Humboldt State University ® Department of Chemistry

Richard A. Paselk
Chem 109

General Chemistry

Spring 2009

Lecture Notes:: 17 February

© R. Paselk 2002
  
     
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Molarity and the Composition of Solutions, cont.

There are a number of different kinds of problems.

Making molar solutions

Make up a 1.000 L solution of 0.25 M NaCl (note that water is the "default" solvent).

First weigh out 0.25 moles of NaCl

= (0.25 mole)(22.99 g + 35.45 g)/mole = 14.61 g

Next we add water until the total volume is 1.000 L.

Determining Molar Concentration

What is the concentration of a solution made by dissolving 10.00 g of KI in enough water to make 1.000 L

First need to find the number of moles of KI:

(10.00 g) / ({39.10 g+ 126.9 g}/mole) = 6.024 x 10-2 mole

Thus the concentration will be 6.024 x 10-2 M

Dilution problems

What is the concentration of a solution resulting from adding 25.0 mL of 0.60 M CaCl2 to 475 mL of water. (Note - both aqueous, so the volumes are additive.)
First, I like to keep in mind that Moles = Moles, that is we have conservation of mass.

With this in mind, Moles1 = (volume1)(Molarity1) = Moles2 = (volume2)(Molarity2)

(Note the units of volume will cancel, so we don't HAVE to convert to L, though it doesn't hurt.)

Can solve as a ratio: (0.025 L) (0.60 M) = (0.475 L) (x)

x = (0.60 M)(0.025 L)/(0.500 L) = 3.00 x 10 -2

= 3.0 x 10-2M 

What is the concentration of chloride ion in this solution?

As an inorganic salt, we know it dissociates completely, thus from the formula = CaCl2

we see that there are (2 Cl-) / (CaCl2)

Therefore [Cl-] = 6.0 x 10 -2M

How much 1.000 M MgSO4 is needed to make 500.0 mL of a 0.25 M solution.

First convert volumes to Liters: 500.0 mL = 0.5000 L

Can solve as a ratio: (0.5000 L) (0.25 M) = (1.0000 M) (x)

x = (0.25M)(0.5000 L)/(1.0000 M) = 0.125 L = 0.13 L

Acid-Base Reactions

Neutralization

When we combine equal numbers of moles of hydrogen ion and hydroxide ion a neutralization occurs. That is, there is no reactive component left, all of the acid has been consumed by all of the base, and water has been synthesized.

Consider the reaction of 50.0 mL of 0.25 M hydrochloric acid with 25.0 mL of 0.50 M sodium hydroxide.

H+ + Cl- + Na+ + OH- H2O + Cl- + Na+

Giving: H+ + OH- H2O

Consider the reaction of a weak acid and strong base: 50.0 mL of 0.25 M acetic acid is reacted with 18.0 mL of 0.50 M sodium hydroxide. Find the number of moles of each of the reactants and products after reaction.

CH3COOH + OH- H2O + CH3COO-

  • First find moles of each reactant:
    • acid = (50.0 mL)(1 L/1000 mL)(0.25 mole/L) = 1.25 x 10-2moles
    • base = (18.0 mL)(1 L/1000 mL)(0.50 mole/L) = 0.900 x 10-2moles
  • After reaction all of the base is consumed, so:
    • base = 0
    • acetate = 0.900 x 10-2moles
    • acid = 1.25 x 10-2moles - 0.900 x 10-2moles = 0.35 x 10-2moles
  • Water synthesized = 0.900 x 10-2moles.

Oxidation Numbers

For simple elemental ions it is easy to determine the charge on an atom, but in many other circumstances this is not the case. In order to name compounds and understand reactions we frequently need this information which is obtained from oxidation numbers.

Oxidation numbers are in essence an electronic accounting method in which electrons are assigned to a particular atom in a bond or interaction. As such they give an approximate picture of where electrons actually reside in compounds. We will find this information very useful later when we look at particular types of chemical reactions. Oxidation numbers are essential for nomenclature.

For simple elemental ions it is easy to determine the charge on an atom, but in many other circumstances this is not the case. In order to name compounds and understand reactions we frequently need this information which is obtained from oxidation numbers.

Oxidation numbers are most readily assigned using a simple set of rules:

In the formula for any substance the sum of the oxidation numbers of all the atoms in the formula is equal to the charge shown. Thus:

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Last modified 16 February 2009