Page 1
University of Pennsylvania Department of Chemistry
Pre-Lab Notes - EQUL-308 :
Solubility Product Constant of Lead (II) Iodide
•
Laboratory Separate: Modular Laboratory Program in Chemistry: EQUL-308:
Solubility Product Constant of Lead (II) Iodide. This experiment is a lab separate.
WORK IN PAIRS. ACH PERSON SHOULD PERFORM THE MEASURMENTS
FOR TWO (2) OF THE FOUR TRIALS.
Questions, Introductory Remarks, & an Example:
Some questions that you should be able to answer before beginning this experiment (as a
result of the material after the questions, and some careful thinking) are below. Prepare the
solutions to these questions and submit them AT THE BEGINNING OF THE
LABORATORY PERIOD. You may want to read the sections following these questions
- as well as your lab separate - before you attempt to answer them.
The Questions follow. Please complete the following.
PRE-LAB QUESTIONS
1. (a) Write down the relationship between the activity of an aqueous species and its
concentration (in M). What is the purpose of the activity coefficient (γ)?
(b) Explain why the equilibrium ion concentrations by themselves are not sufficient to
directly determine the Ksp of a sparingly soluble salt.
(c) What property of the activity coefficient (γ) allows one to determine the Ksp value
without actually measuring and/or calculating γ ? Briefly - but carefully - explain the
computational technique that you will specifically use to determine the Ksp value.
2. A student, following a procedure similar to EQUL-308, sets out to determine the Ksp of the
sparingly soluble ionic salt, M2X3(s). This salt dissociates into the ions M+3(aq) and
X-2(aq). A 100.00 mL aliquot of a 0.0030 M M(NO3)3(aq) solution is titrated by a
0.0020 M solution of K2X(aq). A precipitate begins to form when 4.50 mL of K2X(aq)
have been added. Based on this data, determine the Ksp of M2X3 . As usual, assume that
aqueous solution volumes are additive. Also, neglect the role of the activity coefficient (γ)
when calculating the Ksp value. Show all work.
In EQUL-308, we will be studying the equilibrium that arises when a sparingly soluble ionic
solid dissolves, and subsequently dissociates as a result of reaction with water. You have actually
been introduced to this equilibrium reaction before (cf. pre-lab for Experiment # 1041).
Specifically, we will measure the solubility product (equilibrium) constant (Keq = Ksp) of a
particular ionic solid - lead (II) iodide (PbI2). Before we write down the “now-familiar”
equilibrium reaction and accompanying law of chemical equilibrium, we must state what we
mean by a sparingly soluble solid and its equilibrium condition, i.e., a saturated solution.
Page 2
Recall, when a solid is placed into a solvent (such as water) - depending on factors such
as: temperature, how much solid is placed into the solvent, how much solvent is present, and
other specific “reaction conditions” - some of the solid will dissolve and some will remain as
solid (also termed precipitate). If no more solid will dissolve at the prevailing conditions - such
that a small amount of solid remains undissolved (i.e., as a precipitate) - the solution is termed
saturated. For a “generic” solid, “X”, - dissolved in water - we can write the equilibrium as
X(s) <------> X(aq)
For an ionic solid (such as PbI2(s)), whatever amount of the solid that does dissolve, dissociates
readily (we presume completely) into its ions. Thus, for a hypothetical ionic solid,
“ MmZn(s)”, - dissolved in water - we can write the equilibrium between the undissolved solid
and the saturated (equilibrated) aqueous solution as an equilibrium between the solid and its
aqueous ions. Specifically, we can write:
MmZn(s) <------> m M+n(aq) + n Z-m(aq)
The equilibrium constant that specifies the equilibrium between an ionic solid and a saturated
aqueous solution of its ions is termed the solubility product (equilibrium) constant and is
denoted by Ksp . Remember, for a sparingly soluble ionic solid - which has limited solubility
- the Ksp value is much less than one. Prove for yourself that this makes sense. Remember
also that all equilibrium constants depend on temperature; thus, when we refer to a particular
Ksp value we should specify the temperature. Usually, Ksp values are tabulated at 25oC.
In EQUL-308, the sparingly soluble solid that we will investigate will be lead(II) iodide
(PbI2(s)). Once it dissolves, it dissociates readily (we presume completely) into Pb+2(aq)
and I-(aq). The specific equilibrium is:
PbI2(s) <------> Pb+2(aq) + 2 I-(aq) .
From the background in the pre-lab for Experiment # 1041, it seems “obvious” that the
equilibrium constant known as the Ksp should be written as:
2
Ksp = [Pb+2]eq•[I-]eq.
It turns out that the above relationship is NOT QUITE CORRECT. In fact, even the equilibrium
expression for a saturated solution for the general situation above:
m
n
Ksp = [M+n]eq•[Z-m]eq,
is not quite correct. In fact, this ion product, i.e., the multiplication of the molar
concentrations of the product ions - each product concentration raised to its stoichiometric
coefficient in the balanced reaction - will not yield the Ksp - even when those concentrations
represent equilibrium (saturated solution) concentrations. For a solution that is not very dilute
Page 3
- such as a saturated solution - the interactions among the solute particles and between the solute
and solvent particles become important. Therefore, some additional parameter(s) must be
included in the ion product - besides the number of (moles of) solute ions per Liter of solution
- to take these interactions into account. According to work accomplished mostly by G. N. Lewis
and M. Randall, such situations, i.e., solutions of high concentration (and non-ideal gases), a
parameter termed the activity (a) of a substance was deemed necessary to correctly express
this substance’s contribution to the law of chemical equilibrium. Actually, the activity
⎛1⎞
of a substance “X” (aX) in aqueous solution is defined as the exponential of the quantity ⎜⎝RT⎟⎠
times the difference between the free energy per mole of the solvated species and the free
energy per mole of X in its standard “reference” state. In symbols - for a substance “X”,
we have:
⎛ ∆G ⎞
aX = activity of “X(aq)” = exp⎜⎝ RT ⎟⎠ ;
where ∆G is the free energy difference mentioned above, R is the ideal gas constant, T is
the kelvin temperature, and exp(...) means e(...) . The activity can further be re-expressed as
the product of the molar concentration and an “interaction parameter” - termed the
activity coefficient (γ). Once again, for our hypothetical substance X(aq), we have:
⎛ ∆G ⎞
aX = activity of “X(aq)” = γX•[X] = exp⎜⎝ RT ⎟⎠ .
Therefore, for our “generic” equilibrium above, the Ksp would be more correctly expressed in
terms of the activity of M+n (“a” of M+n) and the activity of Z-m (“a” of Z-m). We
should write:
Ksp =
(a ) • (a )
m
M
+n
Z
eq
−m
(
n
= γ M + n • [M + n ]
eq
) • (γ
m
eq
Z
−m
• [Z − m ]
)
n
eq
It turns out that only the mean (average) activity coefficient of the entire ionic solution can be
measured. This average value is often symbolized by γ± or simply as γ. We will use γ to
express the mean activity coefficient of the ionic solution. Prove for yourself that - in terms
of γ - the Ksp expression above can be rewritten as:
Ksp =
(a ) • (a )
m
M+ n
eq
Z− m
n
eq
(
) (
= γ m+ n • [M + n ]
) • ([Z ])
m
eq
−m
n
eq
.
Also prove for yourself that for the equilibrium between PbI2(s) and the Pb+2(aq) and
I-(aq) ions:
PbI2(s) <------> Pb+2(aq) + 2 I-(aq) ;
the Ksp equilibrium expression can be rewritten as:
2
Ksp = γ3•[Pb+2]eq•[I-]eq.
Page 4
Therefore, besides being able to determine the equilibrium concentrations of the ions , i.e., the
concentrations of the ions in the saturated solution, we must also be able to determine the mean
activity coefficient. It turns out that γ depends on the specific concentration of the ions, i.e.,
it is concentration dependent. Thus, one would have to determine γ for each trial of the
experiment. Luckily, when the solution is “infinitely dilute” (i.e., an “infinite amount of
solvent (H2O(l))), the mean activity coefficient is unity. Such a hypothetical solution would
have “zero concentration” of any solute species. Symbolically,
γ = 1 for an infinitely dilute solution (zero concentration of solute species).
Unfortunately, by necessary design of the experiment, our solutions are saturated - definitely
not at “zero concentration”.
So, if we could only determine the Ksp at “zero concentration” of the solute ions, we could
once again determine the Ksp value from the ion product of the equilibrated system. Since
“zero concentration” is “outside” the concentration range of our experiment, we must
extrapolate our data. The problem now becomes: what function of the data must be measured
and/or calculated in order to validate this yet-to-be-specified extrapolation procedure? Also,
how will this yet-to-be-specified extrapolation procedure allow us to actually determine the
desired Ksp value? If we plotted, for example, the equilibrium ion product
2 ⎞
⎛ +2
⎜[Pb ]eq•[I-]eq⎟ versus the [Pb+2]eq , and extrapolated this curve to “zero concentration
⎝
⎠
+2
of Pb (aq)”, i.e., determine the y-intercept; we could determine the equilibrium ion product at
unit activity - and hence the Ksp value. The problem with this approach - as Figure 1 and the
accompanying text on page 3 of the EQUL-308 lab separate suggest - is two-fold. First, when
the data is plotted in this way, there is considerable curvature in the region of extrapolation,
which reduces confidence in the extrapolated value of the Ksp . Second, the value of the Ksp
- plotted in this way - is too small to be read accurately from the graph. Remember, the Ksp of
a sparingly soluble salt - due to this limited solubility - is much less than one. Plotting the
logarithm of the equilibrium ion product would increase the precision with which we could read
values from the ordinate (y-axis) of our graph. Justify this for yourself. This resolves the
second issue mentioned above. But, how about the first issue, i.e., how do we reduce the
curvature of the extrapolation region? The solution is to find a suitable parameter that
could be plotted on the abscissa (x-axis) - for example - that would transform our graph to a
linear graph. Such a linear graph would give us more confidence in the accuracy of the
extrapolation procedure. The answer is that we must plot a different parameter on the abscissa,
i.e., a different function of the [Pb+2]eq . A detailed analysis of the ion-ion solute interactions in
the solvated system was carried out by P. Debye and E. Hückel - in an effort to develop a model
that would allow one to theoretically determine activity coefficients (γ) and, therefore, activities
(a = γ•concentration). A careful application of the Debye-Hückel model to the saturated
Page 5
PbI2(s)-Pb+2(aq)-I-(aq) system reveals that when log10(Equilibrium Ion Product)
⎛ [Pb+2]eq ⎞
2
⎛ +2
⎞
= log10⎜⎝[Pb ]eq•[I-]eq⎟⎠ (as the ordinate) is plotted against ⎜
⎟,
+2]
1 + [Pb
⎝
eq⎠
(as the abscissa) the resultant plot will be a line. Hence, extrapolation of this line to an infinitely
dilute solution (i.e., “zero ion concentration”) can be done with greater accuracy. Therefore, the
linearly-extrapolated y-intercept ( log10(Ksp)) will yield - by definition - the desired Ksp value.
Before we proceed to the Example problem, I would like to discuss a particular application of
graphical error analysis. Specifically, we will mention a way to graphically estimate the error in
our linearly-extrapolated y-intercept, once we place “error limits” on our dependent variable
(ordinate or “y”). In general, both the dependent variable (ordinate or “y”) and the independent
variable (abscissa or “x”) have error associated with them. Usually, the error in the dependent
(“y”) variable is what is desired. If the dependent variable is plotted - as is customarily done
- on the vertical axis, then it becomes necessary to place “vertical error bars” on each of our
plotted data points. This requires, of course, some way of first determining - or at least estimating
- the error in our dependent variable. In EQUL-308, the dependent variable for the linear plot is
2 ⎞
⎛
log10⎜⎝[Pb+2]eq•[I-]eq⎟⎠ . Hence, we want to estimate the error in the
log10(Equilibrium Ion Product). To do this rigorously, one would have to use a technique
known as propagation-of-errors. Briefly, this technique utilizes the algebraic relationship
between the dependent variable and those independent variable(s) that were directly measured in
the lab. Then, by assigning “reasonable” error estimates for these directly measured variable(s),
one mathematically “propagates” the error(s) through the algebraic relationship that connects
dependent variable with independent variable(s). The technique actually uses the relationship
between error (or variance) and the derivative. The technique basically involves differentiation
of the algebraic relationship of the dependent variable with respect to the independent variable(s).
Then, by interpreting the differentials as the “errors”, one may assign errors to these terms and
“typical experimental values” to the non-differential terms. From this, the error in the dependent
variable can be determined. We will not require you to actually perform this detailed analysis.
(If you’re interested, feel free to ask.). However, it might be useful to get an idea of the algebraic
2 ⎞
⎛
relationship between the log10⎜⎝[Pb+2]eq•[I-]eq⎟⎠ and the directly measured volumes. First,
some notation. We denote the initial concentration of Pb+2(aq) - before mixing with the KI(aq)
- as [Pb+2]o and we denote the initial concentration of I-(aq) - before mixing with the
Pb(NO3)2(aq) - as [I-]o . Since one (1) mole of Pb(NO3)2(aq) generates one (1) mole of
Pb+2(aq), it is easy to see that [Pb+2]o = [Pb(NO3)2]o . Also, since one (1) mole of KI(aq)
generates one (1) mole of I-(aq), it is easy to see that [I-]o = [KI]o . Following the notation of
EQUL-308, we denote the volume of Pb(NO3)2(aq) as Vo and the volume of the added KI(aq)
Page 6
(i.e., the titrant) required to yield a saturated (equilibrium) solution as V. As usual, the
aqueous solution volumes are assumed additive, Vtotal = Vo + V. Verify for yourself that the
equilibrium concentrations, [Pb+2]eq and [I-]eq , are given by:
[Pb+2]eq =
[Pb+2]o•Vo
Vo + V
and
[I-]o•V
[I ]eq = V + V.
o
2 ⎞
⎛
Thus, the log10(Equilibrium Ion Product) = log10⎜⎝[Pb+2]eq•[I-]eq⎟⎠ is given by:
⎛ ⎛[Pb+2] •V ⎞ ⎛ [I-] •V ⎞2 ⎞
2
⎜⎜
o o⎟ ⎜ o ⎟ ⎟
⎛ +2
⎞
log10⎜⎝[Pb ]eq•[I-]eq⎟⎠ = log10⎜ ⎜ V + V ⎟•⎜V + V⎟ ⎟.
o
⎝⎝
⎠⎝ o
⎠ ⎠
Thus, by estimation of the errors in measuring Vo , V, etc., one could - by use of the
propagation-of-errors technique - determine the error in the log10(Equilibrium Ion Product).
2 ⎞
⎛
This error would correspond to the vertical error bars for the plot of log10⎜⎝[Pb+2]eq•[I-]eq⎟⎠
⎛ [Pb+2]eq ⎞
versus ⎜
⎟.
+2]
1 + [Pb
⎝
eq⎠
See the diagram that follows.
For the purposes of this experiment, assume that the estimated error in the
2 ⎞
⎛
log10⎜⎝[Pb+2]eq•[I-]eq⎟⎠values, i.e., the “y”-variable for the linear graph, is
± 0.10.
Page 7
2 ⎞
⎛
log10⎜⎝[Pb+2]eq•[I-]eq⎟⎠
Extrapolation to ordinate
−
•
−
−
•
−
Ordinate error bar = ±0.10,
i.e., 0.10 above data point
& 0.10 below data point.
−
•
−
−
•
−
⎛ [Pb+2]eq ⎞
⎜
⎟
+2
⎝1 + [Pb ]eq⎠
We can then estimate our error in the linearly-extrapolated y-intercept, i.e., the log10(Ksp),
by the following procedure. By connecting the lower error bound of the data point with the
lowest y-value with the upper error bound of the data point with the highest y-value, a line of
steepest (“maximum”) slope can be drawn. Similarly, by connecting the upper error bound
of the data point with the lowest y-value with the lower error bound of the data point with the
highest y-value, a line of least steep (“minimum”) slope can be drawn. Subsequently, if
the line of maximum slope and the line of minimum slope are each linearly extrapolated to obtain
the corresponding y-intercepts, error bounds on log10(Ksp) - and therefore on Ksp - can be
determined. Thus, the error in the graphically determined Ksp value can be found from the largest
and smallest Ksp values obtained by the above-mentioned graphical error technique. The
algebraic relationship is:
Ksp,largest - Ksp,smallest
Error in Ksp value =
.
2
The Example below should help make this clear. As always, be sure to also read your lab separate
for EQUL-308.
Before we display the Example, we note the following change in specific concentrations of the
reagents and in the number of trials (experiments) that are to be performed. First of all, there will be
four (4) samples of Pb(NO3)2(aq) instead of five. The concentrations will be: 0.250 M, 0.100 M,
0.0250 M, and 0.0100 M. The 0.250 M Pb(NO3)2(aq) solution and the 0.100 M Pb(NO3)2(aq)
Page 8
solution will be provided as stock solutions. The 0.0250 M and 0.0100 M solutions will be
made by a 10-fold dilution, respectively, of each of the above two stock solutions.
This Table - Table 1 - is meant to replace the first three filled in lines of Data Sheet 1
- on page 7 of EQUL-308.
•
Table 1 - Amended Laboratory Assignments for EQUL-308 (To replace
first three lines of Data Sheet 1 - on page 7 of EQUL-308)
[Pb(NO3)2]o
(initial)
(in M)
[KI]o
(initial)
(in M)
0.250 M
0.100 M
0.0250 M
0.0100 M
0.0500 M
0.0500 M
0.0500 M
0.0500 M
Vo of
Pb(NO3)2(aq)
(in mL)
100.00 mL
100.00 mL
100.00 mL
100.00 mL
The example below will hopefully enable you to understand how some of the above procedure is
accomplished.
Example:
A certain sparingly soluble ionic salt, of unknown Ksp , is analyzed using the procedure of
EQUL-308. The chemical formula of the salt is MZ2 . The relevant equilibrium is:
MZ2(s) <------> M+2(aq) + 2 Z-(aq)
The Ksp expression for the above equilibrium is, at unit activity coefficient (i.e., γ = 1):
2
Ksp = [M+2]eq•[Z-]eq.
All measurements are made at 25oC. Solutions of M(NO3)2(aq) of known initial
concentrations were titrated with 0.200 M KZ(aq) solution until the first appearance of a
precipitate. The data are tabulated below. As usual, assume that all aqueous solution
volumes are additive.
Initial [M(NO3)2]
(before mixing)
0.226 M
0.101 M
0.0452 M
0.0118 M
Volume (Vo)
of M(NO3)2(aq)
100.00 mL
100.00 mL
100.00 mL
100.00 mL
Initial [KZ],
(before) mixing)
0.200 M
0.200 M
0.200 M
0.200 M
Volume (V) of
KZ(aq) titrant - to
give indicated
precipitation
12.90 mL
12.40 mL
13.00 mL
18.30 mL
Page 9
(a) For each of the above four (4) experiments, determine the [M+2]eq (in M), the [Z-]eq
2
(in M), and the ion product [M+2]eq•[Z-]eq.
2 ⎞
⎛
(b) For each experiment, tabulate the log10⎜⎝[M+2]eq•[Z-]eq⎟⎠ and
⎛ [M+2]eq ⎞
⎜
⎟.
+2]
1 + [M
⎝
eq⎠
2 ⎞
⎛
(c) A suitable plot of log10⎜⎝[M+2]eq•[Z-]eq⎟⎠ - as the ordinate - versus
⎛ [M+2]eq ⎞
⎜
⎟
+2]
1 + [M
⎝
eq⎠
- as the abscissa, gives a best-fit line that has an extrapolated y-intercept of -5.46. Using the
above data as needed, determine the Ksp of MZ2 at 25oC.
2 ⎞
⎛
(d) The estimated error on each plotted ordinate data point , i.e., log10⎜⎝[M+2]eq•[Z-]eq⎟⎠ ,
is ±0.10. Using the ordinate error bars on the points plotted in (c) above, the y-intercept
of the extrapolated line with the maximum slope is found to be: -5.51 and the y-intercept
of the extrapolated line with the minimum slope is found to be: -5.41. Determine
Ksp,largest and Ksp,smallest from this data, and hence, the estimated error of the Ksp
value determined in (c).
Solution to Example:
(a) For each of the above experiments, convince yourself that K+(aq) and NO3-(aq) are
spectator ions. Since one (1) mole of M+2(aq) is produced for each one (1) mole of
M(NO3)2(aq) introduced, it is easy to see that - in the above experiments - the
concentrations of M+2(aq) before mixing, [M+2]o , are: 0.226 M, 0.101 M, 0.0452 M,
and 0.0118 M, respectively. The volume of the M+2(aq) solutions in each of the above
experiments (Vo) is 100.00 mL. Since one (1) mole of Z-(aq) is produced for each one (1)
mole of KZ(aq) introduced, it is easy to see that - in the each of the above experiments
- the concentration of Z-(aq) before mixing, [Z-]o , is 0.200 M. The titrated volumes of
Z-(aq) solution - at the endpoint - in each of the above experiments (V) are: 12.90 mL,
12.40 mL, 13.00 mL, and 18.30 mL, respectively. The total solution volume - assumed
additive - is given by Vo + V. Convince yourself that equilibrium (i.e., endpoint)
concentrations of M+2(aq), [M+2]eq , and of Z-(aq), [Z-]eq , are given by the
following relationships.
[M+2]eq =
[M+2]o•Vo
Vo + V
and
[Z-]
[Z-]o•V
eq = Vo + V .
Page 10
• First solution: [M+2]o = 0.226 M, [Z-]o = 0.200 M, Vo = 100.00 mL, and V = 12.90 mL
[M+2]eq =
[Z-]
[M+2]o•Vo
(0.226 M)•(100.00 mL)
Vo + V = (100.00 mL + 12.90 mL) = 0.200 M. Also,
[Z-]o•V
(0.200 M)•(12.90 mL)
eq = Vo + V = (100.00 mL + 12.90 mL)
= 0.0229 M.
2
The ion product for the first solution = [M+2]eq•[Z-]eq = (0.200)•(0.0229)2
= 1.05 x 10-4 .
• Second solution: [M+2]o = 0.101 M, [Z-]o = 0.200 M, Vo = 100.00 mL, and V = 12.40 mL
[M+2]eq =
[M+2]o•Vo
(0.101 M)•(100.00 mL)
=
Vo + V
(100.00 mL + 12.40 mL) = 0.0899 M. Also,
[Z-]o•V
(0.200 M)•(12.40 mL)
[Z ]eq = V + V = (100.00 mL + 12.40 mL)
o
= 0.0221 M.
2
The ion product for the second solution = [M+2]eq•[Z-]eq = (0.0899)•(0.0221)2
= 4.39 x 10-5 .
• Third solution: [M+2]o = 0.0452 M, [Z-]o = 0.200 M, Vo = 100.00 mL, and V = 13.00 mL
[M+2]eq =
[M+2]o•Vo
(0.0452 M)•(100.00 mL)
=
Vo + V
(100.00 mL + 13.00 mL) = 0.0400 M. Also,
[Z-]o•V
(0.200 M)•(13.00 mL)
[Z ]eq = V + V = (100.00 mL + 13.00 mL) = 0.0230 M.
o
2
The ion product for the third solution = [M+2]eq•[Z-]eq = (0.0400)•(0.0230)2
= 2.12 x 10-5 .
Page 11
• Fourth solution: [M+2]o = 0.0118 M, [Z-]o = 0.200 M, Vo = 100.00 mL, and V = 18.30 mL
[M+2]eq =
[M+2]o•Vo
(0.0118 M)•(100.00 mL)
Vo + V = (100.00 mL + 18.30 mL) = 0.00997 M. Also,
[Z-]o•V
[Z ]eq = V + V
o
(0.200 M)•(18.30 mL)
= (100.00 mL + 18.30 mL) = 0.0309 M.
2
The ion product for the fourth solution = [M+2]eq•[Z-]eq = (0.00997)•(0.0309)2
= 9.52 x 10-6 .
(b) We tabulate the results below.
Ion Product =
2
[M+2]eq[Z-]eq
(from (a))
+2
2
log10([M+2]eq[Z-]eq) [M ]eq
(from (a))
[M+2]eq
1 + 1.05 x 10-4
4.39 x 10-5
-3.979
0.200 M
0.309
-4.357
0.0899 M
0.231
2.12 x 10-5
9.52 x 10-6
-4.674
0.0400 M
0.167
-5.021
0.00997 M
0.0908
[M+2]eq
2 ⎞
⎛
(c) As stated in the given information, when a plot of log10⎜⎝[M+2]eq•[Z-]eq⎟⎠
- versus
⎛ [M+2]eq ⎞
⎜
⎟
⎝1 + [M+2]eq⎠
is performed, the best-fit line has an extrapolated y-intercept
of -5.46. The Ksp value is the antilog10 of the extrapolated y-intercept - as explained above.
Ksp = antilog10(-5.46) or 10-5.46 = 3.47 x 10-6 .
(d) As explained above, the estimated error is determined by drawing lines of “maximum”
slope and “minimum” slope through the ordinate error bars. The extrapolated y-intercepts
of these maximum and minimum sloped lines give the lower (“smallest”) and the upper
(“largest”) bounds of the Ksp value. Prove this relationship for yourself. These are termed
“Ksp,smallest” and “Ksp,largest”, respectively. The estimated error is then given by:
Error in Ksp value =
Ksp,largest - Ksp,smallest
.
2
Page 12
Thus, the extrapolated y-intercept of the line with maximum slope = -5.51
Thus, the extrapolated y-intercept of the line with minimum slope = -5.41
Ksp,smallest = antilog10(-5.51) or 10-5.51 = 3.09 x 10-6 . Also,
Ksp,largest = antilog10(-5.41) or 10-5.41 = 3.89 x 10-6 .
The estimated error is therefore given by:
Ksp,largest - Ksp,smallest
3.89 x 10-6 - 3.09 x 10-6
Error =
=
2
2
Error = 0.40 x 10-6 .
Thus, from the above data, one may express the Ksp value as (prove for yourself):
Ksp = (3.47 ± 0.40 ) x 10-6 .
(Next Page for Write-up Details)
Page 13
University of Pennsylvania Department of Chemistry
EQUL-308: Solubility Product Constant of Lead (II) Iodide
Laboratory Procedure Reminders & Report Write-up Directions
• PAY CAREFUL ATTENTION TO YOUR T.A., CONCERNING ALL COMMENTS,
DEMONSTRATIONS AND/OR CHANGES IN PROCEDURE FOR EQUL-308.
• WORK IN PAIRS.
• EACH PERSON SHOULD PERFORM THE TITRATION MEASUREMENTS FOR
TWO (2) OF THE FOUR TRIALS.
•
BE SURE TO PROPERLY DISPOSE OF ALL LEAD (Pb) WASTE –
PRECIPITATES & ANY AQUEOUS SOLUTIONS CONTAINING LEAD (Pb)
IN THE DESIGNATED (LABELED) WASTE CONTAINERS.
•
REMEMBER: FILL IN YOUR MEASURED DATA ON THE POSTED
“RAW DATA” SHEET BEFORE YOU LEAVE LAB.
•
REMEMBER: THE WRITE-UP FOR EXP. EQUL-308 IS DUE AT THE
BEGINNING OF THE NEXT LABORATORY PERIOD IN WHICH YOU MEET.
•
The write-up for EQUL-308 will consist of the following items - in the order indicated:
1.
The solutions to the Pre-lab Questions. NOTE: YOU HAVE ALREADY
HANDED THESE IN. They need not be re-submitted.
2.
Data Sheet 1 - on page 7 - is not to be filled in. However, Data Sheet 2 - on page 8
of the lab separate for EQUL-308 - must be completely filled in. Instead of “Average
Class Data”, include your individual data. NOTE: THE CALCULATIONS MUST
BE SHOWN COMPLETELY FOR ONLY ONE (1) OF THE TRIAL SOLUTIONS
- SPECIFY THE TRIAL CLEARLY - ON SEPARATE NOTEBOOK PAPER.
THEN - JUST LIST THE RESULTS OF THE OTHER TRIALS
THE FOLLOWING TWO (2) GRAPHS - CLEARLY LABELED:
2
• GRAPH # 1: PLOT OF [Pb+2]eq•[I-]eq (“y-axis”) VERSUS [Pb+2]eq
(“x-axis”). THE EXTRAPOLATED Y-INTERCEPT (Ksp) MUST BE
CLEARLY LABELED ON YOUR GRAPH. THIS GRAPH IS CONSIDERED
PART OF YOUR CALCULATIONS.
Page 14
•
GRAPH # 2:
2 ⎞
⎛
PLOT OF log10⎜⎝[Pb+2]eq•[I-]eq⎟⎠ (“y-axis”) VERSUS
[Pb+2]eq
1 + [Pb+2]eq
(“x-axis”). THE EXTRAPOLATED Y-INTERCEPT
(log10(Ksp)) MUST BE CLEARLY LABELED. ALSO, THE ± 0.10 VERTICAL
ERROR BARS MUST BE CLEARLY LABELED AND THE “MAX” SLOPE
& “MIN” SLOPE LINES MUST BE CLEARLY DRAWN AND LABELED.
THE EXTRAPOLATED Y-INTERCEPTS (log10(Ksp,smallest)) &
log10(Ksp,largest)) MUST BE CLEARLY LABELED. THIS GRAPH IS
CONSIDERED PART OF YOUR CALCULATIONS. ALL CALCULATIONS
THAT ARE PERFORMED ON DATA EXTRACTED FROM GRAPH # 2
ABOVE SHOULD BE DONE ON SEPARATE NOTEBOOK PAPER (SEE # 3,
BELOW).
3.
REMEMBER: I WOULD LIKE YOU TO DETERMINE THE ERROR IN YOUR
Ksp, AS DISCUSSED IN THE PRE-LAB NOTES.
4. The carefully worked out and explained answer to the following problem - labeled “Problem”
below. Submit the explained solutions to the parts on separate notebook paper.
•
The completed laboratory report is considered to be items 1-4 above. However, you
already handed in the pre-lab questions (Item 1 - above). Thus, you must hand in
items 2, 3, & 4 above - in the order listed - stapled together with your NAME (Printed),
CHEM COURSE # (54), & LAB SECTION # clearly displayed at the top of the first page
of the report.
Problem:
A certain sparingly soluble salt has the chemical formula AZ3(s). In aqueous solution
- upon dissolution - it dissociates into A+3(aq) and Z-(aq).
(a) Write down the balanced chemical equilibrium reaction for a saturated solution of
AZ3(s), A+3(aq), and Z-(aq).
(b) Write down the algebraic expression for the equilibrium constant described by
the reaction in (a) above, i.e., the Ksp expression. The expression should be a function
of: the [A+3]eq , the [Z-]eq , and the mean activity coefficient (γ).
(Problem continued on next page)
Page 15
(c) Following a procedure similar to EQUL-308, a student performs a series of titrations - with
Z-(aq) as the titrant - to achieve saturated solutions for a variety of initial A+3(aq)
3 ⎞
⎛
concentrations at 25oC. Subsequently, a plot of log10⎜⎝[A+3]eq •[Z-]eq⎟⎠ versus
[A+3]eq
1 + [A+3]eq
is constructed and the extrapolated y-intercept of the resultant best-fit line
is determined. The y-intercept is found to be -12.35. What is the Ksp (at 25oC) of AZ3(s)
according to this student’s data? Show all work.
(d) For one of the saturated solutions in part (c) above, the student’s specific data (at 25oC)
were as follows: [A+3]initial (before titration) = 2.50 x 10-2 M, [Z-]initial (before titration)
= 1.95 x 10-2 M, volume of A+3(aq) solution before titration = 100.00 mL, and volume
of Z-(aq) solution (titrant) added to achieve a saturated solution = 9.65 mL. Using this data
and any previous results (as needed), determine the mean activity coefficient (γ)
- at 25oC - for this specific saturated solution. Assume that all aqueous solution
volumes are additive. Show all work.
•
The total possible points for EQUL-308 are 65 points. The points will be awarded in
the following manner.
• Pre-lab Questions:
• Data & Calculations:
(This includes filled-in data sheet and
accompanying calculations.)
• Graph # 1 - (described above)
• Graph # 2 - (described above)
• Accuracy (comparison with actual Kspvalue):
• Solution to “Problem” above:
TOTAL
=
12 points total
20 points total
6 points total
6 points total
9 points
12 points total
65 points