Transcript Topic C – Part II: Acid Base Equilibria and Ksp

Topic 1C – Part II
Acid Base Equilibria and Ksp
• Applying Equilibrium
• Titrations
• Buffers
• Solubility Product
Titrations
• Titration is an experimental technique used to perform a neutralization
reaction (see unit 3).
• Used to determine the unknown concentration of an acid or base.
• As a base or acid is added to an acid or base in a titration, there is
very little change in pH initially
• When the moles of titrant (from the buret) are in the exact
stoichiometric proportion with the analyte (in the flask)
• (100% of the acid or base has been neutralized)
• This is the equivalence point has been reached.
• At this point there is a rapid change in pH
• These changes can be summarized using titration curve
plots.
• Most acid-base titrations involve the addition of one
colorless solution to another colorless solution
• We need a method of determining when the equivalence
point has been reached.
• We use an indicator, a chemical that changes colors at
various pH’s.
• Indictors are often weak acids,
• the ionized and the unionized form have different colors.
• will change color over a small, given range of pH, for example;
• The equivalence point is located in the middle of the
vertical portion of the titration curves.
• We need to choose an indicator that changes color at a
pH value as close to the equivalence point as possible
• This observable color change of the indicator is called the
end point,
• Should correspond to the equivalence point as closely as
possible.
• Using the titration plots and the table on the previous
slide, suitable indicators can be chosen;
• Strong acid - Strong base  most indicators
• Weak acid - Strong base  phenolphthalein
• Strong acid - Weak base  methyl orange
Titration Calculations
• Millimole (mmol) = 1/1000 mol
• Molarity = mmol/mL = mol/L
• Makes calculations easier because we will rarely add
Liters of solution.
• Adding a solution of known concentration until the
substance being tested is consumed.
• This is called the equivalence point.
• Graph of pH vs. mL is a titration curve.
Strong acid with Strong Base
•
•
•
•
Do the stoichiometry.
There is no equilibrium .
They both dissociate completely.
The titration of 50.0 mL of 0.200 M HNO3 with 0.100 M
NaOH
• Calculate the pH after the following ml of NaOH are
added:
• A. no NaOH
• B. 10ml of NaOH
• C. 20ml of NaOH
• D. 50ml of NaOH
• E. 100ml of NaOH
• F. 150ml of NaOH
Weak acid with Strong base
 There is an equilibrium.
 Do stoichiometry.
 Then do equilibrium.
 Titrate 50.0 mL of 0.10 M HC2H3O2
(Ka = 1.8 x 10-5) with 0.10 M NaOH. Calculate the pH
at the following points.
 A. No NaOH
 B. 10ml of NaOH
 C. 25ml of NaOH
 D. 40ml of NaOH
 E. 50ml of NaOH
 F. 60ml of NaOH
Titration Curves
•
Strong acid with strong Base
Equivalence at pH 7
•
pH
7
mL of Base added


Weak acid with strong Base
Equivalence at pH >7
pH
>7
mL of Base added


Strong base with strong acid
Equivalence at pH 7
pH
7
mL of Base added

Weak base with strong acid
 Equivalence at pH <7
pH
<7
mL of Base added
Summary: Titration of a Weak Acid with
a Strong Base
• ** At 0 ml base find pH with ICE and Ka
• ** Weak acid before equivalence point
– Stoichiometry first
– Then Ka Equation or Henderson-Hasselbach
• ** Weak acid at ½ equivalence pt
•
[H+] = [Ka] or pH=pKa
• ** Weak acid at equivalence point find pH with ICE and
Kb
• ** Weak base after equivalence - leftover strong base.
Indicators
• Weak acids that change color when they become
bases.
• weak acid written HIn
• Weak base
• HIn
H+ + Inclear
red
• Equilibrium is controlled by pH
• End point - when the indicator changes color.
Indicators
• Since it is an equilibrium the color change is gradual.
• It is noticeable when the ratio of
-
[HIn]/[In ] is 1/10
• Since the Indicator is a weak acid, it has a Ka.
• pH the indicator changes at is.
-
• pH=pKa +log([In ]/[HIn]) = pKa +log(1/10)
• pH=pKa - 1 on the way up (adding a base)
[In-]/[HIn] or
Indicators
-
• pH=pKa + log([HIn]/[In ]) = pKa + log(10)
• pH=pKa+1 on the way down (adding an acid)
• For strong acid base titrations, the indicator color
change is sharp and a wide range of indicators ,
might be suitable.
• For titrations of weak acids choose the indicator with
a pKa as close to equivalence point as possible.
The Common Ion Effect
 When the salt with the anion of a weak acid is added to




that acid,
It reverses the dissociation of the acid.
Lowers the percent dissociation of the acid.
The same principle applies to salts with the cation of a
weak base.
The calculations are the same as last chapter.
• HF (aq)
•
•
•
•
H+ (aq) + F- (aq)
If NaF is added to the solution
[F-] increases and the equilibrium shifts left
This makes the solution less acidic
Soln of NaF + HF less acidic than soln of HF
Buffered solutions
• A solution that resists a change in pH.
• Either a weak acid and its salt or a weak base and its
salt.
• We can make a buffer of any pH by varying the
concentrations of these solutions.
• Same calculations as before.
15.2 – The pH of a Buffered Solution
• A buffered solution contains 0.50 M acetic acid (HC2H3O2,
Ka = 1.8 x 10-5) and 0.50 M sodium acetate (NaC2H3O2).
• Calculate the pH of the solution.
Adding a strong acid or base
• Do the stoichiometry first.
• A strong base will grab protons from the weak acid
reducing [HA]0
• A strong acid will add its proton to the anion of the salt
reducing [A-]0
• Then do the equilibrium problem.
15.3 – pH Changes in Buffered Solutions
• Calculate the change in pH that occurs when 0.010 mol
solid NaOH is added to 1.0 L of the buffered solution in
sample exercise 15.2.
Buffers – How do they work?
• A buffered solution contains large quantities of a weak
acid (HA) and its conjugate base (A-) from the salt.
• When OH- ions are added they pull off the H+ from the
HA
• OH- + HA
A- + H2O
• OH- cannot accumulate. It reacts with HA to produce A-
General equation
+
-
 Ka = [H ] [A ]
[HA]
+
 so [H ] = Ka [HA]
[A-]
+
-
 The [H ] depends on the ratio [HA]/[A ]
 taking the negative log of both sides
-
 pH = -log(Ka [HA]/[A ])
-
 pH = -log(Ka)-log([HA]/[A ])
-
 pH = pKa + log([A ]/[HA])
This is called the HendersonHasselbach equation
-
• pH = pKa + log([A ]/[HA])
• pH = pKa + log(base/acid)
15.4 – The pH of a Buffered Solution II
• Calculate the pH of a solution containing 0.75 M lactic
acid (Ka = 1.4 x 10-4) and 0.25 M sodium lactate.
15.5 – The pH of a Buffered Solution III
• A buffered solution contains 0.25 M NH3 (Kb = 1.8 x 10-5)
and 0.40 M NH4Cl. Calculate the pH of this solution.
15.6 – Adding Strong Acid to a Buffered Solution I
• What would the pH be if 0.10 mol of HCl is added to 1.0 L
of the buffered solution from Ex. 15.5.
Buffering Capacity
• The pH of a buffered solution is determined by the ratio
-
[A ]/[HA].
• As long as this doesn’t change much the pH won’t change
much.
• The more concentrated these two are the more H
OH- the solution will be able to absorb.
• Larger concentrations bigger buffer capacity.
+ and
15.7 – Adding Strong Acid to a Buffered Solution II
• Calculate the change in pH that occurs when 0.010
mol of HCl(g) is added to 1.0L of each of the following:
• 5.00 M HAc and 5.00 M NaAc
• 0.050 M HAc and 0.050 M NaAc
-5
• Ka= 1.8x10
Buffer capacity
• The best buffers have a ratio [A-]/[HA] = 1
• This is most resistant to change
-
• True when [A ] = [HA]
• Make pH = pKa (since log1=0)
15.8 – Preparing a Buffer
• A chemist needs a solution buffered at pH 4.30
and can choose from the following acids (and
their sodium salts):
• a. chloroacetic acid (Ka = 1.35 x 10-3)
• b. propanoic acid (Ka = 1.3 x 10-5)
• c. benzoic acid (Ka = 6.4 x 10-5)
• d. hypochlorous acid (Ka = 3.5 x 10-8)
• Caclulate the ratio [HA] / [A-] required for each to
yield a pH of 4.30. Which system will work best?
Solubility Equilibria
(Equilibria between solids and solutions)
• All dissolving is an equilibrium.
• If there is not much solid it will all dissolve.
• As more solid is added the solution will
become saturated.
• Solid
dissolved
• The solid will precipitate as fast as it
dissolves .
• Equilibrium
Consider the following equilibrium
Ca2+ (aq) + 2F- (aq)
• Ultimately this reaches equilibrium
• Solubility Product Constant (Ksp) is
• CaF2 (s)
2+] [F-]2
• Ksp = [Ca
• Called the solubility product.
Solubility is not the same as solubility product
 Solubility product is an equilibrium constant.
 it doesn’t change except with temperature.
 Solubility is an equilibrium position for how much can
dissolve.
 A common ion can change this.
15.12 - Calculating Ksp from Solubility I
• Copper (I) Bromide has a measured solubility of 2.0 x
10-4 mol/L at 25oC.
• Calculate its Ksp value
15.13 - Calculating Ksp from Solubility II
• Calculate the Ksp value for bismuth sulfide (Bi2S3),
which has a solubility of 1.0 x 10-15 mol/L at 25oC.
15.14 - Calculating Solubility from Ksp
• The Ksp value for copper (II) iodate, Cu(IO3)2, is 1.4 x
10-7 at 25oC. Calculate its solubility at 25oC.
Relative solubilities
• Ksp will only allow us to compare the solubility of solids
the that fall apart into the same number of ions.
• The bigger the Ksp of those the more soluble.
• If they fall apart into different number of pieces you have
to do the math.
Common Ion Effect
• If we try to dissolve the solid in a solution with either the
cation or anion already present less will dissolve.
• 15.15 - Solubility and Common Ions
• Calculate the solubility of solid CaF2 (Ksp = 4.0 x 10-11) in a 0.025
M NaF
pH and solubility
-
• OH can be a common ion.
• More soluble in acid or less in base.
Mg2+ (aq) + 2OH- (aq)
• For other anions if they come from a weak acid they are
more soluble in a acidic solution than in water.
• H+ + PO43HPO42• Ag3PO4 (s)
3Ag+ (aq) + PO43- (aq)
• Mg(OH)2 (s)
Precipitation and Qualitative Analysis
 Ion Product, Q =[M+]a[N-]b
 If Q>Ksp a precipitate forms.
 If Q<Ksp No precipitate.
 If Q = Ksp equilibrium.
 15.17 – Determining Precipitation Conditions
 A solution is prepared by mixing 150.0 mL of 1.00 x 10-2 M
Mg(NO3)2 and 250.0 mL of
1.00 x 10-1 M NaF. Calculate the concentrations of Mg2+ and F- at
equilibrium with solid MgF2 (Ksp = 6.4 x 10-9).
Salts as acids and bases
• Salts are ionic compounds.
• Salts of the cation of strong bases and the anion
of strong acids are neutral.
• for example:
• NaCl (salt) – NaOH or HCl
• KNO3 (salt) – KOH or HNO3
• There is no equilibrium for strong acids and
bases.
• Na+, K+ and Cl-, NO3- have no affinity for OH- ions
or H+ ions.
Salts that Produce a Basic Soln
• If the anion of a salt is the conjugate base of a weak acid
•
•
•
•
•
– this produces a basic solution.
In an aqueous solution of NaF (Salt)
F- (anion) / HF (weak acid)
The major species are Na+, F-, and H2O
F- + H2O
HF + OHKb = [HF][OH-]
[F- ]
Ka tells us Kb
• Ka x Kb = KW = 1.0 x 10-14
• The anion of a weak acid is a strong base.
• Calculate the pH of a solution of 0.30 M NaF solution.
• (Ka of HF is 7.2 x 10-4)
• (Ex. 14.18)
• The F- ion competes with OH- for the H+
Acidic salts
• A salt with the conj. acid of a weak base and the anion of
a strong acid will be acidic.
• NH4Cl (salt)
• NH4+ (conj. acid of weak base – NH3)
• Cl- (anion of a strong acid - HCl)
• Calculate the pH of a solution of 0.10 M NH4Cl (K
b of
NH3 1.8 x 10-5) (Ex. 14.19)
• Other acidic salts are those of highly charged metal ions.
Anion of weak acid, cation of weak
base
• If the salts in solution produce both cations and anions
with acidic and basic properties.
•K
a > Kb
• K < Kb
a
• K = Kb
a
acidic
basic
Neutral
• Example: NH4C2H3O2 or NH4CN
The Lewis Acid-Base Model
Model
Definition of Acid
Definition of Base
Arrhenius
H+ producer
OH- producer
Bronsted-Lowry
H+ donor
H+ acceptor
Lewis
Electron-pair acceptor
Electron-pair donor
• Lewis Acid
• Has an empty orbital that can accept an electron pair
• Lewis Base
• Has a lone pair of electrons that can be shared
Formation of Hydronium Ion
H
+
ACID
••
•• O—H
H
BASE
H
•• +
O—H
H
Formation of Metal Ions
Co
2+
ACID
•• ••
O—H
H
BASE
Co
2+
•• ••
O—H
H