CH. 17 ACID -- BASE EQUILIBRA

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CH. 17ACID -- BASE EQUILIBRA BUFFERS
17.1 Common ion effect
17.1 Calculation Henderson- Hasselbalch
eqn Buffers how works ion effect pH range
17.3 Titration Curves indicators SA/SB -
SA/WB WA/SB - Poly
Calculation pH
17.5 Solubility pH Complex Ions
17.6 .7 Ion Groups Separation
17.4 Solubility Calculation Ksp (Qsp)
precipation
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COMMON ION EFFECT
The shift in the position of an equilibrium on
addition of a subst that provides an ion in
common w/ one of the ions already involved in
the equilibrium process. Also, decr solubility of
soluble aqueous portion of ionic cmpd
LeChatilier
H3O lowers pH shifts away NO2- already formed
add NO2- from NaNO2
Dissolve HNO2 in NaNO2 soln (H2O)
common ion
H3O NO2-
3
resist ?es in pH composed of WA CB or WB CA
BUFFERS
Add sm amt base to buffer soln Acidic component
neutralize base added
Add sm amt acid to buffer soln Basic component
neutralize acid added
BIO SYS blood pH 7.4 control buffer of H2O/HCO3-
4
Fig. 16.7 pg 664
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HENDERSON -- HASSELBALCH
To prep buffer soln select WA w/ pKa close to
?1 pH unit of buffer soln
How pH affects dissoc WA pH of buffer soln not
depend on soln vol but on pKa molar amt WA-CB
EX. Titrant 0.100 M NaOH and neutralize 40.0 ml
of 0.100 M butanoic acid (HC4H7O2) Ka
1.5410-4 Find pH add 20.00 ml of titrant
Find pH at equivalence pt determine best
indicator to use
6
1st determine mmol HC4H7O2 present
mol HC4H7O2 40.0 mL (0.100 mmol/1 mL) 4.00
mmol need 4.00 mmol NaOH to reach
equiv pt means need 40.0 ml
of 0.100 M NaOH
a) No base, NaOH, added butanoic WA Ka
x2/HA x2 (1.5410-4)(0.100)
.00392 pH -Log(0.00392) 2.41
b) Add 20.00 ml NaOH, also midpoint so pH pKa
moles WA 4.00 mmol mol NaOH added 20.00
ml0.100 M 2.00 mmol
2.0 mmol/60 ml 0.0033 M ratio of
HC4H7O2/C4H7O2- 0.033/0.033 1
7
c) Add 40.0 ml NaOH _at_ equiv pt
All HC4H7O2 neutralized, find C4H7O2-
C4H7O2- mmol/tot mL
mmol C4H7O2- 40.00 ml (0.100 mmol/ml) 4.0
mmol
C4H7O2- 4.0 mmol/80.0 ml 0.05 M
pH -Log(5.5610-9) 8.25
indicator PHENOLPHTHALEIN
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TITRATION CURVES
Monoprotic Acids
SA - SB
pH
7
Vol SB
9
TITRATION CURVES
Monoprotic Acids
WA - SB
9
pH
Vol SB
10
TITRATION CURVES
Monoprotic Acids
SA - WB
pH
3.5
Vol SA
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TITRATION CURVES
Polyprotic Acids
pH
SO4-2 pKa2
HSO4- pKa1
H2SO4
Vol Base
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Problem 25.0-mL of 0.145 M HCl soln is titrated
by 0.200 M KOH. How many mL of base must be
added to reach the end point.
Want mL KOH Data 25.0-mL HCl _at_ 0.145
mol/L 0.200 mol/L KOH HCl KOH
---? H2O KCl 1 mol HCl 1 mol KOH
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Problem What is the pH at each point in the
titration of 25.00-mL of 0.100 M HAc with 0.100 M
NaOH? a) before NaOH added b) after
10.00-mL c) after 12.50-mL d) after
25.00-mL (these are volumes added to
original 25 ml)
a. Find initial H3O use Ka (1.810-5)
HC2H3O2 H2O ?-? H3O C2H3O2- Eq 0.100-x
x x
1.310-3
pH -log(1.310-3) 2.89
b. 35-mL tot vol amt neutralize 0.025 L(0.100
mol/L) 0.0025 mol base added 0.01
L(0.100 M) 0.001 mol HC2H3O2
OH ?-? H2O C2H3O2- I
0.0025
0 C -0.001
0.001
E 0.0015
0.001 Convert to M using vol
.0015/.035 .0429 M .001/.035
.0286 M
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c. 37.5-mL tot vol NaOH added0.0125 L(0.100
mol/L) 0.00125 mol
HC2H3O2 OH ?-? H2O C2H3O2- I
0.0025
0 C -0.00125
0.00125
E 0.00125
0.00125 Convert to M
using vol .00125/.0375 .0333 M
.00125/.0375 .0333 M
d. Add 25-mL at eq.pt, neutralization complete
50.00-mL tot vol amt C2H3O2- present 0.0025
mol/0.050 L 0.05 M
C2H3O2- H2O ?-? HC2H3O2 OH-
Eq 0.05-x
x x Kb
Kw/Ka 5.610-10 Kb x2/0.05 x
2.5310-6 pOH 5.28 then pH 8.72
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e. Follow up. What happens beyond the equivalence
pt? pH is determined from excess base
present. - add 26.0 mL, what is the pH?
tot vol 51.00 mL (0.026
L)(0.100 M) 0.0026 mol 0.0026 0.0025
0.0001 mol OH- excess M 0.0001/.051
1.9610-3
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Problem 25.0-mL of 0.145 M HCl soln is titrated
by 0.200 M KOH. How many mL of base must be
added to reach the end point.
Want mL KOH Data 25.0-mL HCl _at_ 0.145
mol/L 0.200 mol/L KOH HCl KOH
---? H2O KCl 1 mol HCl 1 mol KOH
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Problem What is the pH at each point in the
titration of 25.00-mL of 0.100 M HAc with 0.100 M
NaOH? a) before NaOH added b) after
10.00-mL c) after 12.50-mL d) after
25.00-mL (these are volumes added to
original 25 ml)
a. Find initial H3O use Ka (1.810-5)
HC2H3O2 H2O ?-? H3O C2H3O2- Eq 0.100-x
x x
1.310-3
pH -log(1.310-3) 2.89
b. 35-mL tot vol amt neutralize0.025L(0.100
mol/L) 0.0025 mol base added 0.01
L(0.100 M) 0.001 mol HC2H3O2
OH ?-? H2O C2H3O2- I
0.0025
0 C -0.001
0.001
E 0.0015
0.001 Convert to M using vol
.0015/.035 .0429 M .001/.035
.0286 M
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c. 37.5-mL tot vol NaOH added0.0125 L(0.100
mol/L) 0.00125 mol
HC2H3O2 OH ?-? H2O C2H3O2- I
0.0025
0 C -0.00125
0.00125
E 0.00125
0.00125 Convert to M
using vol .00125/.0375 .0333 M
.00125/.0375 .0333 M
d. Add 25-mL at eq.pt, neutralization complete
50.00-mL tot vol amt C2H3O2- present 0.0025
mol/0.050 L 0.05 M
C2H3O2- H2O ?-? HC2H3O2 OH-
Eq 0.05-x
x x Kb
Kw/Ka 5.610-10 Kb x2/0.05 x
2.5310-6 pOH 5.28 then pH 8.72
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SOLUBILITY EQUILIBRA
Principals examines quantitative aspects of
solubility ppt process
Ksp solubility constant
Cr2(SO4)3(s) --------gt 2 Cr3(aq) 3 SO4-2(aq)
Qc solid Ksp
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Ksp 1. temp dependant 2. used to calc
solubility of cmpd
Comparing Ksp cmpds w/ same total ions
formed, then, higher Ksp is greater the
solubility
CALCULATIONS Find
Ksp Solubility of silver dichromate _at_ 15oC is
8.310-3 g/100 mL
Write dissoc eqn, mols each ion Write Ksp
expression Find Molarity - convert solubility to
molar
Ag2Cr2O7(s) --------gt 2 Ag1(aq) Cr2O7-2(aq)
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Cr207-2 2 Ag 0.0192 20.0192
0.0384
Ksp 0.03842 0.0192 2.8310-5
Find Solubility
What is the molar solubility of iron III
hydroxide in water? Ksp 6.310-10
Write dissoc eqn, ion expression, ICE table
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Ksp Fe3 OH-3
Fe(OH)3(s) --------gt Fe3(aq) 3 OH-1(aq)
I ------ 0
0 C -x x
3x E -x
x 3x
Ignore solid
Ksp x 3x3 27x4 Ksp 27x4
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PPT FORMATION -- will or not form ???
Qsp Ksp soln satur, no ? Qsplt Ksp soln
unsat, no ppt Qsp gt Ksp ppt form till soln satur
Will ppt form, if so, what is it? 0.20 L of 0.050
M Na3PO4 0.10 L of 0.20 M Ca(NO3)2
Ions present Na1 PO4-3 Ca2 NO3-1
solubility rules Ca3(PO4)2 (s) the ppt,
insoluble Ca3(PO4)2 (s) lt----gt 3 Ca2 (aq) 2
PO4-3 (aq) Ksp Ca23PO4-32
Ksp 1.210-29
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mols Ca2 0.10 L 0.20 M 0.020 mols Ca2
0.020 mols/0.30 L 0.067 M
mols PO4-3 0.20 L 0.05 M 0.10 mols PO4-3
0.10 mols/0.30 L 0.033 M
Qsp gt Ksp, so ppt will form Ca3(PO4)2 ppt till
Qsp 1.210-29
Qsp Ca23 PO4-32 (0.067)3(0.033)2
3.2810-7
If Qsp 6.710-35 What then??? Qsp lt Ksp, no
ppt
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COMPLEX IONS
Metal ion acts as Lewis Acid, Ligand acts as Base
Central METAL ion covalent bonded to 2
LIGANDS
Can separate metal from its ore, isolate
remove toxic metal, convert metal ion to diff
form
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ION GROUPS
Separation of 2 competing ppting metal ions
thru diff properties of solubility of cmpds
Add ppting ion till Qsp of more soluble ion close
to its Ksp This allows max amt of less soluble
cmpd ppt out, while none of the more soluble
Mixture of many diff metal ions, use various
techniques to sep ions into characteristic
groups pg 737
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Group 1 Insol Chlorides ppt w/ HCl Ag
Hg22 Pb2
Group 2 Acid-insol Sulfides ppt w/ H2S Cu2
Cd2 As3 Sb3 Bi3 Sn2 Sn4 Hg2
Pb2
Group 3 Base-insol S-2 OH-1 ppt w/ NH4-NH3
buffer Zn2 Mn2 Ni2 Fe2 Co2 as
S-2 Al3 Cr3 as OH-
Group 4 Insol PO4-3 ppt w/ (NH4)2HPO4,
NaCO3 Mg2 Ca2 Ba2
Group 5 Alkali NH4 Ions generally speaking,
whats left
Use various techniques to ppt out specific ions
from each group
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Calculate the pH of a soln that is 0.350 M
pyridinium chloride (C5H5NHCl) and 0.210 M
pyridine (C5H5N).
WB C5H5N C5H5NHCl contain common ion effect
C5H5NH
C5H5N(aq) ) H2O(l) ?? C5H5NH(aq) OH-(aq)
I 0.210
0.350 0.0 C
-x x
x E 0.210-x
0.350x x

assume 5 rule for WA
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Kb 1.710-9
x OH- 1.0210-9 pOH
-Log(1.0210-9) 8.99 pH 14
8.99 5.01
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The addition of bromide ion will decrease the
water solubility of which of the following salts?

• a.BaSO4 b. Li2CO3 c. PbS d. AgBr

AgBr
Which pair of compounds will form a buffer
solution when dissolved in water in equimolar
amounts?
a. HCl and KCl b. HNO3 and NaNO3 c. HCl
and NH4Cl d. NH3 and NH4Cl
NH3 and NH4Cl
31
The Ka of HCN is 4.9 x 10-10. What is the pH of
a buffer solution that is 0.100 M in both HCN and
KCN?

• a. 4.7 b. 7.0 c. 9.3 d. 14.0

9.3
same concentrations, -log(4.910-10)
The Ka of HCN is 4.9 x 10-10. What is the pH of
a buffer solution that is 0.100 M in HCN and
0.200 M in KCN?
a. 7.0 b. 9.0 c. 9.3 d. 9.6
9.6
pH pKa logKCN/HCN
32
The Ka of HCN is 4.9 x 10-10. What is the pH of
a buffer solution that is 1.00 M in HCN and 0.100
M in KCN?

• a. 7.0 b. 8.3 c. 9.0 d. 9.3

8.3
pH pKa logKCN/HCN
33
In titrating a weak base with a strong acid, the
best indicator to use would be

1. methyl red (changes color at pH 5).
2. bromothymol blue (changes at pH 7).
3. phenolphthalein (changes at pH 9).
4. none of the above.

methyl red (changes color at pH 5)
Titrating a weak acid with a strong base, best
indicator to use would be

1. methyl red (changes color at pH 5).
2. bromothymol blue (changes at pH 7).
3. phenolphthalein (changes at pH 9).
4. none of the above.

phenolphthalein (changes at pH 9)
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The Ksp of BaCO3 is 5.0 x 10-9. What is the
concentration of barium ion in a saturated
aqueous solution of BaCO3?

• a. 7.1 x 10-5 M b. 2.5 x 10-9 M c. 5.0 x
  10-9 M d. 1.0 x 10-8 M

7.1 x 10-5 M
5.010-9 xx 5.010-9 x2
The Ksp of BaF2 is 1.7 x 10-6. What is the
concentration of barium ion in a saturated
aqueous solution of BaF2?
a. 1.7 x 10-6 M b. 3.4 x 10-6 M c. 7.5 x
10-3 M d. 1.5 x 10-2 M
7.5 x 10-3 M
Ksp Ba2F-2 (x)(2x)2 1.710-6
4x3
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The Ksp of BaF2 is 1.7 x 10-6. What is the
concentration of fluoride ion in a saturated
aqueous solution of BaF2?

• a. 1.7 x 10-6 M b. 5.7 x 10-5 M c.
  7.6 x 10-3 M d. 1.5 x 10-2 M

1.5 x 10-2 M
BaF2 --? Ba2 2 F- F- 2x 2(7.510-3)
36

• Which of the following substances will be more
  soluble in acidic solution than in basic
  solution
• Ni(OH)2(s), (b) CaCO3(s), (c) BaF2(s), (d)
  AgCl(s)?
• So, which is more soluble at low pH than high pH

Ni(OH)2 more soluble in acidic, basicity of
OH- CaCO3 dissolves in acidic, as CO3- basic
anion BaF2 dissolves, as F- basic anion
(d) The solubility of AgCl is unaffected by
changes in pH because Cl is the anion of a
strong acid and therefore has negligible basicity.
37
Addition of _____ will increase the solubility of
MgCO3.

• MgCl2
• Na2CO3
• NaOH
• HCl
• KHCO3

HCl
38
Predict the order of precipitation of Ba2, Pb2,
Ca2 with the addition of NaF.

• Ca2 then Ba2 then Pb2
• Pb2 then Ba2 then Ca2
• Pb2 then Ca2 then Ba2
• Ca2 then Pb2 then Ba2
• Ba2 then Pb2 then Ca2

Ksp
BaF2 1.7 x 106
PbF2 3.6 x 108
CaF2 3.9 x 1011
Ca2 then Pb2 then Ba2
39
Predict the order of precipitation of Cl, CO32,
Br upon the addition of AgNO3.

• Br then CO32 then Cl
• Br then Cl then CO32
• Cl then CO32 then Br
• Cl then Br then CO32
• CO32 then Cl then Br

Ksp
AgCl 1.8 x 1010
Ag2CO3 8.1 x 1012
AgBr 5.0 x 1013
Br then Cl then CO32
40
A solution contains 1.0 10-2 M Ag and 2.0
10-2 M Pb2. When Cl is added to the solution,
both AgCl (Ksp 1.8 10-10) and PbCl2 (Ksp
1.710-5) precipitate from the solution. What
concentration of Cl is necessary to begin the
precipitation of each salt? Which salt
precipitates first?
We are given Ksp values for the two possible
precipitates. Using these and the metal ion
concentrations, we can calculate what
concentration of Cl ion would be necessary to
begin precipitation of each. The salt requiring
the lower Cl ion concentration will precipitate
first.