18 Additional Aspects of Acid – Base Equilibria
Common ion effect
solutions containing acids, bases, salts, solvent
Buffer solutions
solutions containing a weak acid and its salt
Indicators (acid-base)
colored acids and bases
Titration curve
chemistry and pH during titration
Solutions of salts and polyprotic acids (hydration)
reaction of solvent with ions
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Exam paper return
The exam papers are piled according to the last
digit of your ID number from 0 to 9.
All papers are now in front of the Chem123 Lab on
the shelf. You may retrieve yours from the
hallway in ESC on the first floor.
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Ka Kb and Kw
H+ + Base = Conjugate_acid of Base+
Acid = H+ + Conjugate_base of AcidFor example:
A- + H2O = HA + OH[HA] [OH-] [H+]
Kb = ————— ———
[A-]
[H+]
NH3 + H2O = NH4+ + OHKa for NH4+ = Kw / Kb for NH3
HA = H+ + AKb for A- = Kw / Ka for HA
Thus, Ka Kb = Kw for conjugate pairs.
18 Acid-Base Equilibria
[HA]
= ———— [OH-] [H+]
[A-] [H+]
1
= —— Kw
Ka
3
Ka Kb and Kw - comparison
A- + H2O = HA + OH[HA] [OH-] [H+]
Kb = ————— ———
[A-]
[H+]
HA = H+ + A[H+] [A-] [OH-]
Ka = ———— ———
[HA] [OH-]
[HA]
= ———— [OH-] [H+]
[A-] [H+]
[A-]
= ————— [H+] [OH-]
[HA] [OH-]
1
= —— Kw
Ka
1
= —— Kw
Kb
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Ka Kb and Kw – another way
+)
HA + H2O (l) = H3O+ (aq) + A– (aq)
A– + H2O (l) = OH–(aq) + HA (aq)
2 H2O (l) = H3O+ (aq) + OH– (aq)
Ka of HA
Kb of A–
Kw = Ka Kb
Review qn:
When you add two equations to get a third, what are the
relationship between the Ks?
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Ka, Kb of
A
& Kw
Two ways to show their relationship
A- + H2O = HA + OH[HA] [OH-] [H+]
Kb = ————— ———
[A-]
[H+]
[HA]
= ———— [OH-] [H+]
[A-] [H+]
HA = H+ + A-
Ka
A- + H2O = HA + OH-
Kb
-- add them together - H2O = H+ + OH-
Kw = Ka Kb
1
= —— Kw
Ka
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Properties of salt solutions
The spectator ions in acid-base reactions form salts. Salts completely
ionize in their aqueous solutions. A salt NH4A, the ionizes
NH4A = NH4+ + AThe ions of salts interact with water (called hydration),
A- + H2O = HA + OHIf the anions are stronger base than H2O, the solution is basic.
NH4+ + H2O = NH3 + H3O+
If the cations are stronger acid than H2O, the solution is acidic.
These competitive reactions make the solution acidic or basic
depending on the strength of the acids and bases.
Are the solutions of the following salts neutral, basic or acidic
NaCl?
NaAc?
NH4Cl?
NH4Ac?
KClO4?
KNic?
CH3NH3ClO4?
modified
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Hydration problems
What is the pH and concentrations of various species of a 0.100 M
KA salt solution if Ka for HA = 1.0e-5?
(K = potassium, HA a general acid)
Solution:
KA = K+ + AA– + H2O (l) = OH–(aq) + HA (aq)
0.10-x
x
x
Kb = 1e–14/1.0e–5 = 1e–9
x2
Kb = ———— = 1.0e–9
0.10 – x
(0.10 – x)  0.10 = [A–]
x = 1e-9*0.1 = 1e–5 = [OH–] = [HA]
pOH = 5 (= pKb / 2=10/2); pH = 14 – 5 = 9 (basic)
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Acidity of Salt Solutions
What is the pH of a 1.0-M NaHCO3 solution? For H2CO3, Ka1 = 4.3e-7
and Ka2 = 4.8e-11.
Solution: In the solution, [Na+] = 1.0 M and [HCO3–] = 1.0 M. Two
competitive reaction takes place:
H2O + HCO3– = H2CO3 + OH–
HCO3– = H+ + CO32–
Kb = Kw / Ka1 = 2.33e-8
Ka2 = 4.8e-11
Thus, HCO3 – is a stronger base than an acid. The solution is basic.
Discussion: The overall dissociation constant,
Koverall = Ka1* Ka2 = 2.1e-17
Generalization: if Ka2 < Kb or Ka1*Ka2 < Kw, the solution of NaHA is
basic for the diprotic acid H2A.
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Since salts, acids and
bases ionize in their
solutions, their common
Calculate the degree of ionization of HA
ions such as H+, OH–,
in a solution containing 0.10 M each of
other cations, anions, will
HCl and HA, if Ka = 1.0e–5 for HA.
affect the equilibria.
Solution: (assume [A-] = x and work out the relations as shown)
HCl = H+ + Cl–
0.10+x 0.10
(0.10+x) x
HA = H+ + A– Ka = 1.0e–5 = ——————
0.10-x 0.10+x
x
0.10-x (0.1-x=0.1+x=0.1)
Common ion effect
x = 1.0e–5 M
degree of ionization = 1.0e–5/0.1 = 1.0e–4 = 0.01%
ReDo if [HA] = 0.01 M and [HCl] = 0.10 M?
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Mass and Charge Balance Equations
In a solution of electrolytes, the stoichiometry leads to mass or material
balance (mbe) and charge balance equations (cbe).
The merit is to consider the various ion species and equilibria in the
solution when we write these equations.
In a 0.10 M NaHCO3 solution,
mbe: 0.10 = [Na+] = [H2CO3] + [HCO3 –] + [CO32– ]
cbe: [Na+] + [H+] = [HCO3 –] + 2 [CO32– ] + [OH–]
Consider these equilibrium eqn
for mbe and cbe:
NaHCO3 = Na+ + HCO3HCO3- + H2O = H2CO3 + OHHCO3- + H2O = CO32- + H3O+
New See Week 8 Part b problem #1.
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Buffers
A buffer contains a weak acid and a salt of the same acid.
A buffer contains a weak base and a salt of the same base.
Concentrations of various species for a solution containing 0.10 M each of
KA and HA, Ka = 1.0e–5.
KA =
K+ + A–
0.10
0.10+x
x (0.10+x)
HA = H+ + A– Ka = 1.0e–5 = ——————
0.10-x
x
0.10+x
0.10-x (0.1-x=0.1+x=0.1)
x = 1.0e–5 = [H+]
pH = pKa in this case, because [HA] = [A–]
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pH of buffers
For a weak acid, HA, we have
HA =
H+
+
A–
[H+] [A–]
Ka = ————
[HA]
[A–]
– log Ka = – log [H+] – log ———
[HA]
[A–]
pKa = pH – log ——
[HA]
Adding acid or base only
affect the ratio [A–] / [HA],
the pH changes little.
[A–]
pH = pKa + log ——
[HA]
18 Acid-Base Equilibria
[salt]
HendersonHasselbach
equationn
[acid]
13
A buffer solution
A buffer solution is made up using 10.0 mL0.10 M acetic acid (HA,
Ka = 1.7e-5) and 20.0 0.10 M sodium acetate (NaA). Evaluate its pH.
Hint:
[A–] = 20.0*0.1 / (20.0+10.0) = 0.067 M
[HA] = 10.0*0.1 / (20.0+10.0) = 0.033 M
–]
[A
HendersonHasselbach pH = pKa + log ——
equationn
[HA]
= – log (1.7e-5) + log (2/1)
= 4.77 + 0.30 = 5.07
Find ways to see
[A-] /
2/
=
[HA]
1
pH of a sol’n by mixing
10.0 mL 0.1 M HCl and
10.0 mL 0.3 M NaA
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Making a buffer of certain pH
Find a weak acid with pKa close to the desirable pH
Find out the desirable [A-] / [HA] ratio
Measure appropriate amount (moles) of acid and its salt
Dissolve in appropriate amount of water (volume does not matter)
Use a strong acid and a salt or a weak acid and a strong base
instead of acid and salt.
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Indicators
Indicators are substances
whose solutions change
color due to changes in pH.
HIn = H+ + In-,
and define the equilibrium constant as Kai,
[H+][In-]
Kai = —————
[HIn]
Kai
[In-]
—— = ——
[H+] [HIn]
The color varies according to the ratio Kai / [H+]
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Common indicators
Acid color pH range
Base color
Methyl violet
Yellow
0 – 1.6
Blue-violet
Methy orange
Red
3.2-4.4
Yellow
Litmus
Red
5-8
Blue
Phenolphthalein
Colorless
Thymolphthalein
Colorless 9.4-10.6
Name
8.2 - 10.0
18 Acid-Base Equilibria
Pink
Blue
17
Phenolphthalein – a indicator
C20H14O4 (MW = 318.33)
Formal name: 3,3-bis(4-hydroxyphenyl)-1(3H)-isobenzofuranone,
3,3-bis(4-hydroxyphenyl)phthalide)
Colorless in acidic solution
Pink in basic solution,
C20H12O42–
pH ~ 10
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Titration calculation
When a strong acid is titrated with a strong
base, consider:
The amount of acid present = Va * Ca
The amount of base NaOH added = Vb * Cb
The amount of acid left = Va * Ca - Vb * Cb
The concentration of acid and thus
Va * Ca - Vb * Cb
[H+] = —————————
Va + Vb
Please plot the curve from the data
18 Acid-Base Equilibria
Titration of 10.0 mL 1 M
HCl using 1 M NaOH
Base add
0
5
9
9.5
9.9
9.95
[H+]
1.0
5/15
1/19
.5/19.5
0.1/19.9
0.05/19.95
pH
0
0.48
1.28
1.59
2.30
2.60
10
NaCl soln
7
10.5
10.10
11
15
20
[OH]
0.05 / 20.05 11.40
0.5 / 20.1 11.70
1 / 21 12.68
5 / 25 13.3
20 / 30 13.92
19
Titration curve
Buffers–slide 10-11
Weak HA
slide 7
Hydration of
salts
Slides 6 & 7
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Acid-base equilibria - summary
Know names, formula, & properties of some acids & bases: HA, NH3
Evaluate Ka and Kb and apply them to calculate [ ]’s of various species
using approximation
using quadratic equation
Theory of polyprotic acids
evaluate concentrations of various species
Derive and apply the relation Ka Kb = Kw
Explain why some salt solutions are acidic or basic
Evaluate pH of salts (hydration reacting with water)
Explain theory of buffer
evaluate pH of buffer
make a buffer solution
Explain indicators as weak acids and bases
Plot a titration curve (weak acid titrating with strong base)
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Review
In solving x in problems involving equilibrium constant, how do
we know which method to use.
When can we neglect x?
Also, when can we use something like Successive Approxiamation?
In the example given in class, K was quite small, so as done
in other examples, I probably would have neglected x.
Then there is Newton's method where we plug in trial values for x.
How do we know where a good place is to start?
Technically x could be any value!
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Review 1 – pH of weak acid solutions
Solutions of a weak acid HA, Ka = 1.0e–6 with concentrations of
1.0 M, 0.10 M, 0.00010 M. What are their pH?
Solution:
C = 1.0 M
HA = H+ + A–,
Ka = 1.0e-6
x = Ka*C = 1e–3
C-x
x
x
pH = -log0.001 = 3
2
x
C = 0.10 M
Ka = ———
x = Ka*C = 3.2e–4
C–x
pH = -log3.2e-4=3.49
C = 0.00010 M
– Ka + (Ka2 + 4 Ka*C)
x = Ka*C = 1e–5, 10%of C
x = ————————— = 1.9e–5
2
Do not approximate use
pH = -log1.9e-5 = 4.72
not 5
x2 + Ka x – Ka*C = 0
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Review 2 – pH of buffer
25.00 mL of 0.10 M HAc (acetic acid, pKa = 4.75) is titrated with 0.20 M
NaOH solution. What is the pH when 5.00 mL NaOH has been added?
Analysis
The solution containing 25.00 mL 0.10 M HAc
and 5.00 mL 0.20 M NaOH actually contains
5.00 mL * 0.20 mol/L = 1.0 mmol NaAc,
and (25.00*0.10 – 1.0) mol = 1.50 mmol HAc.
This solution is a buffer
Solution: For a buffer, apply
[A-]
1.0 / total volume
pH = pKa + log ------- = 4.75 + log --------- = 4.75 – 0.18 = 4.57
[HA]
1.5/ total volume
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Review 3 – pH of buffer
25.00 mL of 0.10 M HAc (acetic acid, pKa = 4.75) is titrated with 0.20 M
NaOH solution. What is the pH when 6.25 mL NaOH has been added?
Analysis
The solution containing 25.00 mL 0.10 M HAc
and 6.25 mL 0.20 M NaOH actually contains
6.25 mL * 0.20 mol/L = 1.25 mmol NaAc,
and (25.00*0.10 – 1.25) mol = 1.25 mmol HAc.
This solution is a buffer
Solution: For a buffer, apply
[A-]
1.25 / total volume
pH = pKa + log ------- = 4.75 + log --------- = 4.75 – 0.00 = 4.75
[HA]
1.25 / total volume
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Review 4 – pH of buffer
25.00 mL of 0.10 M HAc (acetic acid, pKa = 4.75) is titrated with 0.20 M
NaOH solution. What is the pH when 7.00 mL NaOH has been added?
Analysis
The solution containing 25.00 mL 0.10 M HAc
and 7.00 mL 0.20 M NaOH actually contains
7.00 mL * 0.20 mol/L = 1.4 mmol NaAc,
and (25.00*0.10 – 1.4) mol = 1.10 mmol HAc.
This solution is a buffer
Solution: For a buffer, apply
[A-]
1.4 / total volume
pH = pKa + log ------- = 4.75 + log --------- = 4.75 + 0.10 = 4.85
[HA]
1.1/ total volume
not 4.57
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Review 5 – pH of salt a solution
25.00 mL of 0.10 M HAc (acetic acid, pKa = 4.75) is titrated with 0.20 M
NaOH solution. What is the pH when 12.50 mL NaOH has been added?
Analysis
The solution containing 25.00 mL 0.10 M HAc
and 12.5 mL 0.20 M NaOH actually contains
12.50 mL * 0.20 mol/L = 2.5 mmol NaAc,
and (25.00*0.10 – 2.5) mol = 0.00 mmol HAc.
[NaA] = 2.5 mmol / (25.0+12.5)mL = 0.067 M
Solution: hydration problem
A- + H2O = HA + OH0.067-x
x
x
x2
Kw
10-14
------------ = Kb = ----- = -------- = 5.62e-10
0.067 – x
Ka
10-4.75
x = (5.62e-10)*0.067 = 1.36e-6
pOH = -log(1.36e-6) = 5.2
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Review 6 – pH of salt and base
25.00 mL of 0.10 M HAc (acetic acid, pKa = 4.75) is titrated with 0.20 M
NaOH solution. What is the pH when 15.0 mL NaOH has been added?
Analysis The solution containing
25.0 mL * 0.10 mol/L = 2.5 mmol NaAc,
[NaAc] = 2.5 mmol /40 mL = 0.0625 M
and
(15.0 m* 0.20 – 2.5) = 0.5 mmol NaOH
[NaOH] = 0.5 mmol / 40 mL = 0.0125 M
Solution: The pH is dictated by the NaOH
NaOH = Na+ + OH0.0125 + x
A- + H2O = HA + OHReview 5; x = 1.36e-6 M
0.0625-x
x 0.0125+x
The extend of this reaction is small, x << 0.0125,
pOH = -log(0.0125) = 1.90
[OH] = 0.0125 M
pH = 14.00 – 1.90 = 12.10
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Buffer solution sites
Buffer solution preparation
csudh.edu/oliver/chemdata/buffers.htm
biochem.mcw.edu/~simont/java/BufferMaker.html
Suppliers:
postapplescientific.com/reagents/buffersoln.html
caledonlabs.com/cgi-bin/products.cgi?category=K
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