Equilibrium and Acid-Base Problems in Chemistry Lecture
Chem. 31 – 11/27 Lecture
Announcements
•
Labs Due Wednesday
–
GC
•
Today’s Lecture
–
Chapter 8 – Advanced Equilibrium
•
The systematic method and its six steps – more practice
•
Generalities about using the systematic method
–
Chapter 9 – Acid/Base Equilibria
•
The “weak acid problem” (pH of weak acid in water)
•
The weak base problem
The Systematic Method
2
nd
Example
•
An aqueous mixture of CdCl
2
and NaSCN is
made
–
Initial concentrations are [CdCl
2
] = 0.0080 M and
[NaSCN] = 0.0040 M
–
Cd
2+
reacts with SCN
-
to form CdSCN
+
K = 95
–
HSCN is a strong acid
–
Ignore any other reactions (e.g. formation of CdOH
+
)
–
Ignore activity considerations
–
Go through steps 1 through 5
The Systematic Method
3
rd
Example
•
A student prepares a solution that
contains 0.050 mol of AgNO
3
and 0.0040
mol NH
3
in water with a total volume of
1.00 L. The AgNO
3
is totally soluble, NH
3
is a weak base, and Ag
+
reacts with NH
3
to form Ag(NH
3
)
2
+
. Assume the Ag
+
does not react with water or OH
-
. Go
through the first 5 steps of the
systematic method.
The Systematic Method
Stong Acid/Strong Base Problems
•
When do we need to use the systematic
approach?
–
when more than 1 coupled reaction occur
(unless coupling is insignificant)
–
examples: 4.0 x 10
-3
M HCl. 7.2 x 10
-3
M
NaOH
–
Key point is the charge balance equation:
-
for strong acid HX, [H
+
] = [X
-
] + [OH
-
]
-
If [X
-
] >> [OH
-
], then [H
+
] = [X
-
]
–
for strong base NaOH, [H
+
] + [Na
+
] = [OH
-
]
The Systematic Method
General Comments
•
Effects of secondary reactions
–
e.g. MgCO
3
dissolution
–
Additional reactions increase solubility
–
Secondary reactions also can affect pH (CO
3
2-
+ H
2
O will produce OH
-
while Mg
2+
+ H
2
O will
produce H
+
)
•
Software is also available to solve these
types of problems (but still need to know
steps 1
→
5 to get problems solved)
Acid – Base Equilibria (Ch. 9)
•
Weak Acid Problems:
–
e.g. What is the pH and the concentration of major
species in a 2.0 x 10
-4
M HCO
2
H (formic acid, K
a
=
1.80 x 10
-4
) solution ?
–
Can use either systematic method or ICE method.
–
Systematic method will give correct answers, but full
solution results in cubic equation
–
ICE method works most of the time
–
Use of systematic method with assumptions allows
determining when ICE method can be used
Acid – Base Equilibria
•
Weak Acid Problem – cont.:
–
Systematic Approach (HCO
2
H = HA to make problem
more general where HA = weak acid)
•
Step 1 (Equations) HA
↔ H
+
+ A
-
H
2
O ↔ H
+
+ OH
-
•
Step 2: Charge Balance Equation: [H
+
] = [A
-
] + [OH
-
]
2 assumptions possible: ([A
-
] >> [OH
-
] – assumption used in ICE
method or [A
-
] << [OH
-
])
•
Step 3: Mass Balance Equation: [HA]
o
= 2.0 x 10
-4
M = [HA]
+ [A
-
]
•
Step 4: K
w
= [H
+
][OH
-
] and K
a
= [A
-
][H
+
]/[HA]
•
Step 5: 4 equations (1 ea. steps 2 + 3, 2 equa. step 4), unk.:
[HA], [A
-
] [H
+
], [OH
-
]
Acid – Base Equilibria
•
Weak Acid Problem – cont.:
–
Assumption #1: [A
-
] >> [OH
-
] so [A
-
] = [H
+
]
–
Discussion: this assumption means that we expect
that there will be more H
+
from formic acid than from
water. This assumption makes sense when [HA]
o
is
large and K
a
is not that small (valid for [HA]
o
>10
-6
M
for formic acid)
–
ICE approach (Gives same result as systematic
method if assumption #1 is made)
–
(Equations) HA
↔ H
+
+ A
-
Initital 2.0 x 10
-4
0 0
Change - x +x +x
Equil. 2.0 x 10
-4
– x x x
Acid – Base Equilibria
•
Weak Acid Problem – Using ICE Approach
–
K
a
= [H
+
][A
-
]/[HA] = x
2
/(2.0 x 10
-4
– x)
x = 1.2 x 10
-4
M (using quadratic equation)
Note: sometimes (but not in this case), a 2
nd
assumption can be made that x << 2.0 x 10
-4
to
avoid needing to use the quadratic equation
[H
+
] = [A
-
] = 1.2 x 10
-4
M; pH = 3.92
[HA] = 2.0 x 10
-4
– 1.2 x 10
-4
= 8 x 10
-5
M
Note:
= fraction of dissociation = [A
-
]/[HA]
total
= 1.2 x 10
-4
/2.0 x 10
-4
= 0.60
Acid – Base Equilibria
•
Weak Acid Problem –
cont.:
–
When is Assumption #1
valid (in general)?
–
When both [HA]
o
and K
a
are high or so long as
[H
+
] > 10
-6
M
–
More precisely, when
[HA]
o
> 10
-6
M and
Ka[HA]
o
> 10
-12
–
See chart (shows region
where error < 1%)
–
Failure also can give [
H
+
]
< 1.0 x 10
-7
M
Assupmption #1
Works
Fails
Acid – Base Equilibria
•
Weak Base Problem:
–
As with weak acid problem, ICE approach can
generally be used (except when [OH
-
] from base is
not much more than [OH
-
] from water)
–
Note: when using ICE method, must have correct
reaction
–
Example: Determine pH of 0.010 M NH
3
solution
(K
a
(NH
4
+
) = 5.7 x 10
-10
, so K
b
= K
w
/K
a
= 1.75 x 10
-5
)
–
Reaction
NH
3
+ H
2
O
NH
4
+
+ OH
-
–
Go over on board
Acid – Base Equilibria
•
Weak Acid/Weak Base Questions:
–
A solution is prepared by dissolving 0.10 moles of
NH
4
NO
3
into water to make 1.00 L of solution. Show
how to set up this problem for determining the pH
using the ICE method.
–
A student is solving a weak base problem for a weak
base initially at 1.00 x 10
-4
M using the ICE method
and calculates that [OH
-
] = 2.4 x 10
-8
M. Was the
ICE method appropriate?
–
The pH of an unknown weak acid prepared to a
concentration 0.0100 M is measured and found to be
3.77. Calculate
and K
a
.
In this lecture, topics such as Advanced Equilibrium, Acid/Base Equilibria, Systematic Method for solving chemical problems, Strong Acid/Strong Base scenarios, and General Comments on reactions are discussed. Examples using the systematic method are provided for practical understanding. Key points on utilizing the systematic approach, coupled reactions, and effects of secondary reactions are detailed, along with a focus on Weak Acid Problems and pH calculations. Various methods such as ICE method and software solutions are covered for solving chemical equilibrium problems.
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Presentation Transcript
Announcements Labs Due Wednesday GC Today s Lecture Chapter 8 Advanced Equilibrium The systematic method and its six steps more practice Generalities about using the systematic method Chapter 9 Acid/Base Equilibria The weak acid problem (pH of weak acid in water) The weak base problem
The Systematic Method 2ndExample An aqueous mixture of CdCl2and NaSCN is made Initial concentrations are [CdCl2] = 0.0080 M and [NaSCN] = 0.0040 M Cd2+reacts with SCN-to form CdSCN+K = 95 HSCN is a strong acid Ignore any other reactions (e.g. formation of CdOH+) Ignore activity considerations Go through steps 1 through 5
The Systematic Method 3rdExample A student prepares a solution that contains 0.050 mol of AgNO3and 0.0040 mol NH3in water with a total volume of 1.00 L. The AgNO3is totally soluble, NH3 is a weak base, and Ag+reacts with NH3 to form Ag(NH3)2+. Assume the Ag+ does not react with water or OH-. Go through the first 5 steps of the systematic method.
The Systematic Method Stong Acid/Strong Base Problems When do we need to use the systematic approach? when more than 1 coupled reaction occur (unless coupling is insignificant) examples: 4.0 x 10-3M HCl. 7.2 x 10-3M NaOH Key point is the charge balance equation: - for strong acid HX, [H+] = [X-] + [OH-] - If [X-] >> [OH-], then [H+] = [X-] for strong base NaOH, [H+] + [Na+] = [OH-]
The Systematic Method General Comments Effects of secondary reactions e.g. MgCO3dissolution Additional reactions increase solubility Secondary reactions also can affect pH (CO32- + H2O will produce OH-while Mg2++ H2O will produce H+) Software is also available to solve these types of problems (but still need to know steps 1 5 to get problems solved)
Acid Base Equilibria (Ch. 9) Weak Acid Problems: e.g. What is the pH and the concentration of major species in a 2.0 x 10-4M HCO2H (formic acid, Ka= 1.80 x 10-4) solution ? Can use either systematic method or ICE method. Systematic method will give correct answers, but full solution results in cubic equation ICE method works most of the time Use of systematic method with assumptions allows determining when ICE method can be used
Acid Base Equilibria Weak Acid Problem cont.: Systematic Approach (HCO2H = HA to make problem more general where HA = weak acid) Step 1 (Equations) HA H2O Step 2: Charge Balance Equation: [H+] = [A-] + [OH-] 2 assumptions possible: ([A-] >> [OH-] assumption used in ICE method or [A-] << [OH-]) Step 3: Mass Balance Equation: [HA]o= 2.0 x 10-4M = [HA] + [A-] Step 4: Kw= [H+][OH-] and Ka= [A-][H+]/[HA] Step 5: 4 equations (1 ea. steps 2 + 3, 2 equa. step 4), unk.: [HA], [A-] [H+], [OH-] H++ A- H++ OH-
Acid Base Equilibria Weak Acid Problem cont.: Assumption #1: [A-] >> [OH-] so [A-] = [H+] Discussion: this assumption means that we expect that there will be more H+from formic acid than from water. This assumption makes sense when [HA]ois large and Kais not that small (valid for [HA]o>10-6M for formic acid) ICE approach (Gives same result as systematic method if assumption #1 is made) (Equations) HA H++ A- Initital 2.0 x 10-4 0 0 Change - x +x +x Equil. 2.0 x 10-4 x x x
Acid Base Equilibria Weak Acid Problem Using ICE Approach Ka= [H+][A-]/[HA] = x2/(2.0 x 10-4 x) x = 1.2 x 10-4 M (using quadratic equation) Note: sometimes (but not in this case), a 2nd assumption can be made that x << 2.0 x 10-4to avoid needing to use the quadratic equation [H+] = [A-] = 1.2 x 10-4 M; pH = 3.92 [HA] = 2.0 x 10-4 1.2 x 10-4= 8 x 10-5M Note: = fraction of dissociation = [A-]/[HA]total = 1.2 x 10-4 /2.0 x 10-4= 0.60
Acid Base Equilibria Weak Acid Problem cont.: When is Assumption #1 valid (in general)? When both [HA]oand Ka are high or so long as [H+] > 10-6M More precisely, when [HA]o> 10-6M and Ka[HA]o> 10-12 See chart (shows region where error < 1%) Failure also can give [H+] < 1.0 x 10-7M Where ICE Method Works 1.00E+00 1.00E-01 Assupmption #1 Works 1.00E-02 1.00E-03 [HA]o 1.00E-04 1.00E-05 Fails 1.00E-06 1.00E-07 1E-10 1E-08 1E-06 0.0001 0.01 1 Ka
Acid Base Equilibria Weak Base Problem: As with weak acid problem, ICE approach can generally be used (except when [OH-] from base is not much more than [OH-] from water) Note: when using ICE method, must have correct reaction Example: Determine pH of 0.010 M NH3 solution (Ka(NH4+) = 5.7 x 10-10, so Kb = Kw/Ka = 1.75 x 10-5) Reaction NH3 + H2O NH4+ + OH- Go over on board
Acid Base Equilibria Weak Acid/Weak Base Questions: A solution is prepared by dissolving 0.10 moles of NH4NO3 into water to make 1.00 L of solution. Show how to set up this problem for determining the pH using the ICE method. A student is solving a weak base problem for a weak base initially at 1.00 x 10-4 M using the ICE method and calculates that [OH-] = 2.4 x 10-8 M. Was the ICE method appropriate? The pH of an unknown weak acid prepared to a concentration 0.0100 M is measured and found to be 3.77. Calculate and Ka.
