CHM 101 GENERAL CHEMISTRY

FALL QUARTER 2008

Section 2

 

Lecture Notes – 10/22/2008

(last revised: 10/22/08, 3:45 PM)

 

 

4.1          Water, the Common Solvent

 

H2O

 

NaCl (s)

——>

Na+ (aq) + Cl- (aq)

In the equation, Na+ (aq) represents a hydrated sodium ion, or a sodium ion surrounded by water molecules, with the partially negatively charges oxygen atoms oriented inward toward the positively charged sodium ions. Similarly, Cl- (aq) represents a chloride (ion) surrounded by water molecules with their hydrogens oriented inward. Figure 4.2 from the text gives this pictorial representation of the dissolution process. In the upper panel, we see two water molecules “tugging” at a chloride with their hydrogens and two other water molecules tugging on a sodium ion with their oxygens. The second panel shows two hydrated ions, one each of sodium ion and chloride, and it shows other water molecules tugging at still more ions from the solid lattice:

4.2          Strong and Weak (and non) Electrolytes

A device for detecting electrical conductivity in solutions is shown in Figure 4.4 in the text:

o       Soluble Salts

o       Strong Acids

o       Strong Bases

 

H2O

 

HCl

——>

H+(aq) + Cl-(aq)

 

H2O

 

HNO3

——>

H+(aq) + NO3-(aq)

 

H2O

 

H2SO4

——>

H+(aq) + HSO4-(aq)

Sulfuric acid (H2SO4) deserves a closer look. When sulfuric acid dissolves in water, the first hydrogen dissociates completely to form protons in aqueous solution, but the second hydrogen remains bonded to the sulfate. Thus aqueous sulfuric acid contains mostly protons (H+(aq)) and hydrogen sulfate (HSO4-(aq)) ions.

 

H2O

 

NaOH (s)

——>

Na+(aq) + OH-(aq)

 

H2O

 

KOH (s)

——>

K+(aq) + OH-(aq)

 

o       Weak acids

o       Weak bases

 

H2O

 

HC2H3O2(aq)

——>
<——

H+(aq) + C2H3O2-(aq)

 

 

H2O

 

NH3 (aq) + H2O (l)

——>
<——

NH4+(aq) + OH-(aq)

 

o       Ethanol (C2H5OH)

o       Sucrose (C12H22O11), also known as cane sugar.

4.3           The Composition of Solutions: Many important chemical reactions occur in solution. We need to be able to perform stoichiometric calculations on these reactions.

In the previous chapter, we learned to count the molecules in a reaction by weighing them, but when we deal with solutions, we need a different way of counting. The most convenient way to measure a liquid is to measure its volume. Thus, for example, if we had prepared a 1.000 liter solution containing 1.000 mol of sodium nitrate and then drew out a 50.00 mL portion of that solution, we would know that the 50.00 mL portion contains 0.05000 mol of sodium nitrate.

 

We can quantitatively describe our example solution as a 1.000 molar aqueous solution of sodium nitrate, and we can abbreviate that to 1.000 M NaNO3.

 

 

H2O

 

Na2SO4 (s)

——>

2Na+(aq) + SO42-(aq)

 

 

H2O

 

ZnCl2 (s)

——>

Zn2+(aq) + 2Cl-(aq)

 

 

Moles of solute before dilution = Moles of solute after dilution

In a dilution problem, you will know or be given information to calculate 3 of the 4 quantities in the line above.

4.4          Types of Chemical Reactions in Solution

These topics take up the rest of Chapter 4. We will cover them next week.