Solutions And Solubility

An Introduction to Solutions and Solubility 1. Solute: a substance that is dissolved in another substance, thus forming a solution. 2. Solvent: a substance that is capable of dissolving another substance. 3. Solution: a mixture of one or more solutes dissolved in a solvent. 4. Dilute: to reduce a mixture’s strength by adding water or another substance. 5. Concentrated: of or relating to a solution whose dilution has been reduced. 6. Saturated: when a substance will no longer dissolve in a solution. 7. Unsaturated: a substance that can still be dissolved in water or another substance. 8. Electrolyte: a substance that, when in solution, will conduct an electric current. 9. Non-Electrolyte: a solution that does not conduct electricity and thus is not ionized.

Questions for Solutions and Solubility UNIT

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1. Potassium nitrate is added to four flasks of pure water until no more will dissolve. Each mixture is sealed and stirred at different temperatures. The same volume of each solution is removed and evaporated to crystallize the solid. The results are shown on the following graph.

Solubility Of Potassium Nitrate At Various Temperatures TEMPERATURE

VOLUME OF SOLUTION

MASS OF EMPTY BEAKER

MASS OF BEAKER, PLUS SOLID

0.0

10.0

92.74

93.99

12.5

10.0

91.75

93.95

23.0

10.0

98.43

101.71

41.5

10.0

93.37

100.15

(a). Plot a graph of solubility (g/ml) VS. Temperature. (b). What can you generalize about the effect of temperatures on solubility?

1(a). You can find the graph for solubility (g/ml) VS. Temperature below.

1(b). The generalization that one can perceive from the effect that temperature has on solubility is quite clear: as temperatures rise, an increase in solubility will result; however, an increase in temperature may decrease solubility.

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Molar Concentration (Molarity) Molarity is the number of mols of solute dissolved per litre of solution.

Molarity =

Number of moles of solute Volume of solution

Where “M” is the molarity, “moles” is the number of mols, and L is the volume in litres.

The Molarity Diagram Cover what you are looking for. If you cover “moles”, then you are going to multiply “M” by “L”, but if you cover “M” or “L”, then you will be dividing “moles” by “M” or “L”, depending on what is not covered.

Things to Remember: •

Molarity is expressed in units of mols/L, or M.

• •

You will sometimes have to use “KHD[ ]dcm” when dealing with molarity. When dealing with molarity problems, you will only use grams when it is asking for the mass of the chemical formula, so you may need to convert from mols to grams, or vice-versa.

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Questions for Molar Concentration (Molarity) 1. If 0.0877 mols of Copper (II) Sulfate, CuSO4, are dissolved in 0.07 L of a solution, then find the molarity of the solution.

mols = 0.0877 mols L = 0.07 L M = ?  1.3 mols/L

Molarity =

Molarity =

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Number of moles of solute Volume of solution

0.0877 mols

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0.07 L

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= 1.3 mols/L

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Dilutions Often, you are required to lessen the concentration of a solution before starting a lab. You can accomplish this by adding more solvent to the solution.

V1

V2

M1

M2

M = concentration

V = volume

Since the amount of solute remains constant on both solutions, they both contain the same number of mols.

M1 = moles/V1

M2 = moles/V2

moles = M1V1

moles = M2V2

Therefore:

M1V1

=

M2V2

This above equation is known as the dilution formula and it is very useful when doing labs.

Things to Remember: •

I = initial and F = final

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M1 and V1 are of the original solution, while M2 and V2 are the new products of dilution.

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You will sometimes have to use “KHD[ ]dcm” when dealing with Dilutions, as you must use only “L” and not “mL”.

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Questions for Dilutions 1. Determine the concentration of the solution when 25 mL of 3.0 M HCl (aq) is diluted to 65.0 mL. The formula: M1V1 = M2V2 M1 = 3.0 M (initial) V1 = 25 mL (initial) M2 = ?  1.15 M (final) V2 = 65.0 mL (final)

M2 =

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(3.0 M)(0.025 L) (0.065 L)

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= 1.15 M

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