CHM 101 - LECTURE NOTES

CHM 101 GENERAL CHEMISTRY

FALL QUARTER 2008

Section 2

Lecture Notes – 10/27/2008

(last revised: 10/27/08, 9:30 PM)

4.5 Precipitation Reactions: In this section and in the remainder of Chapter 4, we will be concerned mostly with what happens, if anything, when two solutions, each containing a different electrolyte, are mixed. The most simple possibility is that nothing happens. The next most simple is that a substance forms that is not soluble in water, and it precipitates out of the mixture.

An Example of a Precipitation Reaction: The first example from the text is a nicely illustrated demonstration of what happens when an aqueous solution of potassium chromate (K₂CrO₄ (aq)) is added to an aqueous solution of barium nitrate (Ba(NO₃)₂ (aq)). In Figure 4.13, we see that when the yellow solution of K₂CrO₄ (aq) in the graduated cylinder is added to the colorless solution of Ba(NO₃)₂ (aq) in the beaker, a yellow precipitate forms.
What is the Chemistry? As aspiring chemists, we cannot content ourselves with a verbal description of what we see when we combine these two solutions. We need to determine the chemistry that takes place, and we need to be able to write an equation for chemical reaction that has obviously happened.
The Reactants: Let us start by considering what species are present in the two solutions before they are mixed. We know that solid barium nitrate is a salt containing barium ions (Ba²⁺) and nitrate ions (NO₃^-) arranged in a crystal lattice and held together by electrostatic forces. When it is dissolved in water, the lattice breaks up, the ions become hydrated, and they disperse evenly throughout the solution.

	H₂O
Ba(NO₃)₂ (s)	——>	Ba(NO₃)₂ (aq)	——>	Ba²⁺ (aq) + 2NO₃^-(aq)

Similarly, we know that the potassium chromate is a salt containing potassium ions (K⁺ (aq)) and chromate ions (CrO₄^2- (aq)) in a crystal lattice held together by electrostatic forces. When it is dissolved in water, its lattice breaks up, the ions become hydrated, and they also disperse evenly throughout the solution.

	H₂O
K₂CrO₄ (s)	——>	K₂CrO₄ (aq)	——>	2K⁺ (aq) + CrO₄^2-(aq)

Figure 4.14 from your text pictures the two solutions and diagrams the ions contained in each:

The Reaction and its Products: When we mix the two solutions, we can identify the four different kinds of ions as possible reactants and we can describe the products as a yellow solid in contact with an aqueous phase:

2K⁺ (aq) + CrO₄^2-(aq) + Ba²⁺ (aq) + 2NO₃^-(aq) ——>
yellow solid + aqueous phase

The question is, what actually reacts with what to form the yellow solid. We know that it must contain both cations and anions, because it cannot have any net charge. There are 4 different combinations of a cation with an anion from the above set of reactants. Two of them, Ba(NO₃)₂ and K₂CrO₄ can be eliminated because they were the soluble salts in the two original solutions. This leaves the following candidates:

o BaCrO₄

o KNO₃

Which of these is most likely to precipitate out of aqueous solution as a yellow solid? We know that our initial Ba(NO₃)₂ solution had no color, and if we have worked at all with potassium salts, we know that most of them are also colorless. Thus the yellow color of the K₂CrO₄ solution must have been from its chromate ions, so by elimination, the yellow solid must be BaCrO₄. Another thing we should know is that nearly all potassium salts and nitrate salts are water soluble, so we can identify the colorless aqueous phase that is in contact with the solid BaCrO₄ as a solution of KNO₃. Thus we can now write the equation for the reaction:

2K⁺ (aq) + CrO₄^2-(aq) + Ba²⁺ (aq) + 2NO₃^-(aq) ——>
BaCrO₄ (s) + 2K⁺ (aq) + 2NO₃^-(aq)

Notice, however, that we have two potassium ions and two nitrate ions on each side of the equation. These ions do not take part in the reaction, so we can leave them out. (We call them spectator ions.) This allows us to simplify the equation to:

CrO₄^2-(aq) + Ba²⁺ (aq) ——> BaCrO₄ (s)

Figure 4.15 from the text give us a good illustration of this precipitation reaction. Parts a and b show what happens on the ionic level, while part c is a photograph after the reaction has taken place:

The Reaction between Silver Nitrate and Potassium Chloride: The second example in the text is the reaction that takes place when silver nitrate (AgNO₃ (aq)) and potassium chloride (KCl (aq)) are mixed. Figure 4.16 show a beaker of KCl solution just after the addition of some AgNO₃ solution. So let us figure out what is happening. We start as we did for the first example, with the equation:

AgNO₃ (aq) + KCl (aq) ——>
white solid + aqueous phase

We rewrite the equation to show the reactant ions:

Ag⁺ (aq) + NO₃^- (aq) + K⁺ (aq) + Cl^- (aq) ——> white solid + aqueous phase

The white solid is one of the following two salts:

o AgCl

o KNO₃

But it cannot be KNO₃, because we identified that as the soluble salt that forms in the reaction of K₂CrO₄ (aq) with Ba(NO₃)₂ (aq). Therefore it must be AgCl, and the ionic reaction for its formation can be written:

Ag⁺ (aq) + Cl^- (aq) ——> AgCl (s)

We have omitted the spectator ions, K⁺ (aq) and NO₃^- (aq), because they do not participate in the reaction.

Solubility Rules for Salts: You may have noticed that we used information from the first example to help us determine the nature of the reaction in the second example. We cited the fact that KNO₃ is water soluble. Chemists have studies the water solubility (or lack thereof) for a great number of salts and have compiled some simple rules whether particular salts of the more common ions are water soluble. Table 4.1 in your text is such a set of rules. In CHM 101, we will use the following version of these rules. (This is also posted on Marty Wallace’s website as a PDF file.):

Most nitrate (NO₃^-) salts are soluble. (Memorize this!)
Most salts containing alkali metal ions (Li⁺, Na⁺, K⁺, Rb⁺, & Cs⁺) or the ammonium ion (NH₄⁺) are soluble. (Memorize this!)
Most chloride, bromide, and iodide (Cl^-, Br^-, & I^-) salts are soluble. Exceptions for CHM 101: The chloride, bromide, and iodide salts of Ag⁺, Pb²⁺, and Hg₂²⁺ are insoluble.
Most sulfate (SO₄^2-) salts are soluble. Exceptions for CHM 101: BaSO₄, PbSO₄, Hg₂SO₄, and CaSO₄ are insoluble.
Most hydroxide (OH^-) “salts” are insoluble, unless Rule #2 applies.
Most sulfide, carbonate, chromate, and phosphate (S^2-, CO₃^2-, CrO₄^2-, & PO₄^3-) salts are insoluble, unless Rule #2 applies.

As with many other sets of rules, these have a number of exceptions. Moreover, the absolutes, soluble and insoluble, are extremes on a continuum from very, very soluble to somewhat soluble, to hardly soluble at all. Nevertheless, we will use them for now, but as my Freshman Chemistry professor at Harvard used to say, we will have to take them with a grain of salt.

Predicting Reaction Products: Now we turn to Sample Exercise 4.8 (pp. 144-5) from your text. What are the reaction products (if any)? We will work them on the whiteboard:

a) KNO₃ (aq) & BaCl₂ (aq)

b) Na₂SO₄ (aq) & Pb(NO₃)₂ (aq)

c) KOH (aq) & Fe(NO₃)₂ (aq)

4.5 Describing Reactions in Solution

Let us briefly return to the first example from the previous section. Recall that when we determined that the precipitate was BaCrO₄, we could have written the formula equation for the reaction as:

K₂CrO₄ (aq) + Ba(NO₃)₂ (aq) ——> BaCrO₄ (s) + 2KNO₃ (aq)

We actually did write the formulas for the three soluble salts in terms of their component ions, obtaining the complete ionic equation:

2K⁺ (aq) + CrO₄^2-(aq) + Ba²⁺ (aq) + 2NO₃^-(aq) ——>
BaCrO₄ (s) + 2K⁺ (aq) + 2NO₃^-(aq)

Then we eliminated the spectator ions, K+ (aq) and NO3-(aq), two of each on each side of the equation:

CrO₄^2-(aq) + Ba²⁺ (aq) ——> BaCrO₄ (s)

What remains is the net ionic equation, including only those components directly involved in the reaction. Let’s summarize these three types of equations:

o Formula Equation: The Formula Equation gives the overall stoichiometry for the reaction, but it does not necessarily show the actual forms of the reactants and products.

o Complete Ionic Equation: The Complete Ionic Equation shows all the ions present in the reactants, all the solid reaction products, and all ions that remain after the reaction takes place.

o Net Ionic Equation: The Net Ionic Equation shows only the ions that react plus the products that form from them.

Writing Equations for Reactions: Now we turn to Sample Exercise 4.9 (p.146) from your text. Write the formula equation, the complete ionic equation, and the net ionic equation for the following reactions. We will work them on the whiteboard:

a) Aqueous potassium chloride is added to aqueous silver nitrate to form a precipitate of silver chloride and a solution of potassium nitrate.

b) Aqueous potassium hydroxide is mixed with aqueous iron (III) nitrate to form a precipitate of iron (III) hydroxide and aqueous potassium nitrate.

4.6 Stoichiometry of Precipitation Reactions: This isn’t really a new topic. The good news is that you already know how to stoichiometric calculations on precipitation reactions.

In Section 3.9 you learned how to start with the mass (usually in grams) of component A, convert that to the number of moles of A, convert that to the number of moles of B (according to the balanced equation), and finally convert that to the mass of B.
In Section 3.10 you learned how to work limiting reactant problems, i. e., how to determine which of two reactants limited the amount of products that would form.
In Section 4.3 you learned about concentrations of reagents in solution, and how if you multiplied molar concentration (molarity) by volume, your result would be the number of moles contained in that volume.
Finally in Sections 4.5 and 4.6 you learned how to determine the nature of a precipitation reaction and how to write a net ionic equation for the reaction.
Determining the Mass of Product Formed: Now you can apply the foregoing skills to determine masses of products formed in precipitation reactions. We will jump immediately into some sample exercises:

o Sample Exercise 4.10 (p. 147): Calculate the mass of solid NaCl that must be added to 1.50 L of a 0.100 M AgNO₃ solution to precipitate all the Ag⁺ ions as AgCl (s).

o The first step is to list the ions that are initially present when solid NaCl is dissolved in a solution of AgNO₃. The AgNO₃ solution already contains Ag⁺ and NO₃^- ions prepared by dissolving AgNO₃ in water:

	H₂O
AgNO₃ (s)	——>	Ag⁺ (aq) + NO₃^-(aq)	(1)

o And the reaction for the dissolution of NaCl in water can be written:

	H₂O
NaCl (s)	——>	Na⁺ (aq) + Cl^-(aq)	(2)

o Thus the ions are:

Na⁺, Cl^- (from NaCl), Ag⁺, NO₃^- (from AgNO₃)

o The next step is to use Table 4.1 to determine that AgCl is insoluble in water, while NaNO₃ is water soluble and remains in solution as (unchanged) Na⁺ and NO₃^- ions. From this information we can write the net ionic equation:

Ag⁺ (aq) + Cl^- (aq) ——> AgCl (s) (3)

o Since we know the molarity and the volume of the silver nitrate solution, we can determine the number of moles of AgNO₃ it contains:

(4)

o How do we use this information to determine the mass of NaCl (s) needed to precipitate the Ag⁺ ions from the solution? We know from Eq. (1) that the number of moles of Ag⁺ ions in the silver nitrate solution is:

(5)

o Equation (3) gives us:

(6)

o And Equation (2) gives us:

(7)

o Combining Eqs. (4), (5), (6), & (7), we get the number of moles of NaCl required for the complete precipitation of the AgCl:

(8)

o All we need to do to finish is to convert moles of NaCl to mass:

o As you gain familiarity with this type of calculation, you will be able to write equations like Eq. (8) directly, without explicitly writing the intermediate steps (5), (6), & (7).

A System for Solving Stoichiometry Problems for Reactions in Solution: The text gives the following step by step procedure for solving solution stoichiometry problems:

Identify the species present in the combined solution, and determine what reaction occurs.
Write the balanced net ionic equation for the reaction.
Calculate moles of reactants.
Determine which reactant is limiting.
Calculate the moles of product(s), as required.
Convert to grams or other units, as required.

Determining the Mass of Product Formed (Sample Exercise 4.11(p. 148):

When aqueous solutions of Na₂SO₄ and Pb(NO₃)₂ are mixed, PbSO₄ precipitates. Calculate the mass of PbSO₄ formed when 1.25 L of 0.0500 M Pb(NO₃)₂ and 2.00 L of 0.0250 M Na₂SO₄ are mixed. We will work this out on the whiteboard.

4.7 Acid-Base Reactions:

Definitions – Acids and Bases: We will start by revisiting the definitions of acids and bases.

Arrhenius’ Definitions:

1. Acid: An acid produces H⁺ ions when dissolved in water.

2. Base: A base produces OH^- ions when dissolved in water.

These definitions are fundamentally correct, and they serve us well for dealing with strong acids and strong bases. However, they are somewhat inconveniently limiting when we consider weak acids and weak bases. Consider the weak base, ammonia (NH₃). What is the source of the OH^- ions that it generates when dissolved in water? It is not to be found in gaseous ammonia. The fact is that an ammonia molecule must combine with a water molecule in order to generate a hydroxide:

NH₃ (aq) + H₂O (l) = NH₄⁺ (aq) + OH^- (aq)

In other words, in order for an ammonia molecule to generate a hydroxide ion in aqueous solution, it must accept a proton from a water molecule. Then what is left of the water molecule is the desired hydroxide ion.

The problem is not so severe for weak acids. Consider the weak acid, acetic acid (HC₂H₃O₂). Acetic acid will generate protons when dissolved in water:

HC₂H₃O₂ (aq) = H⁺ (aq) + C₂H₃O₂^- (aq)

However, as we learned earlier, only a small fraction of acetic acid molecules will form ions. In 0.1 M acetic acid, about 99% of the acetic acid remains as undissociated neutral acetic acid molecules; only about 1% will ionize to form hydrogen ions and acetate ions.

Picturing the Reaction of a Weak Acid with a Strong Base: If equimolar aqueous solutions of a weak acid and a strong base are mixed, the reaction will go to completion. All of the base will consume all of the acid. For example, if acetic acid is reacted with sodium hydroxide, we can write the overall reaction as:

HC₂H₃O₂ (aq) + NaOH (aq) = NaC₂H₃O₂ (aq) +H₂O (l)

If we strictly apply the Arrhenius definition of an acid, we must imagine that the initial 1% of hydrogen ions from the acetic acid react with hydroxide ions. Then more acetic acid molecules ionize, and those hydrogen ions react with more hydroxides, and so on until all the acetic acid has ionized and all the resulting hydrogen ions have reacted with hydroxides. Wouldn’t it be simpler if we said that acetic acid reacts directly with hydroxide ion by donation of a proton to hydroxide?

HC₂H₃O₂ (aq) + OH^- (aq) = C₂H₃O₂^- (aq) +H₂O (l)

Brønsted and Lowry’s Definitions:

1. Acid: An acid is a proton donor.

2. Base: A base is a proton acceptor.

These definitions are more general than those of Arrhenius and they resolve the difficulties we have just discussed.

Completeness of Acid – Base Reactions: For reactions of equimolar quantities:

1. A strong acid will react completely with a strong base.

2. A strong acid will react completely with a weak base.

3. A weak acid will react completely with a strong base.

Stoichiometry Calculations for Acid – Base Reactions: The text gives this step-by-step procedure for performing acid – base calculations. It is very similar to the procedure that we just applied to precipitation reactions:

1. List the species present in the combined solution before any reaction occurs, and determine what the reaction will be.

2. Write the balanced net ionic equation for this reaction.

3. Calculate the moles of reactants. For reactions in solution, use the volumes of the original solutions and their molarities.

4. Determine the limiting reactant where appropriate.

5. Calculate the moles of the required reactant or product.

6. Convert to grams of volume (of solution) as required.

Neutralization Reactions: These are reactions where just enough acid is added to some base (or just enough base is added to some acid) to react completely. Since there is no excess either of acid or base, we say the resulting solution is neutralized.

o Sample Exercise 4.12 (p. 150): What volume of 0.100 M HCl solution is required to neutralize 25.0 mL of 0.350 M NaOH?

After the solutions are mixed, but before any reaction takes place, the ions in solution are:

H⁺ (aq), Cl^- (aq) (from HCl), Na⁺ (aq), OH^- (aq) (from NaOH)

Since NaCl is water soluble, we do not expect any reaction between the Na⁺ (aq), and the Cl^- (aq), but we do expect that H⁺ (aq) and OH^- (aq) will react:

H⁺ (aq) + OH^- (aq) ——> H₂O (l)

The reaction is already balanced, but if it weren’t, we would balance it now.

Since this is an exact neutralization reaction, the numbers of moles of hydrogen ion and hydroxide ion (before reaction) are equal, and there is no need to ask which is the limiting reactant:

But these are the same as the numbers of moles of HCl and of NaOH:

We can rewrite this in terms of the molarities and volumes of the NaOH and HCl solutions:

Since we know the molarities of the two solutions and the volume of the NaOH solution, we can rearrange this equation to solve for the volume of HCl:

o Sample Exercise 4.13 (p. 151): In a certain (peculiar?) experiment, 28.0 mL of 0.250 M HNO₃ and 53.0 mL of 0.320 M KOH are mixed. Calculate the amount of water formed in the resulting reaction. What is the concentration of H⁺ or OH^- ions (whichever is in excess) after the reaction goes to completion?

After the solutions are mixed, but before any reaction takes place, the ions in solution are:

H⁺ (aq), NO₃^- (aq) (from HNO₃), K⁺ (aq), OH^- (aq) (from KOH)

Since KNO₃ is water soluble, we do not expect any reaction between the K⁺ (aq), and the NO₃^- (aq), but we do expect that H⁺ (aq) and OH^- (aq) will react:

H⁺ (aq) + OH^- (aq) ——> H₂O (l)

The reaction is already balanced, but if it weren’t, we would balance it now.

We have data to calculate the numbers of moles of both starting reagents. It is very likely that one or the other will be in excess. The number of moles of HNO₃ is:

And the number of moles of KOH is:

Thus HNO₃ is the limiting reactant. The net ionic reaction shows that 1 mole of H₂O will be generated for each 1 mole of H⁺ that reacts. And we know that each 1 mole of HNO₃ will supply 1 mole of H⁺. Thus the number of moles of water generated is:

The mass of this water is:

The number of moles of unreacted OH- is:

Now we need to be careful. The volume of the final solution is the combined volumes of the HNO₃ solution and the KOH solution. (We neglect the volume of product water that forms in the reaction.)

So the molarity of OH^- in the end solution is:

Acid – Base Titrations: In volumetric analysis, one determines the amount of an unknown substance by reacting it with a carefully measured volume of a solution of a known substance with a known concentration.

For example, suppose one had a solution of a weak acid with an unknown concentration. And let us also suppose one had a solution of a strong base whose concentration is known. One could determine the concentration of the weak acid solution by measuring a precise volume of it (with a transfer pipette) into an Erlenmeyer flask and adding strong base solution to it from a burette until one reaches the stoichiometric point (i. e., the end point) of the reaction where the number of moles of added base exactly matches the number of moles of weak acid present in the original sample of weak acid. An indicator added to the solution in the flask signals the end point by changing its color when the end point is achieved.
The titration of a sample of an unknown weak acid with a strong base is pictured in Figures 4.18a-c from the text:

The process is called titration, and the solution of known substance is called the titrant. The solution of the unknown is called the analyte. The reaction of the known with the unknown is well characterized, so from the stoichiometry of the reaction, one can determine the amount of unknown present in the original unknown sample.