity
mass percent concentration
mass of solute

g solute

× 100% = m/m %

mass of solution

× 100% = m/m %

g solute + g solvent

Mass Percent Concentration
  • Mass percent concentration compares the mass of solute to the mass of solvent.
  • The mass/mass percent (m/m %) concentration is the mass of solute dissolved in 100 g of solution.
calculating mass mass percent
5.50 g NaCl

× 100% = m/m %

5.00 g NaCl + 97.0 g H2O

5.00 g NaCl

× 100% = 4.90 %

102 g solution

Calculating Mass/Mass Percent
  • A student prepares a solution from 5.00 g NaCl dissolved in 97.0 g of water. What is the concentration in m/m %?
mass percent unit factors
100 g solution

100 g solution

4.90 g NaCl

95.1 g water

100 g solution

100 g solution

4.90 g NaCl

95.1 g water

95.1 g water

4.90 g NaCl

95.1 g water

4.90 g NaCl

Mass Percent Unit Factors
  • We can write several unit factors based on the concentration 4.90 m/m% NaCl:
mass percent calculation
100 g solution

25.0 g sucrose ×

= 500 g solution

5.00 g sucrose

Mass Percent Calculation
  • What mass of a 5.00 m/m% solution of sucrose contains 25.0 grams of sucrose?
  • We want grams solution, we have grams sucrose.
molar concentration
moles of solute

= M

liters of solution

Molar Concentration
  • The molar concentration, or molarity (M), is the number of moles of solute per liter of solution, is expressed as moles/liter.
  • Molarity is the most commonly used unit of concentration.
calculating molarity
1 mol NaOH

18.0 g NaOH

×

= 4.50 M NaOH

40.00 g NaOH

0.100 L solution

Calculating Molarity
  • What is the molarity of a solution containing 18.0 g of NaOH in 0.100 L of solution?
  • We also need to convert grams NaOH to moles NaOH (MM = 40.00 g/mol).
molarity unit factors
1 L solution

4.50 mol NaOH

1 L solution

4.50 mol NaOH

1000 mL solution

4.50 mol NaOH

1000 mL solution

4.50 mol NaOH

Molarity Unit Factors
  • We can write several unit factors based on the concentration 4.50 M NaOH:
molar concentration problem
0.100 mol K2Cr2O7

294.2 g K2Cr2O7

×

250.0 mL solution ×

1000 mL solution

1 mol K2Cr2O7

Molar Concentration Problem
  • How many grams of K2Cr2O7 are in 250.0 mL of 0.100 M K2Cr2O7?
  • We want mass K2Cr2O7, we have mL solution.

= 7.36 g K2Cr2O7

molar concentration problem26
1 mol HCl

1000 mL solution

7.30 g HCl ×

×

36.46 g HCl

12.0 mol HCl

Molar Concentration Problem
  • What volume of 12.0 M HCl contains 7.30 g of HCl solute (MM = 36.46 g/mol)?
  • We want volume, we have grams HCl.

= 16.7 mL solution

dilution of a solution
Dilution of a Solution
  • Rather than prepare a solution by dissolving a solid in water, we can prepare a solution by diluting a more concentrated solution.
  • When performing a dilution, the amount of solute does not change, only the amount of solvent.
  • The equation we use is: M1 × V1 = M2 × V2
    • M1 and V1 are the initial molarity and volume and M2 and V2 are the new molarity and volume
dilution problem
(0.10 M) × (5.00 L)

V1 =

= 0.083 L

6.0 M

Dilution Problem
  • What volume of 6.0 M NaOH needs to be diluted to prepare 5.00 L if 0.10 M NaOH?
  • We want final volume and we have our final volume and concentration.

M1 × V1 = M2 × V2

(6.0 M) × V1 = (0.10 M) × (5.00 L)

solution stoichiometry
balanced

equation

solution

concentration

molar mass

Solution Stoichiometry
  • In Chapter 10, we performed mole calculations involving chemical equations, stoichiometry problems.
  • We can also apply stoichiometry calculations to solutions.

molarity known  moles known 

moles unknown  mass unknown

solution stoichiometry problem
187.77 g AgBr

0.100 mol AlBr3

3 mol AgBr

37.5 mL soln ×

×

×

1 mol AgBr

1000 mL soln

1 mol AlBr3

Solution Stoichiometry Problem
  • What mass of silver bromide is produced from the reaction of 37.5 mL of 0.100 M aluminum bromide with excess silver nitrate solution?

AlBr3(aq) + 3 AgNO3(aq) → 3 AgBr(s) + Al(NO3)3(aq)

  • We want g AgBr, we have volume of AlBr3

= 2.11 g AgBr

conclusions
Conclusions
  • Gas solubility decreases as the temperature increases.
  • Gas solubility increases as the pressure increases.
  • When determining whether a substance will be soluble in a given solvent, apply the like dissolves like rule.
    • Polar molecules dissolve in polar solvents.
    • Nonpolar molecules dissolved in nonpolar solvents.
conclusions continued
Conclusions Continued
  • Three factors can increase the rate of dissolving for a solute:
    • Heating the solution
    • Stirring the solution
    • Grinding the solid solute
  • In general, the solubility of a solid solute increases as the temperature increases.
  • A saturated solution contains the maximum amount of solute at a given temperature.
conclusions continued33
moles of solute

= M

liters of solution

mass of solute

× 100% = m/m %

mass of solution

Conclusions Continued
  • The mass/mass percent concentration is the mass of solute per 100 grams of solution:
  • The molarity of a solution is the moles of solute per liter of solution.
conclusions continued34
Conclusions Continued
  • You can make a solution by diluting a more concentrated solution:

M1 × V1 = M2 × V2

  • We can apply stoichiometry to reactions involving solutions using the molarity as a unit factor to convert between moles and volume.
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