Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2011

Lecture Notes 12: 14 February

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Balancing Redox Equations, cont.

Basic Solution:

Example: Balance the equation above in basic solution (I don't believe this reaction actually occurs in basic solution, but due to our time constraints we'll pretend it does.)

First we balance as above to give:

16 H+ + 2 MnO4- + 10 Cl- right arrow 2 Mn2+ + 8 H2O + 5 Cl2

Next add 16 OH- to each sides to cancel H+

16 OH- + 16 H+ + 2 MnO4- + 10 Cl- right arrow 2 Mn2+ + 8 H2O + 5 Cl2 + 16 OH-

Combine OH- and H+ to give 16 H2O

16 H2O + 2 MnO4- + 10 Cl- right arrow 2 Mn2+ + 8 H2O + 5 Cl2 + 16 OH-

canceling waters then gives the final equation:

8 H2O + 2 MnO4- + 10 Cl- right arrow 2 Mn2+ + 5 Cl2 + 16 OH-

(Additional examples for balancing Redox equations in acidic solution can be found in the Discussion Module.)

Molarity and the Composition of Solutions

The most commonly used concentration term in chemistry is molarity = 1 mole of solute dissolved in enough solvent to give 1 L = M = mol/L. This is the most popular concentration unit at least in part because of the convenience of making molar solutions with volumetric flasks (show flask).

Note that molarity has various meanings in describing solutions. Commonly molarity refers to the amount of substance dissolved in one liter of liquid to give a solution. Thus if 158.52 g (1 mole) of strontium chloride is dissolved in enough water to give one liter we will have a 1 L solution of strontium chloride. Note however, that while the solution is 1M in Sr2+, it is 2 M in Cl- and 3 M in terms of particles!

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 10-2

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- right arrow H2O + Cl- + Na+

Giving: H+ + OH- right arrow 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- right arrow 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.

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© R A Paselk

Last modified 14 February 2011