### Ch 17 notes

```More
Acid/Base
Equilibria!!!
17.1 – 17.3
17.1 The Common Ion Effect
Consider the ionization of a weak acid, acetic acid:
HC2H3O2(aq)  H+(aq) + C2H3O2–(aq)
If we increase the [C2H3O2–] ions by adding NaC2H3O2, the
equilibrium will shift to the left. (Le Chatelier)
This reduces the [H+] and raises the pH (less acidic)
This phenomenon is called the common-ion effect.
Common ion equilibrium problems are solved following the same
pattern as other equilibrium problems (ICE charts) EXCEPT
the initial concentration of the common ion must be
considered (it is NOT zero).
Example 1:
Does the pH increase, decrease, or stay the
same on addition of each of the following?
(a) NaNO2 to a solution of HNO2
(b) (CH3NH3)Cl to a solution of CH3NH2
(c) sodium formate to a solution of formic acid
(d) potassium bromide to a solution of hydrobromic acid
(e) HCl to a solution of NaC2H3O2
(a)
(b)
(c)
(d)
(e)
HNO2  H+ + NO2increases
CH3NH2 + H2O  CH3NH3+ + OHdecreases
HCHO2  H+ + CHO2increases
HBr  H+ + Brno change
C2H3O2-1 + H2O  HC2H3O2 + OH-1
decreases
Example 2: Using equilibrium constants from Appendix D,
calculate the pH of the solution containing 0.060 M
KC3H5O2 and 0.085 M HC3H5O2
H C 3 H 5 O 2 (aq) 
in itial
0.085
change:
-x
Equilibrium: 0.085-x
H
+1
-1
(aq) + C 3 H 5 O 2 (aq)
0
0 .060
+x
+x
x
.060+ x
(.060  x )( x )
(.085  x )
 1.3 *10
5
x2+.060013x-(1.105*10-6)=0
x=1.84*10-5
pH = -log(1.84*10-5) = 4.74
K a = 1.3 x 10
-5
17. 2 Buffered Solutions
A buffered solution or buffer is a solution that resists a change in pH
after addition of small amounts of strong acid or strong base.
A buffer consists of a mixture of a weak acid (HX) and its conjugate
base (X– ) or weak base (B) and its conjugate acid (HB+)
Thus a buffer contains both:
an acidic species to neutralize added OH–
When a small amount of OH– is added to the buffer solution, the
OH– reacts with the acid in the buffer solution.
a basic species to neutralize added H+
When a small amount of H+ is added to the buffer solution, the H+
reacts with the base in the buffer solution.
Composition of a Buffer - 4 ways to make a buffer solution:
1.)
Weak acid + salt of the acid
HCN and NaCN
weak acid: HCN weak base: CN-1
2.)
Weak base + salt of the base
NH3 and NH4 Cl
weak acid: NH4+1 weak base: NH3
3.)
EXCESS Weak acid + strong base
2 mol HCN + 1 mol NaOH  1 mol HCN + 1 mol NaCN + H2O
weak acid: HCN weak base: CN-1
2 mol NH4Cl + 1 mol NaOH  1 mol NH4Cl + 1 mol NH3 + NaCl
weak acid: NH4+1 weak base: NH3
4.)
EXCESS Weak base + strong acid
2 mol NH3 + 1 mol HCl  1 mol NH3 and 1 mol NH4Cl
weak acid: NH4+1 weak base: NH3
2 mol NaF + 1 mol HCl  1 mol NaF + 1 mole HF + NaCl
weak acid: HF
weak base: F-1
Example 3: Explain why a mixture of HCl and KCl does not
function as a buffer, whereas a mixture of HC2H3O2 and
NaC2H3O2 does.
HCl is a strong acid - Cl-1 is a negligible base and will
change the pH of the solution
HC2H3O2 and C2H3O2-1 are a weak conjugate acid/base
pair which act as a buffer
leaving the [H+1] and pH relatively unchanged
Buffer Capacity and pH
Buffer capacity is the amount of acid or base that can be
neutralized by the buffer before there is a significant change
in pH.
Buffer capacity depends on the concentrations of the
components of the buffer - the greater the concentrations
of the conjugate acid-base pair, the greater the buffer
capacity.
The pH of the buffer is related to Ka and to the relative
concentrations of the acid and base.
Henderson-Hasselbalch equation – used for buffer solutions
(on AP equation sheet!!)
p H = p K a + lo g
p O H = p K b + lo g
A- 


HA 
= pK a +
HB + 


B 
 b a se 
lo g
 a cid 
= pK b +
 a cid 
lo g
 b a se 
a cid fo rm
b a se fo rm
These equations technically use the equilibrium concentrations of the
acid (base) and the conjugate base (acid).
However, since the acid/base in the buffer is WEAK – the amount of the
conjugate produced by dissociation is generally small compared to the
amount of the conjugate added as a salt. IF this is true (it is for all AP
buffer problems!) we do not need to do an equilibrium problem – just
use the INITIAL concentrations.
Example 4 (Example 2 again!): Using equilibrium constants
from Appendix D, calculate the pH of the solution containing
0.060 M KC3H5O2 and 0.085 M HC3H5O2
H C 3 H 5 O 2 (aq) 
in itial
0.085
H
+1
0
-1
(aq) + C 3 H 5 O 2 (aq)
K a = 1.3 x 10
0 .060
this is a BUF F ER solution - use the H enerdson-H asslebalch equation
use the initial concentrations !
p H = p K a + log
 0.060 
-5
= - log(1.3 x 10 ) + log 
 =
[acid ]
 0.085 
[b ase]
4.73
-5
Example 5: Calculate the pH of a buffer that is 0.12 M in
lactic acid and 0.11 M in sodium lactate
H C H O 2 (aq)
in iti al
H
+1
-1
(aq) + C H O 2 (aq)
0.12
K a  1.4 x 1 0
-4
0. 1 1
Using the H enderson-H asselbalch equation (and initial conc)
pH = pK a
 0.11 
+ log
= - log(1.4 x 10 ) + log 
 = 3.85 - .038 = 3 .81
[acid]
 0.12 
[base]
-4
yo u w o u ld g et th e sam e an sw er d o in g a co m p lete IC E ch art
Example 6: A buffer is prepared by adding 20.0 g of acetic acid, HC2H3O2 and
20.0 g of sodium acetate to enough water to form 2.00 L of solution.
(a) Determine the pH of the buffer
(b) Write the complete ionic equation for the reaction that occurs when a few
drops of hydrochloric acid are added to the buffer
(c) Write the complete ionic equation for the reaction that occurs when a few
drops of sodium hydroxide are added to the buffer
(a)
[H C 2 H 3 O 2 ] =
-1
2
[C 2 H 3 O ] =
20.0 g H C 2 H 3 O 2
in i tial
pH = pK a
2.00 L
x
2.00 L
H
= 0.167 M
60.0 g H C 2 H 3 O 2
20.0 g N aC 2 H 3 O 2
H C 2 H 3 O 2 (aq)
x
1 m ol H C 2 H 3 O 2
1 m ol N aC 2 H 3 O 2
= 0.1 22 M
82.0 g N aC 2 H 3 O 2
+1
-1
(aq) + C 2 H 3 O 2 (aq)
0.167
K a = 1.8 x 10
-5
0.122
 0.122 
+ log
= - log(1.8 x 10 ) + log 
 = 4.74 - .14 = 4.60
[acid]
 0.16 7 
[base]
-5
(b) C2H3O2-1(aq) + H+1(aq) + Cl-1(aq)  HC2H3O2 (aq) + Cl-1 (aq)
(c) HC2H3O2 (aq) + Na+1 (aq) + OH-1 (aq)  C2H3O2-1 (aq) + H2O (l) + Na+1 (aq)
17.3 Acid-Base Titrations – Titration Curves
In an acid-base titration:
A solution of base (or acid) of known concentration
(called standard) is added to an acid (or base).
Acid-base indicators or a pH meter are used to signal
the equivalence point (when moles acid = moles
base).
The plot of pH versus volume during a titration is called
a pH titration curve.
Starts high
Ends low
Equivalence point = 7
equal moles of acid
and base present
end
start
Strong acid added to strong base
Starts low
Ends high
Equivalence point = 7
end
start
Strong base added to strong acid
Starts med-high
Ends low
Equivalence point < 7
Buffer area – in this area
there is weak base and
some salt of the weak base
Actual pH
depends on the
salt formed
but it will be < 7
Strong acid added to weak base
Starts low
Ends med-high
Equivalence point < 7
end
start
Weak base added to strong acid
Starts high
Ends med-low
Equivalence point > 7
Actual pH
depends on the
salt formed
but it will be > 7
end
start
Weak acid added to strong base
Starts med-high
Ends med-low
Equivalence point = 7
end
start
Strong base added to strong diprotic acid (H2SO4)
Example 7: Predict whether the equivalence point of each
of the following titrations is below, above or at pH 7:
a) NaHCO3 titrated with NaOH b) NH3 titrated with HCl
c) KOH titrated with HBr
At the equivalence point, only products are present in solution, so determine the products
of the reaction and then determine if the solution is acidic, basic or neutral
a) NaHCO3 + NaOH  Na2CO3 + H2O
weak acid strong base
pH > 7
CO3-2 is basic, Na+ is neutral, H2O is neutral
b) NH3
+
HCl
 NH4Cl
weak base strong acid
pH < 7
NH4+1 is acidic, Cl- is neutral
c) KOH
+ HBr
 KBr + H2O
strong base strong acid
pH = 7
K+ and Br- are both neutral
Example 8: How many mL of 0.0850 M NaOH solution is
required to titrate 40.0 mL of 0.0900 M HNO3?
? mL
40.0 mL
0.0850 M
0.0900 M
NaOH + HNO3 
1 mole
1 mole
0.0400 L H N O 3 x
0.090 m ol H N O 3
1 L HNO 3
x
H2O
+
1 m ol N aO H
1 m ol H N O 3
NaNO3
x
1 L N aO H
0.0850 m ol N aO H
= 0.0423 L or 42.3 m L
Example 9: A 20.0 mL sample of 0.200 M HBr solution is titrated with 0.200 M
NaOH solution. Calculate the pH of the solution after the following volumes of
(a) 15.0 mL (b) 19.9 mL (c) 20.0 mL (d) 20.1 mL (e) 35.0 mL
(a)
mL
HBr
mL
NaOH
mL
Total
mol H+1
(M) (V)
mol OH-1
(M) (V)
M of
excess ion
(mol / tot vol)
pH
20.0
15.0
35.0
0.00400
0.00300
0.0286 M H+1
1.544
(.200)(.0200)
(.200)(.0150)
(.004  .003) m ol
- log(.0286)
0.0350 L
(b)
20.0
19.9
39.9
0.00400
0.00398
0.0005 M H+1
3.3
(c)
20.0
20.0
40.0
0.00400
0.00400
1 x 10-7 M H+1*
7.0
(d)
20.0
20.1
40.1
0.00400
0.00402
0.0005 M OH-1
10.7
35.0
55.0
0.00400
0.00700
0.0545 M OH-1
12.736
(e)
20.0
When molarity of H+ (or OH-) is less than 10-6 we must consider the autoionization of water!
(H+ = 1.0*10-7)
Example 10: Calculate the pH at the equivalence point for titrating 0.200
M solutions of each of the following bases with 0.200 M HBr: (a) NaOH
(b) NH2OH
(a) strong acid/strong base titration so pH = 7
(b)
HBr + NH2OH
 Br- + NH2OH2+
strong acid weak base
.200M
.200M
all product at equivalence point – no excess & Br- is neutral and will have no affect on pH
Volume doubles (equal molarity and 1:1 stoich ratio) so molarity halves
[NH2OH2+] = 0.200mol / 2 = 0.100 M
I
C
E
NH2OH2+  H+1 + NH2OH Kb = 1.1 x 10-8 (appendix)
.100
0
0
-x
+x
+x
0.100 – x
x
x
Ka = Kw / Kb = 1  10-14 / 1.1  10-8 = 9.1  10-7
(x2) / (0.100-x) = 9.1 x 10-7
x = 3.0 x 10-4 M = [H+1]
pH = - log(3.0 x 10-4) = 3.52
Acid – Base Indicators
The equivalence point of an acid-base titration can be
determined by measuring pH, but it can also be
determined by using an acid-base indicator which
marks the end point of a titration by changing color.
Although the equivalence point (defined by the
stoichiometry) is not necessarily the same as the
end point (where the indicator changes color),
careful selection of the indicator can ensure that the
difference between them is negligible.
Acid-base indicators are complex molecules that are
themselves, weak acids (represented by HIn).
They exhibit one color when the proton is attached
and a different color when the proton is absent.
Acid – Base Indicators
Bromthymol Blue Indicator
In Acid
In Base
Acid – Base Indicators
Methyl Orange Indicator
In Acid
In Base
Acid – Base Indicators
Phenolphthalein color at different pH values
pH values
5
6
7
8
9
Acid – Base Indicators
Consider a hypothetical indicator, HIn, a weak acid with Ka=1.0x10-8.
It has a red color in acid and a blue color in base.
HIn(aq)  H+1(aq) + In-1(aq)
red
blue

Ka =

[H ][In ]
[H In ]


Ka
Ka
[In ]
[In ]
=
or
=


[H ]
[H In ]
[H In ]
[H ]
R e arranging , w e get
S uppose w e add a few drops of indicator to an acid solution

1
w hose pH = 1.0 ([H ] = 1.0 x 10 )

[In ]
[H In ]
=
Ka

[H ]
=
1.0 x 10
1.0 x 10
8
1
= 10
7
=
1
10,000,000
Acid – Base Indicators
HIn(aq)  H+1(aq) + In-1(aq)
red
blue
1
1 0 ,0 0 0 ,0 0 0

=
[In ]
[H In ]
This ratio shows that the predominant form of the
indicator is HIn, resulting in a red solution. As OH-1 is
added (like in a titration) [H+1] decreases and the
equilibrium shifts to the right, changing HIn to In-. At
some point in the titration, enough of the In- form will
be present so we start to notice a color change.
Acid – Base Indicators
It can be shown (using the Henderson-Hasselbalch equation)
that for a typical acid-base indicator with dissociation
constant, Ka, the color transition occurs over a range of pH
values given by pKa ± 1.
For example, bromthymol blue with Ka = 1.0 x 10-7 (pKa = 7),
would have a useful pH range of 7 ± 1 or from 6 to 8.
You want to select an indicator whose pKa value is close to
the pH you want to detect (usually the pH at the equivalence
point)
Acid – Base Indicators
The pH curve for the
titration of 100.0 mL of
0.10 M HCl with 0.10 M
NaOH. Neither of the
indicators shown would
be useful for a
titration.
Bromthymol
blue (pKa=7) would be
useful.
The pH curve for the
titration of 50 mL of
0.1 M HC2H3O2 with
0.1 M NaOH.
Here, phenolphthalein
is the indicator of
choice. It has a pKa
Example 11: Use the following table to determine which of the
following would be the best indicator to use to indicate the
equivalence point of the titrations described in Example 10.
Indicator
Ka
Methyl Yellow
1*10-4
Methyl Red
1*10-5
Bromthymol Blue
1*10-7
Phenolpthalein
1*10-9
a. pH at equivalence point was 7.0
Bromthymol Blue
b. pH at equivalence point was 3.52
Methyl Yellow
Solubility Equilibria & Complex
Ions
17.4 – 17.6
SOLUBILITY GUIDELINES
Soluble Compounds
Exceptions
NOT precipitates PRECIPITATES
Nitrates
None
Acetates
None
Chlorates
None
Chlorides
Ag+1, Hg2+2, Pb+2
Bromides
Ag+1, Hg2+2, Pb+2
Iodides
Ag+1, Hg2+2, Pb+2
Sulfates
Ca+2, Sr+2, Ba+2, Hg2+2, Pb+2
We classify these based on the
Solubility - maximum
amount of solute that dissolves
in water.
Insoluble Compounds
Exceptions
PRECIPITATES
NOT Precipitates
Sulfides
NH4+1, Li+1, Na+1, K+1, Ca+2, Sr+2, Ba+2
Carbonates
NH4+1, Li+1, Na+1, K+1
Phosphates
NH4+1, Li+1, Na+1, K+1
Hydroxides
Li+1, Na+1, K+1, Ca+2, Sr+2, Ba+2
Chromates
NH4+1, Li+1, Na+1, K+1, Ca+2, Mg+2
17.4 Solubility Equilibria
The Solubility-Product Constant, Ksp
Consider a saturated solution of BaSO4 in contact with solid
BaSO4.
We can write an equilibrium expression for the dissolving of the
solid.
BaSO4(s)  Ba2+(aq) + SO42–(aq)
Since BaSO4(s) is a pure solid, the equilibrium expression
depends only on the concentration of the ions.
Ksp = [Ba2+][ SO42–]
Ksp is the equilibrium constant for the equilibrium between an
ionic solid solute and its saturated aqueous solution.
Ksp is called the solubility-product constant
• In general: the solubility product is equal to the
product of the molar concentration of ions
raised to powers corresponding to their
stoichiometric coefficients.
Al2(CO3)3  2 Al+3 + 3 CO3-2
Ksp = [Al+3]2 [CO3-2]3
Solubility and Ksp
Solubility is the amount of substance that
dissolves to form a saturated solution.
This can be expressed as grams of solid that will
dissolve per liter of solution.
Molar solubility - the number of moles of solute
that dissolve to form a liter of saturated
solution.
Solubility can be used to find Ksp and Ksp can be
used to find solubility (see problems)
Example 1:
a. If the molar solubility of CaF2 at 35oC is 1.24  10-3 mol/L, what is Ksp
at this temperature?
CaF2

E .00124M actually dissolves
Ca+2
.00124 M
+
2 F-1
2(.00124) = .00248 M
Ksp = [Ca+2] [F-1]2 = (.00124 M)(.00248 M)2 = 7.63 x 10-9
b. It is found that 1.1  10-2 g of SrF2 dissolves per 100 mL of
aqueous solution at 25oC. Calculate the solubility product of
SrF2.
[SrF2] = (.011 g / 125.6 g/mole) / .100 L = .00088 M
E
SrF2
.00088 M

Sr+2
.00088 M
+
2 F-1
2(.00088) = .00176 M
Ksp = [Sr+2] [F-1]2 = (.00088 M)(.00176)2 = 2.7 x 10-9
c. The Ksp of Ba(IO3)2 at 25oC is 6.0  10-10.
What is the molar solubility of Ba(IO3)2?
Ba(IO3)2
E x

Ba+2
+
2 IO3-1
x
2x
(x) (2x)2 = 4 x3 = 6.0 x 10-10
x = 5.3 x 10-4 M
17.5 Factors That Affect Solubility
Factors that have a significant impact on solubility are:
- The presence of a common ion
- The pH of the solution
Common-Ion Effect
Solubility is decreased when a common ion is added.
This is an application of Le Châtelier’s principle:
Consider the solubility of CaF2:
CaF2(s)  Ca2+(aq) + 2F–(aq)
If more F– is added (say by the addition of NaF), the equilibrium
shifts left to offset the increase.
Therefore, more CaF2(s) is formed (precipitation occurs).
Example 2: Using Appendix D, calculate the molar solubility of AgBr
in (a) pure water (b) 3.0  10-2 M AgNO3 solution (c) 0.50 M NaBr
solution
Br-1
x
Ksp = 5.0  10-13
(b) AgBr  Ag+1 + Br-1
E x
.030 + x
x
5.0  10-13 = (.030+x) x
x = 1.7  10-11 M
Ksp = 5.0  10-13
(c) AgBr  Ag+1 + Br-1
x
x
.50 + x
5.0  10-13 = x (.50+x)
x = 1.0  10-12 M
Ksp = 5.0  10-13
(a) AgBr  Ag+1
E x
x
5.0  10-13 = x2
x = 7.1  10-7 M
+
notice the DECREASED solubility with the common ion in (b) and (c)
pH effects
Consider: Mg(OH)2(s)  Mg2+(aq) + 2 OH–(aq)
If OH– is removed, then the equilibrium shifts right and Mg(OH)2
dissolves.
OH– can be removed by adding a strong acid (lowering the pH):
OH–(aq) + H+(aq)  H2O(aq)
Another example:
CaF2(s)  Ca2+(aq) + 2 F–(aq)
If the F– is removed, then the equilibrium shifts right and CaF2
dissolves.
F– can be removed by adding a strong acid (or lowering pH):
F–(aq) + H+(aq)  HF(aq)
Example 3: Calculate the molar solubility of Mn(OH)2 at (a) pH 7.0 (b) pH 9.5 (c)
pH 11.8
the [OH-1] is set by the pH (or pOH)
(a) pH = 7.0 so pOH = 7.0 so [OH-1] = 1.00  10-7
Mn(OH)2  Mn+2 + 2 OH-1
x
x
1.00 x 10-7
Ksp = [Mn+2][OH-1]2
1.6  10-13 = (x) (1.00  10-7)2
x = 16 M
(b)
Ksp = 1.6  10-13
pH = 9.5 so pOH = 4.5 so [OH-1] = 3.16  10-5
Mn(OH)2  Mn+2 + 2 OH-1
Ksp = 1.6  10-13
x
x
3.16 x 10-5
Ksp = [Mn+2][OH-1]2
1.6  10-13 = (x) (3.16  10-5)2
x = 1.7  10-4 M
pH = 11.8 so pOH = 2.2 so [OH-1] = 6.31  10-3
Mn(OH)2  Mn+2 + 2 OH-1
Ksp = 1.6  10-13
x
x
6.31 x 10-3
Ksp = [Mn+2][OH-1]2
1.6  10-13 = (x) (6.31  10-3)2
x = 4.0  10-9 M
Common ion effect – increasing [OH-] decreases solubility
(c)
Example 4: Which of the following salts will be
substantially more soluble in acidic solution than in pure
water:
(a) ZnCO3 (b) ZnS (c) BiI3 (d) AgCN (e) Ba3(PO4)2
If the anion of the salt is the conjugate base of a weak
acid, it will combine with H+1, reducing the concentration
of the anion and making the salt more soluble
ZnCO3  Zn+2 + CO3-2
the CO3-2 ion will react with the added H+
CO3-2 + H+  HCO3-1
Le Chatelier effect of removing CO3-2
more soluble in acid: ZnCO3, ZnS, AgCN, Ba3(PO4)2
17.6 Precipitation and Separation of Ions
Consider the following:
BaSO4(s)  Ba2+(aq) + SO42–(aq)
At any instant in time, Q = [Ba2+][ SO42– ]
If Q > Ksp, (too many ions) precipitation occurs until Q = Ksp.
If Q = Ksp equilibrium exists (saturated solution)
If Q < Ksp, (not enough ions) solid dissolves until Q = Ksp.
Selective Precipitation of Ions
Removal of one metal ion from a solution of two or more metal
ions is called selective precipitation.
Ions can be separated from each other based on the solubilities of
their salt compounds.
Example: If HCl is added to a solution containing Ag+ and Cu2+,
the silver precipitates (as AgCl) while the Cu2+ remains in
solution
Generally, the less soluble ion is removed first!
Example 5: Will Ca(OH)2 precipitate if the pH of
a 0.050 M solution of CaCl2 is adjusted to 8.0?
if Q > than Ksp then precipitation will occur
pH = 8.0 so pOH = 6.0 so [OH-1] = 1.0  10-6 M
Ca(OH)2  Ca+2 + 2 OH-1
Ksp = 6.5  10-6
.050
1.00 x 10-6
Q = [Ca+2] [OH-1]2
Q = (.050)(1.0  10-6)2 = 5.0  10-14
Q<K
so no precipitation occurs
Example 6: A solution contains 0.00020 M Ag+1 and
0.0015 M Pb+2 . If NaI is added, will AgI or PbI2
precipitate first? Specify the [I-1] needed to begin
precipitation for each cation.
the cation needing the lower [I-1] will precipitate first
AgI

Ag+1 + I-1
.000200
x
Ksp = [Ag+1][x]
8.3  10-17 = (.00020)[x]
4.2  10-13 = x = [I-1]
PbI2

Pb+2 + 2 I-1
0.0015
x
Ksp = [Pb+2][x]2
1.4  10-8 = (.0015)[x]2
3.1  10-3 = x = [I-1]
Ksp = 8.3 x 10-17
Ksp = 1.4 x 10-8
AgI will precipitate first at an [I-1] = 4.2  10-13
Complex Ions
• Complex ion – a metal ion bonded to one or
more Lewis bases. (We saw this with water in
chapter 16)
• It can happen with other Lewis bases (things
that have lone pairs of electrons)
• Rule of thumb: The number of Lewis bases
(ligands) that a metal ion attracts is equal to
double its charge. (Works about 75% of the
time!)
• Extra Stuff Below…
Acid – Base Indicators
How much In- must be present for the human
eye to detect that the color is different? For
most indicators, about 1/10 of the initial form
must be converted to the other form before a
color change is apparent. We can assume
that in the titration of an acid with a base, the
[In ]
1
=
color change will occur at a pH where

[H In ]
10
Acid – Base Indicators
Bromthymol blue, an indicator with a Ka = 1.0 x 10-7, is yellow in its
HIn form and blue in its In- form. Suppose we put some strong
acid in a flask, add a few drops of bromthymol blue and titrate
with NaOH. At what pH will the indicator color change first be
visible?
HIn(aq)  H+1(aq) + In-1(aq)
yellow
blue

K a = 1.0 x 10
7
=

[H ][In ]
[H In ]

w e assum e that the color change is visib le w hen
[In ]
[H In ]
=
1
10

thus K a = 1 x 10
7
=
[H ](1)
(10)

and [H ] = 1.0 x 10
6
or pH = 6.00
Selective Precipitation of Ions (continued)
Sulfide ion is often used to separate metal ions.
Example: Consider a mixture of Zn2+(aq) and Cu2+(aq).
CuS (Ksp= 6 x 10–37) is less soluble than ZnS (Ksp= 2 x 10–25).
Because CuS is LESS SOLUBLE than ZnS, CuS will be
removed from solution before ZnS.
As H2S is bubbled through the acidified green solution, black
CuS forms.
When the precipitate is removed, a colorless solution containing
Zn2+(aq) remains.
When more H2S is added to the solution, a second precipitate of
white ZnS forms.
Formula Type of
acid or
base
HCl
Formula of Hydrolysis equation
conjugate
of the acid
acid or base
Strong acid
Cl-1
HOCl
Weak acid
OCl-1
NH3
Weak base
NH4+1
Ba(OH)2 Strong base
KI
neutral
H2O
K+1 or I-1
Hydrolysis equation
of the base
HCl + H2O  H3O+1 + Cl-1
Cl-1 + H2O  X
HOCl + H2O  H3O+1 + OCl-1 OCl-1 + H2O  HOCl + OH-1
NH4+1 + H2O  H3O+1 + NH3
NH3 + H2O  OH-1 + NH4+1
H2O + H2O  H3O+1 + OH-1
OH-1 + H2O  X
I-1 + H2O  X
K+1 + H2O  X
NaC2H3O2 weak base HC2H3O2 HC2H3O2 + H2O  H3O+1 + C2H3O2-1
C2H3O2-1 + H2O  HC2H3O2 + OH-1
```