Transcript No Slide Title

Chapter 15
Applications of EquilibriumTitrations & Solubility
Section 15.4-15.5: Recall Titrations
What are titrations used for?
– Find an unknown concentration.
End point
– When the indicator changes color.
Equivalence point
– Equal stoichiometric amounts of acid and base
are present.
Titrant vs. Analyte
– Substance whose concentration is known vs.
substance whose concentration is unknown.
Titration/pH Curves
As a titration proceeds, the pH can be
monitored.
No matter what is titrated with what
(strong acid, strong base; weak acid,
strong base; etc.), they all have
characteristic components.
Strong acid-Strong base
*Notice the steep
drop around the
equivalence point.
*Acid reacts with
added base and is in
excess (pH still
acidic) UNTIL
equivalence point.
*Equivalence point
pH = 7; no acid or
base present, and
ions present are
neutral.
Strong base – Strong acid
*Notice the
similarities between
the curves.
*Base reacts with
added acid and is in
excess (pH still
basic) UNTIL
equivalence point.
*Equivalence point
pH = 7; no acid or
base present, and
ions present are
neutral.
Weak acid – Strong base
*Still looks similar
to previous curves,
but initial pH is
higher than with the
strong acid.
*Notice the initial
steep increase of the
pH, until buffer
kicks in.
*Equivalence point
pH = 9. Why?
*Conjugate base
present is weak!
Weak base – Strong acid
*Initial pH is lower
than with the strong
base.
*Again, notice initial
steep decrease of the
pH, until buffer
kicks in.
*Equivalence point
pH = 5. Why?
*Conjugate acid
present is weak!
pH of Titrations
Consider a weak acid that is titrated with a
strong base.
Titration can be broken down into the
following parts when studying pH changes:
(1) Initial pH: find pH of a weak acid.
(2) Acid & base react completely- LR problem.
(3) If weak acid remains, a buffer develops.
(4) At equivalence point, weak acid has all
reacted into the conjugate base. Calculate pH
of the weak base.
(5) After equivalence point, find pH of a strong
base.
Example of Titration Components
Consider a titration where nitrous acid is
titrated with sodium hydroxide.
1)Only weak acid: HNO2  NO2- + H+
2)Some NaOH added:
Na+ has been left
out- spectator!
HNO2 + OH-  NO2- + H2O
*The HNO2 completely reacts with OH-. This
is a stoichiometry problem at first! Find LR.
*Say OH- is LR, so some HNO2 still
present:
HNO2  H+ + NO2- (HNO2 & NO2- present)
*Buffer problem to find pH!
Example of Titration Components
Consider a titration where nitrous acid is titrated
with sodium hydroxide.
3)More NaOH added:
*HNO2 continues to completely react with
NaOH. Calculations are the same as step 2
UNTIL equivalence point is reached.
*At equivalence point, no HNO2 & no NaOH.
*Only conjugate base NO2- left:
NO2- + H2O  HNO2 + OH*Weak base problem to find pH!
Example of Titration Components
Consider a titration where nitrous acid is titrated
with sodium hydroxide.
4)Even more NaOH added:
*All acid has been used up. Now you have a
strong and weak base mixed together.
*Strong base only needs to be considered.
*Find concentration of OH-, then find pOH
and pH.
Homework
Pg. 741 #51
Example Calculations
To demonstrate the components
described, various amounts of strong
base (NaOH) will be added to the weak
acid (HNO2).
Notice that, with the exception of the
initial pH, the same approach is used
initially.
Depending on what remains, various
equations may be used to find the pH.
Example Calculations
(1) Initially, 100.0mL of 0.150M nitrous
acid is present. Ka = 4.5 x 10-4
Set up ICE table and solve!
– Note: cannot ignore x here!
x = 0.0080 = [H+]
pH = 2.10 to start with
Example Calculation
(2) 25.00mL of 0.300M NaOH is added.
Stoichiometry problem- IF table!
– Acid and base react to completion.
Determine which reactant is limiting:
NaOH + HNO2  NO2- + H2O
I 0.00750mol
0.0150mol
F
0.00750mol
0mol
0
-----
0.00750mol
Use new molar amounts to find new
concentrations. You now have a buffer!
Example Calculation
(2) 25.00mL of 0.300M NaOH is added.
Buffer: HNO2  H+ + NO2Henderson-Hasselbalch to find pH.
New concentrations (total V =
125.00mL): [HNO2] = [NO2-] = 0.0600M
pH = 3.35
This buffer will continue until the
equivalence point is reached!
Example Calculation
(3) 50.00mL of 0.300M NaOH is added.
Stoichiometry problem- IF table!
– Acid and base react to completion.
Determine which reactant is limiting:
NaOH + HNO2  NO2- + H2O
I 0.0150mol
F
0mol
0.0150mol
0mol
0
-----
0.0150mol
No buffer this time! Only weak conjugate
base. Solve for pH of a weak base.
Example Calculation
(3) 50.00mL of 0.300M NaOH is added.
Weak base (new [NO2-] = 0.100M):
NO2- + H2O  HNO2 + OHI 0.100M
0M
0M
C
-x
+x
+x
E 0.100 – x
x
x
Solve for x: x = 1.50 x 10-6 = [OH-]
pOH = 5.82; pH = 8.18  Equivalence
point pH!
Example Calculation
(4) 75.00mL of 0.300M NaOH is added.
Stoichiometry problem- IF table!
– Acid and base react to completion.
Determine which reactant is limiting:
NaOH + HNO2  NO2- + H2O
I 0.0225mol
0.0150mol
F 0.0075mol
0
0
-----
0.0150mol
Strong and weak base remain. Strong base
determines pH- solve for pH of a strong base.
pOH = 1.37; pH = 12.63  Past equivalence
point!
Strategy
When asked to find the pH at some
point in a titration:
– Unless it’s the initial pH, ALWAYS set up and
use an IF table. The acid/base reaction will go
to completion based on the LR.
– Determine what species remain and how
much.
– Determine if you have a buffer, are at the
equivalence point (only conjugate species
present), or are past the equivalence point
(strong species & conjugate species present).
– Then calculate the pH appropriately.
Homework
Pg. 742 #53
Section 15.6- Solubility Equilibria
Recall that salts typically dissociate in
water. But do all salts dissolve?
No! Some are only slightly soluble, and
therefore are in equilibrium between
the solid salt and dissolved ions.
Ex: PbSO4 (s)  Pb+2 (aq) + SO4-2 (aq)
It is important in these problems to take
into account how many moles of each
ion are in one mole of the salt!
Solubility Product Constant, Ksp
Ksp is used for the equilibrium
expression. It is known as the solubility
product constant.
– Still found by taking [products]/[reactants] in
equilibrium.
– However, the reactant will always be a solid
salt, so it is not included.
– Thus, Ksp only depends on the concentration of
ions.
From previous ex: Ksp = [Pb+2][SO4-2]
Sample Calculation
The Ksp of magnesium fluoride in water
is 8 x 10-8. How many grams of
magnesium fluoride will dissolve in
0.250L of water?
-8 = 4x3
K
=
8
x
10
First solve for x:
sp
-3
x
=
3
x
10
+2
MgF (s)  Mg + 2F
2
I
C
E
-------------
0
+x
x
0
+2x
2x
Sample Calculation
All we need to solve this problem is one of
the concentrations of the ions.
– Once we have this, we can use stoichiometry/mole
ratios to find moles of the salt from moles of the
ion.
x = 3 x 10-3M= [Mg+2]  use this with the
given volume to find moles:
3 x 10-3mol x 0.250L = 8 x 10-4mol Mg+2
L
Sample Calculation
Recall: MgF2 (s)  Mg+2 (aq) + 2F- (aq)
8 x 10-4 mol Mg+2 x 1mol MgF2 x 62.31g MgF2
1mol Mg+2 1mol MgF2
= 0.05g MgF2
Practice
If the Ksp of lead (II) sulfate is 1.6 x 10-8,
how many grams will dissolve in
0.500L of water?
PbSO4 (s)  Pb+2 (aq) + SO4-2 (aq)
Set up ICE table, solve for x:
1.6 x 10-8 = x2
x = 1.3 x 10-4M = [Pb+2] = [SO4-2]
Use stoichiometry: 0.020g will dissolve
More Practice
What is the value of Ksp if the solubility of
CaC2O4 is 6.1 x 10-3g/L?
Set up ICE table
CaC2O4 (s)  Ca+2 (aq) + C2O2-2 (aq)
I ----0
0
C ----+x
+x
E ----x
x
Convert solubility to mol/L: 4.8 x 10-5
Ksp = (4.8 x 10-5M)2  Ksp = 2.3 x 10-9
Homework
Pg. 743 # 75, 77(b), 79, 81(a)
Recall LeChatelier’s Principle
Remember what this said?
– Any stress added to a system at equilibrium will
cause the equilibrium to respond and be reestablished.
– Ex: N2 (g) + 3H2 (g)  2NH3 (g)
• If more H2 is added, what is the effect?
• Equilibrium shifts right to use up extra H2,
so N2 decreases and NH3 increases.
This principle can be applied to
solubility: common-ion effect.
Common-Ion Effect
Adding a solution with an ion also
present in the slightly soluble salt will
decrease the salt’s solubility.
Ex: PbSO4 (s)  Pb+2 (aq) + SO4-2 (aq)
– If Na2SO4 is added, what is the effect?
– Adds more SO4-2, so equilibrium shifts left.
This means more of the salt remains as a solid,
making it less soluble.
Calculations now involve taking into
account additional ion added.
Common-Ion Calculation
What is the concentration of silver ions if a
1.00M Na2CrO4 solution is added to solid
Ag2CrO4? Ksp = 1.9 x 10-12
Steps are the same as previously practiced,
except now the concentration of CrO4- comes
from the solution AND the salt.
Note that for every mole of Na2CrO4 that
dissolves, one mole of CrO4- ions form:
Na2CrO4 (s)  2Na+ (aq) + CrO4-2 (aq)
[Na2CrO4] = [CrO4-2] = 1.00M add to ICE table
Common-Ion Calculation
Ag2CrO4 (s)  2Ag+ (aq) + CrO4-2 (aq)
I ----0
1.00
C ----+2x
+x
E ----2x
1.00 + x
1.9 x 10-12 = (2x)2(1.00 + x)  x is
negligible!
x = 6.9 x 10-7
[Ag+] = 2x  [Ag+] = 1.4 x 10-6M
AP Practice Question
No
A student wishes to reduce the zinc ion calculators!
concentration in a saturated zinc iodate solution
to 1 x 10-6 M. How many moles of solid KIO3
must be added to 1.00L of solution? Ksp for
Zn(IO3)2 = 4 x 10-6
As usual, write down the equilibrium reaction:
Zn(IO3)2 (s)  Zn+2 (aq) + 2IO3- (aq)
Now, think about what you know!
– Know Ksp and the desired equilibrium [Zn+2]
– Solve for [IO3-], then convert to moles!
AP Practice Question
A student wishes to reduce the zinc ion
concentration in a saturated zinc iodate solution
to 1 x 10-6 M. How many moles of solid KIO3
must be added to 1.00L of solution? Ksp for
Zn(IO3)2 = 4 x 10-6
1mol
b) 0.5mol
c) 2mol
d) 4mol
a)
Homework
Pg. 743 #90
Remember Q?
Recall Q from the beginning of the
equilibrium section.
– Q = reaction quotient
– Tells you where the reaction is with respect to
reaching equilibrium
• Smaller = before equilibrium; equal = at
equilibrium; larger = past equilibrium
This same idea can be used with Ksp!
Ion-product
Technical name for Q with respect to
Ksp.
Found the same way Ksp is calculated,
then compared to Ksp value.
– Before (smaller than Ksp) = no ppt
– After (larger than Ksp) = ppt
Calculation strategy: ONLY difference
is the new [ ]’s must be found
(changing V!), then calculate Q.
Example Calculation
If 10.0mL of 0.100M BaCl2 is added to 40.0mL
of 0.0250M Na2SO4 solution, will BaSO4
precipitate? Ksp for BaSO4 = 1.1 x 10-10.
Write down equilibrium reaction:
BaSO4 (s)  Ba+2 (aq) + SO4-2 (aq)
– Only concentrations you need to worry about are for
Ba+2 and SO4-2! Nothing else is needed to find Q!
Find new concentrations after mixing:
[Ba+2] = 0.0200M
[SO4-2] = 0.0200M
Example Calculation with Q
If 10.0mL of 0.100M BaCl2 is added to 40.0mL
of 0.0250M Na2SO4 solution, will BaSO4
precipitate? Ksp for BaSO4 = 1.1 x 10-10.
Now find Q:
Q = (0.0200)(0.0200) = 0.000400
Compare Q to Ksp:
Q > Ksp, therefore BaSO4 will ppt. (past
equilibrium, so excess ions will form a ppt.)
Warm Up Problem
Will a precipitate form if 100.0mL of
4.0 x 10-4M Mg(NO3)2 is mixed with
100.0mL of 2.0 x 10-4M NaOH? Ksp for
Mg(OH)2 = 8.9 x 10-12.
1) Mg(OH)2 (s)  Mg+2 (aq) + 2OH- (aq)
2) Find new concentrations:
[Mg+2] = 2.0 x 10-4M [OH-] = 1.0 x 10-4M
3) Find Q: Q = 2.0 x 10-12; Q < K, so no ppt.
Homework
Pg. 743 #97 (look on pg. 718 for Ksp
values!)
pH and Solubility
pH can impact the solubility of salts
– This means that adding an acid or base to a salt
can increase or decrease solubility.
Consider the following example:
Mg(OH)2 (s)  Mg+2(aq) + 2OH-(aq)
– What if a base (OH-) is added? Decreases solubility!
– What if an acid is added? Increases solubility!
Why is this seen?
– Consider possible reactions and then apply Le
Chatelier’s Principle/Common Ion Effect
pH and Solubility Cont.
Consider another example:
Ag3PO4(s)  3Ag+(aq) + PO4-3(aq)
What if an acid is added?
– PO4-3 reacts with H+ to form HPO4-2, which is a
weak acid
– As we learned, weak acids don’t dissociate
much. Therefore, some PO4-3 is removed from
solution to form HPO4-2
– Reaction shifts right, making Ag3PO4 more
soluble.
pH and Solubility Cont.
Here’s another example:
AgCl(s)  Ag+(aq) + Cl-(aq)
What if an acid is added?
– Cl- would react with H+, but it forms a strong
acid (HCl).
– As we learned, strong acids dissociate
completely. Therefore, Cl- is removed from
solution.
– Reaction is not affected, and the salt has the
same solubility in acid as it does in water.
pH and Solubility Cont.
From the two examples previously
considered, a general rule can be used:
– Any salt, MX, that forms a weak acid, HX, will
be more soluble in acid than in water.
– The H+ ions in the acid will bond with and
remove anion X- from solution to make the salt
more soluble.
Common anions that are more soluble in
acid: OH-, S-2, CO3-2, C2O4-2, and CrO4-2.
Can you think of anions, besides Cl-, that
are not more soluble in acid? Br-, I-, NO3-, SO4-2,
and ClO4-
AP Practice Question
The addition of nitric acid increases the
solubility of which of the following
compounds?
a) KCl(s)
b) Pb(CN)2(s)
Cu(NO3)2(s)
d) NH4NO3(s)
c)
Complex Ion Equilibria
Complex ion: a metal ion surrounded by
other molecules or ions.
Equilibrium exists for complex ion
formation- Kf is used, called formation
constant or stability constant. Example:
Ag+ + 2NH3 Ag(NH3)2+
Kf = [Ag(NH3)2+]/[Ag+][NH3]2
Complex Ion Equilibria
Complex ions form in a series of steps.
Continuing with the previous ex:
Ag+ + 2NH3 Ag(NH3)2+
Steps are as follows:
Ag+ + NH3  Ag(NH3)+
K1 = 2.1 x 103
Ag(NH3)+ + NH3  Ag(NH3)2+ K2 = 8.2 x 103
Notice an additional NH3 molecule is added to
the silver ion each time!
Overall Kf value?
– Kf = K1 x K2  Kf = (2.1 x 103) x (8.2 x 103) = 1.7 x 107
Complex Ion Equilibria
To determine the steps involved in
forming complex ions, simply look at
the final complex ion formed.
– Look to see how many ions/molecules are
bonded to the metal.
– The number of ions/molecules is how many
steps there are, with each step adding an
additional ion/molecule to the metal.
We will not worry about calculations at
this time!
Homework
Pg. 743 #96, 103 (also write the K
expression for each step and the Kf
expression for the overall formation)