Download Chem. 31 * 9/15 Lecture

Chem. 31 – 4/26 Lecture
Announcements I
• Due today:
– Co/Cr lab
– AA resubmission
• Last Quiz – Next Wednesday
• Exam 2 Results
– Results were pretty typical
– Ave = 70% (broad spread)
Score
90-99
N
4
80s
23
70s
11
60s
19
32-59
16
• Homework Set 3 Posted + solutions to
HW3.1 sub set
Announcements II
• Bonus Points Survey
– Will be posting instructions (Read instructions
BEFORE doing survey) and link to survey soon
• Additional Problem 4 – posted
• Today’s Lecture
• Chapter 9: Acid – Base Chemistry
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–
–
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Strong acid/strong base problems (generalities)
Weak acid problems
Weak base problems
Buffer solutions (if time)
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. MgCO3 dissolution
– 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. 8)
• Weak Acid Problems:
– e.g. What is the pH and the concentration of major
species in a 2.0 x 10-4 M 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 ↔ H+ + AH2O ↔ 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: 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-]
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 Ka 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+ + AInitital
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 10-4 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: a = fraction of dissociation = [A-]/[HA]total
a = 1.2 x 10-4 /2.0 x 10-4 = 0.60
Acid – Base Equilibria
• Weak Acid Problem –
cont.:
1.00E+00
1.00E-01
Assupmption #1
Works
1.00E-02
[HA]o
– When is Assumption #1
valid (in general)?
– When both [HA]o and Ka
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
Where ICE Method Works
1.00E-03
1.00E-04
1.00E-05
Fails
1.00E-06
1.00E-07
1E-10
1E-08
1E-06
0.0001
Ka
0.01
1
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 a and Ka.
Acid – Base Equilibria
• Buffer Solutions:
– A buffer solution is designed so that a small addition
of acid or base will only slightly change the pH
– Most buffer solutions have a weak acid and its
conjugate base both present
– Example: Determine pH of a mix of 0.010 M HCO2H
and 0.025 M Na+HCO2- solution (ignoring activity)
– Go to board to show if ICE approach is needed
Acid – Base Equilibria
• Buffer Solutions:
– Question: Was the ICE Problem set up
needed?
– Answer: No. The assumption of x << [HA],
[A-] is valid for all “traditional” buffers
– Traditional Buffer
• Weak acid (3 < pKa < 11)
• Ratio of weak acid to conjugate base in range 0.1
to 10
• mM+ concentration range
Acid – Base Equilibria
• Buffer Solutions:
– Since ICE not needed, can just use Ka
equation
– Ka = [H+][A-]/[HA] = [H+][A-]o/[HA]o
(always valid)
(valid for traditional buffer)
– But log version more common
– pH = pKa + log([A-]/[HA])
– Also known as Henderson-Hasselbalch
Equation
Acid – Base Equilibria
• Buffer Solutions:
– Why are they needed?/useful?
– The main reason is to keep the pH constant
so that the ratio of species of acids and bases
is constants
– Some examples:
• in water hardness titration, we want [Y4-]/[Y]total
constant so sample pH won’t affect results
• spectroscopy: Beer’s law only applies to single
species (e.g. separate laws for HIn and In-)
• chromatography: at pH = 2, benzoic acid is a
molecule and retained (reversed-phase HPLC)