Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 107	Fundamentals of Chemistry	Fall 2008
Lecture Notes: 21 October	© R. Paselk 2005
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Particles, Moles and Mass, cont.

Masses and particles Examples, cont.

• How many atoms are there in 1.00g of:
  • ethanol? (CH₃CH₂OH)
  • How many Carbon atoms?
• What is the mass of 3.01 x 10²⁰ atoms of carbon in amu? in g?
• What is the mass of iron in 200.0 mmoles of iron(III) sulfate?

 

Solution Concentrations

Solutions: a solution occurs when one chemical is completely dissolved or dispersed in another. We most commonly think of solutions as being liquid, but solid solutions also occur, such as the various metal alloys like steel, brass and bronze.

In a solution the substance present in highest concentration is considered to be the solvent, while components in lesser amounts are considered to be solutes. If you dissolve a sugar cube in water you get a sugar solution, where water is the solvent, and sugar is the solute.

Example:

• What is the solvent in 80 proof rum: 80 proof = 40% alcohol in water, so water is the solvent.
• What is the solvent in 151 proof rum: 151 proof = 75.5% alcohol in water, so alcohol is the solvent.

Concentration Terms:

Molarity:

The most commonly used concentration term in chemistry is molarity, M = 1 mole of solute dissolved in 1 L of solvent.

• Two types of situation arise giving two 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).
    • What is the concentration of a solution made by dissolving 10.00 g of KI in enough water to make 1.000 L
  • Dilution problems. In dilution problems, I always like to remember that the number of moles stays constant (another example of conservation of mass).
    Moles = Moles, or (M₁)(V₁) = (M₂)(V₂)
    • What is the concentration of a solution resulting from adding 25.0 mL of 0.60 M CaCl₂ to 475 mL of water.
    • How much 1.000 M MgSO₄ is needed to make 500.0 mL of a 0.25 M solution.
• Concentrated vs. Dilute: (3M = "dilute") 
• %, ppt (two meanings), ppm, and ppb: generally based on mass/mass, though volume/volume also used particularly for gases.
  • % = parts/hundred (1% = 1 g in 99 mL of water; 5 g solute + 95 g solvent = 5% = 5 g in 100 g total)
  • ppt = parts/thousand (1ppt = 1 g in 999 mL water; 5 g solute and 995 g solvent = 5 ppt = 5 g in 1000g total)
  • ppm = parts/million (1 ppm = 1 mg in 1 L of water; 0.005 g solute and 999.995 g {I am ignoring significant figures} solvent = 5 ppm = 5 mg in 1000g total)
  • ppb = parts/billion (1 ppb = 1 mcg in 1 L of water; 5x10^-6g solute and 999.999995 g {I am ignoring significant figures} solvent = 5 ppb = 5 mg in 1000g total)

Chemical Reactions

I'm not terribly interested in the classification system given in your text in Chapter 10. Though one does occasionally run into some of these terms, you don't have to worry about them for this course.

Instead, I want to look at reactions from the perspective of what is happening at the atomic and molecular levels where reactions occur because of the associations of ions and molecules with each other to form precipitates and complexes, or atoms and/or molecules exchange electrons in oxidation/reduction or redox reactions.

Ionic reactions - dissolving and precipitates

Much of the chemistry around us involves the dissolution of ionic solids in water to give aqueous solutions and the precipitation of ions from aqueous solution to give precipitates (solids). So what I would like to do first is to look a little at the process of dissolving and the nature of aqueous solutions. This means I will be talking about stuff that is discussed in your book in later chapters (13 & 14).

First we need to look a bit a water itself. (models, overhead) The thing we need to keep in mind is that the ions in water are not independent - they dissolve because they substitute interactions with water molecules for interactions with counter ions. And they stay in solution because they are insulated from each other by the water "shells" around each ion. A couple of corollaries

• First, only so many ions will dissolve until we run out of water molecules to make the shells, then they will interact with each other and precipitate out.
• Second, the amount of a given ion which will dissolve in water depends on the competition between water and its counter ion for interaction with it. If the ion-ion interaction is very strong, then they will precipitate out at very low concentrations etc.

Let's consider some chemical processes:

• Dissolve NaCl in water (Figure 14.3 in your text, p 411) NaCl_(s) Na⁺_(aq) + Cl^-_(aq)
• Add NaCl solution to a AgNO₃ solution. First recall that each of the ions in solution is in an aqueous complex, so we can write:

Notice that two ions don't change, so why show them. Instead we write a net ionic equation:

 Ag⁺ + Cl^- AgCl_(s)

• Mix barium chloride and potassium sulfate:

Ba²⁺_(aq) + 2 Cl^-_(aq) + 2 K⁺_(aq) + SO₄^2-  BaSO_4(s) + 2 K⁺_(aq) + 2 Cl^- _(aq)

Ba²⁺ + SO₄^2-  BaSO_{4 (s)}

Notice that net ionic equations are very general expressions. Essentially they are saying that any time we have these species present they will react, regardless of what else happens to be there! (Sometimes folks are confused when they add ions which should react and they don't. This is usually a case where something else reacted first, so the ions of interest really weren't there!).

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Last modified 21 October 2008