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CHM 2730 – SP24
Homework Assignment #7 (out of 50 pts)
Due: 8:00am, Wednesday, March 6, 2024
Name________________________
I certify that the work below is my own, I have not copied the work from someone else, nor have I searched for
these answers on the internet. I have completed all of the work on this assignment.
Signature________________________________________________________
For full credit, show all work (each step) in detail, including cancellation of units. You also need to
express your answer to the proper number of significant figures and show units.
1. (12pts) This question involves the reaction below.
5 2+ ( ) + 4− ( ) + 8 + ( ) → 5 3+ ( ) + 2+ ( ) + 4 2 ( )
Suppose you are given a 0.3800 g sample of iron ore and you are tasked with finding the %Fe in the
sample. A procedure is followed to completely dissolve the sample in acidic solution and reduce all
available iron to 2+ ions. It then takes 37.66 mL of a 0.01548 M solution of 4− ( ) to titrate the
sample to the endpoint (indicated by the appearance of the pink color from a slight excess of 4− ).
Use this information to calculate the %Fe in the ore sample. Show your work. (Hint: If you determine
the number of moles of 2+ titrated you have determined the number of moles of Fe in the sample.)
Note: The reaction stoichiometry is not 1:1 here.
Mol of −
added to reach endpoint:
−
−
= . ( ) ×
Mol of + reacted:

+
= . ×
Mass of + = :
−
. −

= . × − −

−
×
. × − ×
+
−
+
− = . ×

.
= . = .

% :
.
× = . %
.
1
2) In lecture we worked through the example from the Harris 9th ed. text (Section 7-3) involving the
precipitation titration of 25.00 mL of a 0.1000 M solution of − ( ) by a 0.05000 M solution of + (aq).
The [ + ] value was calculated for three different points in this titration:
→After 10.00 mL of + ( ) titrant has been added (but before the equivalence point).
→At the equivalence point.
→Just after the equivalence point.
This question involves a similar situation that should be more familiar to you – the precipitation titration
of a sample of soluble chloride by a known concentration of 3 ( ) solution. As with the text
example for the AgI (s) titration, we will calculate the [ + ] values in the flask at three different points
in the titration:
1. After 10.00 mL of [ + ] titrant has been added (but before the equivalence point).
2. At the equivalence point.
3. Just after the equivalence point.
Suppose you dissolved 0.1488 g of a solid that was 55.78% − in enough water to make 25.00 mL of
solution.
The mass of − in this sample would be:
− = 0.1488 × 0.5578 = 0.083001
The [ − ] in the 25.00-mL solution could then be calculated:
[ − ] =
1 −
35.54 −
= 0.093654
0.02500
0.083001 − ×
If the 25.00 mL of 0.093654 − ( ) solution is quantitatively transferred to
a beaker, and a buret is filled with a standardized 3 ( ) solution where
[ + ] = 0.1075 M, we would have a setup that could be represented by the
crude diagram on the right. (Note: this titration would require monitoring the
[ + ] with an ion-selective electrode (instead of using an indicator). A
simulated titration curve for this experiment is below:
2
a) (5pts) Calculate the volume of 0.1075 M + ( ) titrant needed to reach the equivalence point (aka
the equivalence point volume ). Show your work. (Hint: is the volume of 0.1075 M + ( ) titrant
required to deliver enough moles of + to exactly react with the number of moles of − originally in
the beaker; recall that the precipitation of AgCl (s) results from a simple 1:1 reaction between + and
− .)
# − = . ×
. −
= . −

= . + ( ) . +
×
=
. +
= . +
+ ( )
. +
= . = .
. +
+ ( )
b) (1pt) Find the midpoint of the steep descent in the AgCl titration curve on the previous page. This
midpoint is the titration equivalence point. The volume of + titrant at this midpoint should equal the
equivalent point volume . Use a ruler to estimate from the graph. What volume do you obtain? How
does that compare to your calculated value of in part a)?
Estimated value of from the graph is 21.82 mL, which is in good agreement with the value
calculated in part a.
c) Calculate the value of [ + ] after addition of 10.00 mL of the 0.1075 M + ( ) titrant to the 25.00
mL of 0.09365 4 M − ( ) solution in the beaker. NOTE: For full credit on this question, you need to
show two different methods to obtain a solution: 1) The “conventional” method, which is how you would
have likely approached this using skills obtained in General Chemistry, and 2) The “streamlined”
method, which is introduced in the Harris text for the AgI (s) example on p. 150. For both methods,
show your work. Both methods have step-by-step instructions below:
(4pts) The ”conventional” method (with step-by-step guidance):
Step 1: Calculate the #mol − remaining in beaker after 10.00 mL 0.1075 + ( ) titrant added.
# − = # − − # +
− ( ) = . −
. +
= . ×
= . +

+
# − = . − − . + = .
3
Step 2: Calculate the total volume ( ) now in the beaker.
= . + . = .
Step 3: Calculate the [ − ] now in the beaker. (You have #mol − from Step 1 and from Step 2.)
[ − ] =
.
= . = .
.
Step 4: Calculate the [ + ] in equilibrium with the [ − ] calculated in Step 3 by using the of AgCl.
Include 3 sig figs in your answer.
[ + ] =
= . × − = [ + ][ − ]
. × −
= . × − = . × −
.
Step 5: Determine [ + ] from [ + ] calculated in Step 4. Recall that [ + ] = − log[ + ]
[ + ] = − [ + ] = − ( . × − ) = . = .
(4pts) The ”streamlined” method:
Step 1: Calculate [ − ] in the beaker after addition of the 10.00 mL of 0.1075 M + ( ) titrant using
as an example the single line calculation demonstrated for the AgI (s) in the lecture notes and on p.
150 of the Harris text. (We are essentially combing Steps 1 -3 of the conventional method into one
quick calculation.)
[ − ] = �
.
.
� × . × �
� = . = .
.
.
Step 2: Using the results of the single-line calculation above, calculate [ + ] and [ + ]. Include 3
sig figs in your answer.
[ + ] =
= . × − = [ + ][ − ]
. × −
= . × − = . × −
.
[ + ] = − [ + ] = − ( . × − ) = . = .
4
d) (1pt) Using the titration curve for AgCl on the first page of this problem, estimate the value of [ + ]
when 10.00 mL of 0.1075 M + ( ) titrant has been added. How does the value you obtain compare
to the value you calculated in part c)?
Estimated value of [ + ] from the graph is 8.39, which is in good agreement with the value
calculated in part c.
e) (5pts) Calculate [ + ] at the equivalence point. Show your work. (Hint: Remember at the
equivalence point you can assume that all − originally in the beaker has been reacted with + ( )
to form AgCl (s). However, [ + ] and [ − ] do not go to zero at the equivalence point; there is some
of each from the tiny amount of AgCl (s) that re-dissolves as dictated by its value.) Include 3 sig
figs in your answer.
( ) ⇋ + ( ) + − ( )
= [ + ][ − ] = . × − =
= � . × − = . × − = [ + ]
[ + ] = − [ + ] = − � . × − � = .
f) (1pt) Using the titration curve for AgCl on the first page, estimate the value of [ + ] at the
equivalence point (midpoint on the steep descent part of the curve). How does the value you obtain
compare to the value you calculated in part e)?
Estimated value of [ + ] at the equivalence point from the graph is 4.80, which is in good
agreement with the value calculated in part e.
g) Calculate [ + ] when a total of 25.00 mL of 0.1075 M + ( ) titrant has been added. NOTE:
For full credit on this question, you need to show two different methods to obtain a solution: 1) The
“conventional” method and 2) the “streamlined” method (as in part c above). For both methods, show
your work. (Hint: Past the equivalence point, the [ + ] in the beaker is totally determined by the added
excess + from the buret.) Both methods have step-by-step instructions below:
(4pts) The ”conventional” method (with step-by-step guidance):
Step 1: Calculate the excess volume (beyond the equivalence point) of 0.1075 M + ( ) titrant that
has been added.
0.1075 + ( ) = 25.00 –
+ = . − . = .
5
Step 2: Calculate the #mol + in this excess volume of 0.1075 M + ( ) titrant determined in Step
1.
# + = . ×
. +
= . + = . +

Step 3: Calculate [ + ] in the beaker from the #mol + and the total volume ( ) in the beaker.
[ + ]
. +
. +
=
= . +
=
. + . + .
.
Step 4: Calculate [ + ] from the [ + ] determined in Step 3.
[ + ] = − [ + ] = − ( . + ) = .
(4 pts) The ”streamlined” method:
Step 1: Calculate [ + ] in the beaker after addition of the 25.00 mL of 0.1075 M + ( ) titrant using
as an example the single line calculation demonstrated for the AgI (s) in the lecture notes and on p.
151 of the Harris text. (We are essentially combing Steps 1 -3 of the conventional method into one
quick calculation.)
+
[ + ] = [ + ] × �
�

.
[ + ] = . × �
� = .
.
Step 2: Using the results of the single-line calculation above, calculate [ + ].
[ + ] = − [ + ] = − ( . + ) = .
h) (1pt) Using the titration curve for AgCl on the first page of this problem, estimate the value of [ + ]
when a total of 25.00 mL of 0.1075 M + ( ) titrant has been added. How does the value you obtain
compare to the value you calculated in part g)?
Estimated value of [ + ] when a total of 25.00 mL of titrant have been added is 2.10, which is
in good agreement with the value calculated in part g.
6
i) (5pts) In this question we examined the titration curve for a AgCl precipitation reaction where we
already knew the % − in the sample. Suppose you did NOT know the% − in the sample and that
you had to calculate it from data obtained and graphed in the titration curve on the first page of this
problem. Show how you would find the % − in a 0.1488 g sample of soluble chloride using the
equivalence point volume determined from the graph and the concentration of the standardized
+ ( ) titrant (which was 0.1075 M). You already know the answer (55.78 %); you just need to clearly
show how you would use the information from the titration curve to obtain that value for the ”unknown”.
Show your work.
= .
# + = . ×
= −
. +
= . +

− = . − ×
% − =
. −
= . −
−
. −
× = . %
.
j) (3pts) In lab you determined the endpoint of the titration of a known mass of your chloride unknown
by titration with a standardized solution of 3 ( ) and the indicator dichlorofluorescein. Note that
you determined the endpoint by the color change; not necessarily the equivalence point. Why is this
distinction important? Explain.
The endpoint is where you detect that the titration is complete.
The equivalence point is where the titration is complete.
The difference (in the volume of titrant) between the equivalence point and the endpoint is the
titration error. Depending on the observation of a color change often (but not always) causes
you to “overshoot” the equivalence point. You of course try to minimize this “overshoot” by an
appropriate choice of reaction conditions and experimental set-up.
7
Chapter 6: Chemical Equilibrium
Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved
Chapter 6
Review:
Chemical
Equilibrium
(Ch
6)
2
Example: Reaction of N2(g) and H2(g) to produce ammonia NH3(g)
→This is the Haber‐Bosch Process for
the production of ammonia. The
industrial process requires high pressure,
high temperature, and a catalyst.

3
2
→ Produces ~450 million tons of
nitrogen fertilizer per year.
→ Drove the world’s population from 1.6
billion in 1900 to almost 8 billion today.
Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved
Chapter 6
Review:
Chemical
Equilibrium
(Ch
6)
3
Example: Reaction of N2(g) and H2(g) to produce ammonia NH3(g)
→This is the Haber‐Bosch Process for
the production of ammonia. The
industrial process requires high pressure,
high temperature, and a catalyst.

3
2
→ Produces ~450 million tons of
nitrogen fertilizer per year.
→ Drove the world’s population from 1.6
billion in 1900 to almost 8 billion today.
Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved
Chapter 6
Review:
Chemical
Equilibrium
(Ch
6)
4
Forward Reaction
Reverse Reaction
N2(g) + 3H2(g)
2NH3(g)
reactants
products
double headed arrow
Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved
Chapter 6
Review:
Chemical
Equilibrium
(Ch
6)
5
For general reaction:
aA + bB
cC + dD
The equilibrium equation is:
Kc =
[C]c [D]d
[A]a [B]b
products
reactants
equilibrium
equilibrium
constant,
constant expression
c meaning concentration
Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved
Chapter 6
Review:
Chemical
Equilibrium
(Ch
6)
6
For the Haber- Bosh Process:
The equilibrium equation is:
[NH3]2
at 500C
=
0.058
K=
[N2][H2]3
Equilibrium constants do not have units
Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved
Chapter 6
Review:
Chemical
Equilibrium
(Ch
6)
7
The equilibrium constant, K, is related to the thermodynamics of the
reaction from left to right (Δ , Δ , Δ , T
Kinetics, or reaction rates, are related to how quickly a reaction reaches
its equilibrium state. The equilibrium state may have reactant favored,
product favored, or a mix of both reactants and products).
Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved
Chapter 6
Review:
Chemical
Equilibrium
(Ch
6)
8
If Kc > 103, products predominate over reactants at equilibrium. If Kc is
very large, the reaction is said to proceed to completion.
CO(g) + H2O(g)
Kc =
Reactants
[CO2][H2]
[CO][H2O]
CO2(g) + H2(g)
= 1.0 X 105 at 25C
Products
Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved
Chapter 6
Review:
Chemical
Equilibrium
(Ch
6)
9
If Kc is in the range 103 to 103, appreciable concentrations of both
reactants and products are present at equilibrium.
I2(g) + Cl2(g)
Kc =
Reactants
[ICl]2
[I2][Cl2]
2ICl(g)
= 81.9
at 25C
Products
Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved
Chapter 6
Review:
Chemical
Equilibrium
(Ch
6)
10
If Kc < 103, reactants predominate over products at equilibrium. If Kc is very small, the reaction is said to proceed hardly at all. 2H2S(g) 2H2(g) + S2(g) [H2]2[S2] = 1.7 X 107 at 800 C Kc = [H2S]2 Reactants Products Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved Chapter 6 Example Calculation 11 Example Problem: Calculate [N2], [H2], and [NH3] at equilibrium at 500°C if initially [N2]=0.275 M, [H2]=0.275 M, and [NH3]= 0.400 M I C E N2(g) [H2](g) [NH3] 0.275 0.275 0.400 First, we need to ask, “Is this already at equilibrium?” Based on our K = 0.058 at 500°C, the reaction will occur until it reaches that K value Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved Chapter 6 Example Calculation 12 Example Problem: Calculate [N2], [H2], and [NH3] at equilibrium at 500°C if initially [N2]=0.275 M, [H2]=0.275 M, and [NH3]= 0.400 M 0.058 = K when at equilibrium . . . 27.9 we are not at equilibrium. Therefore, this is the Reaction Quotient, Q. . . When . 27.9 the reaction is at equilibrium. Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved Chapter 6 Example Calculation 13 The reaction will occur until Q = K, at which point the reaction is at equilibrium. . . . 27.9 In our case, this means that the numerator needs to decrease, and the denominator needs to increase for Q to approach K. →Therefore, the reaction will proceed left (toward the reactants) to reach equilibrium. →Let’s return to the ice table and complete it to calculate the concentrations of [N2], [H2], and [NH3] at equilibrium (at 500°C). Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved Chapter 6 Example Calculation 14 Example Problem: Calculate [N2], [H2], and [NH3] at equilibrium at 500°C if initially [N2]=0.275 M, [H2]=0.275 M, and [NH3]= 0.400 M I C E N2(g) [H2](g) [NH3] 0.275 +x 0.275+x 0.275 + 3x 0.275+3x 0.400 -2x 0.400-2x 0.058 0.400 2 0.275 0.275 3 Let’s use Wolfram Alpha to solve this for x . . . Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved Chapter 6 Example Calculation 15 Wolfram Alfa returns the following x values: X = - 2.02494 → not possible, would give negative [N2] and [H2] concentrations at equilibrium. X = 0.151133 (this works – we will test this out) X = 0.337933 → not possible, would give negative [NH3] concentration at equilibrium X = 0.985877 → not possible, would give negative [NH3] concentration at equilibrium Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved Chapter 6 Example Calculations 16 At Equilibrium, X = 0.151133 →Let’s see if this one works [N2] = [0.275+x] = [0.275+ 0.151133] = 0.426 M [H2] = [0.275+3x] = [0.275+ 3(0.151133)] = 0.728 M [NH3] = [0.400-2x] = [0.400-2(0.151133)] = 0.0977 M →Do these concentrations make sense? How can we check? 0.058 0.0977 0.426 0.728 0.058 Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved General Characteristics of K and equilibrium constant 17 expressions Chapter 6 →When the reaction is written in reverse, K 3 ⇌ 2 0.058 When K 1 the product side will be favored. →Both say the same thing: the side with the [N2] and [H2] is favored. Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved General Characteristics of K and equilibrium constant 18 expressions Chapter 6 →Pure solids (s) and pure liquids (l) do not appear in the equilibrium constant expression. ⇌ we will eventually call this 2 ⇌ 1 →The overall equilibrium constant for the sum of two or more reactions is equal to the product of the individual K values. Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved Chapter 6 Example Calculation 19 →Example, page 121 of text: ⇌ ⇌ Calculate K for the reaction: ⇌ 1 ⇌ 1 1.8 10 1 1.0 10 10 ⇌ ⇌ 1.0 10 1.8 5.6 10 Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved Chapter 6 Equilibrium Constant, K 20 Why do K values have no units? →The so-called “thermodynamic equilibrium constant expression” has each quantity expressed as the ratio of the concentration of a species to is concentration in the standard state. →For solutions, the standard state is 1 M →For gases, the standard state is 1 bar (1 bar = 0.9869 atm) →For solids and liquids, the standard state is the pure solid or liquid Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved Chapter 6 Equilibrium Constant, K 21 Example: ⇌ 1 1 1 Example: These are gaseous solutions ⇌ 0.058 In Gen Chem we would call this KP (only = K when # mol gaseous reactants = # mol gaseous products) Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved Chapter 6 Equilibrium Constant, K 22 In actuality, a “true” thermodynamic equilibrium constant, K, will use mixed units. Example: Re-dox reaction: 3 2 ⇌ 1 1 1 1 2 This looks a bit strange, but, luckily, most equilibria we will deal with will involve aqueous solutions Note: The equilibrium constant expression is most rigorously expressed as a ratio of activities, not concentrations or pressures. We will not cover activities in this class (Although the text covers this in Chapter 8). Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved Chapter 6 Thermodynamics and K 23 Recall that the larger the K for a reaction, the greater the tendency to form products; that is, K is related to the thermodynamics Δ , Δ , Δ , of the reaction. Δ Δ Δ If Δ is (‐) → Δ →(‐)[exothermic reaction]; Δ →(+)[increasing entropy] If Δ is (+) → Δ →(+)[endothermic reaction]; Δ →(‐)[decreasing entropy] If Δ is (‐) and Δ (‐) →Δ →negative or positive depending on the magnitudes of Δ , Δ , . If Δ is (+) and Δ (+) →Δ →negative or positive depending on the magnitudes of Δ , Δ , . Copyright © 2023 Tiffany M. Pellizzeri, All Rights Reserved Chapter 6 Thermodynamics and K 24 Recall that the larger the K for a reaction, the greater the tendency to form products; that is, K is related to the thermodynamics Δ , Δ , Δ , of the reaction. If Δ is (‐) → K>1; products are favored
If Δ is (+) → K
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