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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.

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• 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) + CrO42-(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) + CrO42-(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) + CrO42-(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:

CrO42-(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.):

o Most nitrate (NO3-) salts are soluble. (Memorize this!)

o Most salts containing alkali metal ions (Li+, Na+, K+, Rb+, & Cs+) or the ammonium ion (NH4

+) are soluble. (Memorize this!)

o 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 Hg2

2+ are insoluble.

o Most sulfate (SO42-) salts are soluble. Exceptions for CHM

101: BaSO4, PbSO4, Hg2SO4, and CaSO4 are insoluble.

o Most hydroxide (OH-) “salts” are insoluble, unless Rule #2 applies.

o Most sulfide, carbonate, chromate, and phosphate (S2-, CO3

2-, CrO42-, & 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)

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b) Na2SO4 (aq) & Pb(NO3)2 (aq)

c) KOH (aq) & Fe(NO3)2 (aq)

4.6 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) + CrO42-(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:

CrO42-(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.

4.7 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:

3 3AgNO AgNO mol0.100 1.5L 0.150 molLN M V= × = × =

+AgN

(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:

3AgNON= (5)

o Equation (3) gives us:

+Cl AgN − N= (6)

o And Equation (2) gives us:

NaCl ClN N −=

(7)

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

3NaCl AgNO 0.150 molN N= = (8)

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



NaCl NaCl NaCl 0.150 mol 58.45 g/mol 8.77 gm N MM= × = × =

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:

o Identify the species present in the combined solution, and determine what reaction occurs.

o Write the balanced net ionic equation for the reaction.

o Calculate moles of reactants.

o Determine which reactant is limiting.

o Calculate the moles of product(s), as required.

o Convert to grams or other units, as required.

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

o 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.

4.8 Acid-Base Reactions:

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

o 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.

o 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)

o 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:


N N+ -H OH=

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

HCl NaOHN N=

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

HCl HCl NaOH NaOHM V M V× = ×

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:

NaOH NaOHHCl

HCl

0.350 mol/L 0.025 L 0.0875 L0.100 mol/L

M VVM

× ×= = =

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:


HNO

KOH KOH KOH

0.0070 molN N

3 3 3HNO HNO 0.250 mol/L 0.0280 L 0.00700 molN M V= × = × =

And the number of moles of KOH is:

0.320 mol/L 0.0530 L 0.0170 molN M V= × = × =

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:

2 3H O HNO= =

H O

3KOH HNOOH0.0170 mol 0.0070 mol 0.0100 molN N N− = − = − =

KOHV V= +

The mass of this water is:

2 2 2H O H O 0.00700 mol 18.02 g/mol 0.125 gm N MM= × = × =

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.)

3HNO 0.0280 L 0.0530 L 0.0810 LV = + =

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

OHOH

0.0100 mol 0.123 mol/L0.0810 L

NM

V−

− = = =

• 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.

o 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.

o The titration of a sample of an unknown weak acid with a strong base is pictured in Figures 4.18a-c from the text:

o 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.

o In general, the requirements for a successful titration are:

1. The exact reaction between titrant and analyte must be known and rapid.

2. The stoichiometric (equivalence) point must be marked accurately.

3. The volume of titrant required to reach the stoichiometric point must be measured accurately.

o In an acid – base titration, the analyte is a base or an acid. (It may either be strong or weak.) The titrant then is a strong acid or a strong base, respectively. A common indicator for acid – base titrations is phenolphthalein, which is colorless in acidic solutions and pink-purplish in basic solutions.


n

ossible.

• Neutralization Titration – Standardization of a Solution of NaOH (Sample Exercise 4.14, p. 153): Solutions of NaOH are commonly used to titrate unknown acids. One might think that one could prepare a solution of NaOH byaccurately weighing some solid NaOH and dissolving it in water to prepare aaccurately measured volume of solution. However, solid NaOH draws water out of its surrounding atmosphere so fast that an accurate weighing is not p

The actual procedure for preparing a standardized solution of NaOH is first to make an NaOH solution that has approximately the desired concentration. Then one takes an accurately weighed amount of potassium hydrogen phthalate, dissolves it in water, and titrates it to a phenolphthalein end point with the NaOH solution. This works because potassium hydrogen phthalate has one hydrogen (the one bonded to oxygen in the figure) that can be donated to a strong base.

Hence it functions as a weak acid, even though it is nominally a salt.

In this particular example, a student weighs out a 1.3009 g sample of potassium hydrogen phthalate (KHC8H4O4, MM = 204.22 g/mol), dissolves it in water in an Erlenmeyer flask, adds phenolphthalein indicator, and titrates it with the solution of NaOH that she wishes to standardize. At the end point of the titration, she finds that she has added 41.20 mL of NaOH solution. What is the concentration of the NaOH solution?

o The first step is to determine what species are initially present and which of them react. The NaOH contributes Na+ (aq) and OH- (aq) ions, and the potassium hydrogen phthalate contributes K+ (aq) and HC8H4O4

- (aq) ions. The actual reaction that takes place is between HC8H4O4

- (aq) and OH- (aq), even though they are both negative ions:

HC8H4O4- (aq) + OH- (aq) ——> C8H4O4

2- (aq) + H2O (l)

The reaction is between the hydrogen phthalate ion, acting as a weak acid (Brønsted-Lowry definition), and the hydroxide, a strong base.


- 8 4 48 4 4NaOH H OOH HC H O

N N N−= =

NaOHM( )V

KHCm

o The stoichiometry is 1:1, so she can write:

KHCN=

o Now she converts moles of NaOH to molarity and

volume , and moles of KHC8H4O8 to mass

and molar mass :

( )

NaOH

8 4 4H O( )8 4 4KHC H O( )MM

8 4 4

8 4 4

KHC H ONaOH NaOH NaOH KHC H O

mM V N N× = = =

8 4 4KHC H OMM

o Plugging in her data, she gets:

8 4 4

8 4 4

KHC H ONaOH NaOH NaOH

KHC H O

Lm

M M VMM

× = × = =

3

0.04120

1.3009 g 6.3701 10 mol204.22 g/mol

−= = ×

o Thus the molarity of her solution is:

36.3701 10 mol 0.1546 mol/L0.04120 L

−×= =NaOHM

7 5 2HC H O NaOHN N

o She will use this NaOH solution and this molarity result in the next example.

• Neutralization Titration – Analysis of a Sample Containing a Weak Acid (Sample Exercise 4.15, pp. 153-4): Our intrepid chemist now turns to her real problem, the analysis of a sample of the effluent from a waste treatment process. This effluent is known to contain carbon tetrachloride (CCl4) and benzoic acid (HC7H5O2), a weak acid that can donate one proton to a strong base. She weighs out a 0.3518 g sample of the effluent and shakes it with water to dissolve the benzoic acid. Then she titrates it with 0.1546 M NaOH, and the titration requires 10.59 mL of the NaOH solution to reach a phenolphthalein end point. What is the mass percent of benzoic acid in the original sample? (The molar mass of benzoic acid is 122.12 g/mol.)

o The titration reaction is between NaOH and HC7H5O2 to produce benzoate ion (C7H5O2

- (aq)), so she writes the net ionic equation:

HC7H5O2 (aq) + OH- (aq) ——> C7H5O2- (aq) + H2O (l)

o Thus the stoichiometry between benzoic acid and NaOH is 1:1, so the number of moles of benzoic acid in the sample is the same as the number of moles of NaOH used in the titration:

=

o She expresses the number of moles of benzoic acid in terms of its (unknown) mass and its molar mass

, and she expresses the number of moles of

NaOH in terms of the molarity and volume

of the solution:

7 5 2HC H O( )m)

NaOH(V

7 5 2HC H O

NaOH( )M)

(MM

7 5 2

7 5 2

7 5 2

HC H OHC H O NaOH NaOH NaOH

HC H O

mN N M V

MM= = = ×


o Now she can plug in her data:

7 5 2H O 30.1546 mol/L 0.01059 L 1.637 10 mol g/mol

−= × = ×HC

122.12m

7 5 2

3HC H O 1.637 10 mol 122.12 g/mol 0.1999 gm −= × × =

o Thus she finds that the mass of benzoic acid in her effluent sample is:

o Finally, she computes the percentage of benzoic acid that was present in her original sample:

7 5 2

7 5 2

HC H OHC H O% 56.82%

0.3518 gsamplem= = =

0.1999 gm


FALL QUARTER 2008 - LTCC Online · 10/27/2008 · CHM 101 GENERAL CHEMISTRY . FALL QUARTER 2008 :...

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