Lesson 4.8 Quantitative Analysis
Suggested Reading

Zumdahl Chapter 4 Sections 4.7 & 4.8
Essential Question

How is quantitative analysis used to determine the amount of
substance or species present in a material?
Learning Objective



Apply the analytical technique of gravimetric analysis.
Explain how to perform an acid-base titration.
Perform calculations involving neutralization reactions and titrations.
Quantitative Analysis
Analytical chemistry deals with the determination of composition of
materials or the analysis of materials. The materials that one might
analyze include air, water, food, body fluids, pharmaceutical preparation,
and so forth. The analysis of materials is divided into qualitative and
quantitative analysis. Qualitative analysis involves the identification of
substances or species present in a material. For instance, you might
determine that a sample of water contains lead(II) ion. Quantitative
analysis, which we will discuss in this lesson, involves the determination
of the amount of a substance or species present in a material. In a
quantitative analysis, you might determine that the amount of lead(II) ion in
a sample of water is 0.067 mg/L. In this lesson we will look at two common
analytical techniques. In this course, you will carry out experiments using
both of these techniques, so take good notes!
Gravimetric Analysis
Gravimetric analysis is a type of quantitative analysis in which the amount
of a species in a material is determined by converting the species to a
product that can be separated and weighed. Precipitation reactions are
often used to do this, where the amount of an ionic species in a mixture is
determined by precipitating it from solution. The precipitate is then filtered,
dried and weighed. The advantages of gravimetric analysis are its
simplicity (at least in theory) and its accuracy. The chief disadvantage is
that it requires meticulous, time-consuming work.
As an example of gravimetric analysis, consider the problem of
determining the amount of lead in a sample of drinking water. Lead, if it
occurs, in the water probably exists as the lead(II) ion, Pb2+. In order to
separate lead(II) ion from the solution we must apply to solubility rules to
determine what precipitation reaction can occur using lead(II) ion as a
reactant. Recall from the solubility rules that most sulfates are soluble, but
PbSO4 is an exception. Thus, lead(II) sulfate would precipitate out of a
solution containing lead(II) ion and sulfate ion. Adding soluble sulfate
compound, such as sodium sulfate, would supply the sulfate ion and
cause a precipitate of PbSO4 to form. If you assume the lead is present is
solution as lead(II) nitrate, you can write the following equation for the
reaction:
Na2SO4(aq) + Pb(NO3)2(aq) → 2NaNO3(aq) + PbSO4(s)
A white precipitate of lead(II) sulfate will form, which can be separated
from solution by filtration. Then, you dry, weigh the precipitate, and finally
determined how much lead was in the original solution.
Watch this overview of the gravimetric analysis procedure used in another
AP Chemistry class. We will be carrying out a similar procedure in the near
future.
Watch the following YouTube Video:
https://www.youtube.com/watch?v=rZMzGxjvklo
Here is an example of the calculation used to determine the amount of
substance by gravimetric analysis.
Example: Determining the Amount of Species by Gravimetric
Analysis
A 1.000 L sample of polluted water was analyzed for lead(II) ion, Pb2+, by
adding an excess of sodium sulfate to it. The mass of lead(II) sulfate that
precipitated was 229.8 mg. What is the mass of lead in a liter of the water?
Give the answer as milligrams of lead per liter of solution.
Solution
You assume that all of the lead in the water solution is precipitated as
lead(II) sulfate, PbSO4. If you know the percentage of lead in PbSO4, you
can calculate the quantity of lead in the water sample. The mass
percentage of Pb in PbSO4 can be determined by dividing the molar mass
of Pb by the molar mass of PbSO4 and then multiplying by 100% as
follows
and the amount of Pb in the 1.000 L sample of water is
Amount of Pb = 229.8 mg PbSO4 x 0.6832 = 157.0 mg Pb
The water sample contains 157.0 mg Pb per liter.
Molarity as a Conversion Factor
You learned in the previous lesson that molarity can be used as a
conversion factor, and in this way you can calculate the volume of solution
that is equivalent to a given mass of solute (please take initiative and
review lesson 4.7 if you are unsure about this). This means that you can
replace mass measurements in stoichiometry problems involving reactions
in solution with volume measurements. This is usually referred to as
solution stoichiometry.
We use solution stoichiometry in volumetric analysis, which is the second
analytical technique we will discuss, but lets first look at an example of a
solution stoichiometry problem.
Example: Calculating the Volume of Reactant Solution Needed
Consider the reaction of sulfuric acid, H2SO4, with sodium hydroxide,
NaOH (What type of reaction is this? How do you know?).
H2SO4(aq) + 2NaOH(aq) → 2H2O(l) + Na2SO4(aq)
Suppose a beaker contains 35.0 mL of 0.175 M H2SO4. How many
milliliters of 0.250 M NaOH must be added to react completely with
(neutralize) the sulfuric acid?
Solution
Using molarity and the mole ratios from the balanced equation as
conversion factors, the calculation is as follows.
This means that 49.0 mL of 0.250 M NaOH are required to neutralize 35.0
mL of 0.175 M H2SO4.
These problems are treated like any other dimensional analysis problem
where you start with the given and use conversion factors to work your
way toward the answer.
Lets talk about volumetric analysis now, which used calculations just like
the one above.
Volumetric Analysis
Titration is an important method for determining the amount of a particular
substance in a solution. In this method you measure the volume of
reactant solution delivered using a special piece of laboratory glassware
called a buret. Solution stoichiometry is then used to determine the
amount of another substance in the reaction mixture.
Suppose substance A reacts in solution with substance B. If you know the
volume and concentration of a solution of B that just reacts with a sample
of substance A, you can determine the amount of A in that sample.
Titration is a procedure for determining the amount of A by adding a
carefully measured volume of a solution B until the reaction of A and B is
just complete. Volumetric analysis is a method of analysis based on
titration.
Watch the video for an overview of a titration procedure. You will be using
a similar procedure in this course:
Watch the following YouTube Video:
https://www.youtube.com/watch?v=9DkB82xLvNE
Figures a, b, and c show a flask containing and unknown amount of acid
(analyte) being titrated with a sodium hydroxide solution (titrant) of known
molarity. If the unknown acid is HX, the reaction is
NaOH(aq) + HX(aq) → NaX(aq) + H2O(aq)
A few drops of phenolphthalein indicator have been added to the acid.
Recall that an indictor is a substance that undergoes a color change when
a reaction approaches completion (see Lesson 4.4). Phenolphthalein is
colorless in acidic solution but turns pink at the completion of the
neutralization reaction with a base. A solution of sodium hydroxide with a
known concentration (standard solution) is contained in a buret. The
NaOH in the buret is added to the HX in the flask until the phenolphthalein
just changes from colorless to pink for at least 10 seconds. At this point
the reaction is complete and the volume of NOH that reacts with the HX is
read from the buret. If the reaction retains the pink color for more than 10
seconds you have passed the endpoint. The volume of titrant delivered
(NaOH in this case) is then used to obtain the mass of HX in the analyte
solution. Lets look at the calculations involved.
Example: Calculating the Quantity of Substance in a Titrated Solution
A flask contains a solution with an unknown amount of HCl. This solution
is titrated with 0.207 M NaOH. It takes 4.47 mL of NaOH to complete the
reaction. What is the mass of HCl?
You convert the volume of NaOH to moles NaOH using the molarity of
NaOH. Then you convert moles NaOH to moles HCl using mole ratios
from the balanced chemical reaction. Finally, you convert moles HCl to
grams HCl (just like in the previous example.
The reaction is
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(aq)
The calculation is
HOMEWORK: book questions page 172 43-47
Study guide questions (this is the other book you got with your textbook) page 104 questions 1, 4-6, 8,
10 , 12, 14-18, 21-23, 26-29, 31, 34-37, 39, 41, 43, 46, 49, 50-52