Download Chapter 17 Additional Aspects of Aqueous Equilibria I. Solubility

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Chapter 17
Additional Aspects of Aqueous Equilibria
I. Solubility Equilibria
The solubility product constant Ksp
Consider
BaSO4(s) X Ba2+(aq) + SO42-(aq)
for which Ksp = [Ba2+][SO42-]
Ksp is the solubility product. (BaSO4 is ignored because its concentration is constant.)
In general: the solubility product is the molar concentration of ions raised to their stoichiometric powers.
Solubility is the amount (grams) of a substance which dissolves to form a saturated solution.
Molar solubility is the number of moles of solute dissolving to form a liter of solution.
II. Factors that Effect Solubility
A.. Common Ion Effect
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the solubility of a partially soluble salt is decreased when a common ion is added.
Consider the equilibrium established when AgI is added to water
AgI(s) XAg+ (aq) + I- (aq)
Ksp = [Ag+][I-]
If a common ion is added, i.e. I- from NaI then [I-] increases and the system is no longer at
equilibrium.
[Ag+][I-]> Ksp when NaI is added.
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To return to equilibrium (at the given Ksp) Ag must decrease and therefore the reverse reaction
will occur.
Solving common ion equilibrium problems follows the same pattern as other equilibrium
problems. However, there is an initial concentration of the common ion (from the salt)
REMEMBER: Addition of a common ion will cause the equation to shift away from the common ion.
B. Solubility and pH
Applying LeChatelier’s principle:
CaF2 (s) XCa+2 (aq) + 2F- (aq)
If F- is removed, then the equilibrium shifts towards the increase of it and causes solubility of
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caF2 to increase.
How can F- be removed? Addition of a strong acid!
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F- (aq) + H+(aq) X HF (aq)
Therefore, as pH increases [H+] increases and solubility increases. The effect of pH on solubility
is tramatic. (Rememeber in labs why we add acids to get things to dissolve!)
C. Formation of Complex ions
Complex ions occur under certain conditions, usually metal complexes formed under acid-base
interactions. Chapter 24 will deal with this extensively. For the time being, complex ions are recognized
by a metal ion containing more bonds than it normally would form. The bonds are usually to a
polyatomic ion or molecule and carry a charge.
[Fe(H2O)5SCN]2+ , [Ag(NH3)2]+
The attached NH3 is called the ligand. (In this case a Lewis base)
The fomation of [Ag(NH3)2]+ is as follows:
Ag+ (aq) + 2NH3(aq) X [Ag(NH3)2]+ (aq)
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the equilibrium expression constant is called Kf (formation constant)
Kf = [Ag(NH3)2]+
[Ag+][NH3]2
Effect on solubility:
Consider the addition of ammonia to AgCl (white precipitate):
AgCl(s) + NH3 (aq) XAg(NH3)3+ (aq) + Cl-(aq)
Effectively, insoluble AgCl has been removed from solution.
D. Amphoterism
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Ampotheric oxides will dissolve either a strong acid or a strong base.
Examples: hydroxides and oxides of Al3+, Cr3+ Zn2+, and Sn2+
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the hydroxides general form complex ions with FOUR hydroxide ligands attached to the metal.
Al(OH)3 (s) + OH-(aq) XAl(OH)4-(aq)
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Hydrated metal ions act as weak acids. Thus, amphoterism is interrupted.
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Al(H2O)63+ (aq) + OH-(aq) X Al(H2O)52+ (aq) + H2O(l)
Al(H2O)52+ (aq) + OH-(aq)X Al(H2O)4(OH)2+ (aq) + H2O(l)
Al(H2O)4(OH)2+ (aq) + OH-(aq)X Al(H2O)3(OH)3 (s) + H2O(l)
Al(H2O)3(OH)3 (s) + OH-(aq) X Al(H2O)2(OH)4- (aq) + H2O(l)
III. Buffers
Composition and Action of Buffered solutions
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A buffer consists of a mixture oroughly equal amounts f a weak acid (HX) and its conjugate base
(X-)
Example: dissolve both acetic acid (HC2H3O2) and sodium acetate (Na+,C2H3O2- ) in water.
HC2H3O2 X H+ + C2H3O2C
A buffer has the ability to resist a change in pH when either OH- or H+ is added.
I.e. adding C2H3O2- increases the concentration of acetate ion in the equilibrium above. The
reaction shifts to produce more acetic acid until they are in equal ratio again leaving
the Ka
unchanged and therefore pH unchanged.
Buffer capacity and pH
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Buffer capacity is the amount of acid or base neutralized by the buffer before there is a
significant change in pH. Buffer capacity depends on the composition of the buffer. The greater
the amounts of the conjugate acid, base pair, the greater the buffer capacity.
the pH of the buffer depends on the Ka . If ka is sufficiently small ( the equilibrium concentration
of the undissociated acid is close to the initial concentration), then
Ka =
[H+][A-]
[HA]
So, [H+] =( Ka)[HA]/[A-]
Taking negative logs we get -log [H+] = -logKa - log [HA]/[A-]
pH = pKa-log [HA]/[A-]
Or pH pKa+log [base]/[acid] Henderson-Hasselbalch equation
Addition of Strong acids or Bases to Buffers
We break the calculation into two parts: stoichiometric and equilibrium
The amount of strong acid or base added results in a neutralization reaction:
X- + H3O+ 6HX + H2O
HX + OH- 6 X- + H2O
By knowing how much H3O+ or OH- was added (stoichiometry) we know how much HX or X- is
formed. With the concentrations of HX and X- (note the change in volume of solution) we can calculate
pH from the Henderson hasselbalch equation.
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Example 1: What is the pH of a buffer that is 0.12 M in lactic acid, HC3H5O3, and 0.10 M in sodium
lactate? For lactic acid, Ka = 1.4 x 10-4.
solution:
pKa = -log 10Ka
pH = pKa + log ([base]/[acid]) = 3.85 + log (0.10/0.12)
pH = 3.77
Example 2:
A buffer is made by adding 0.300 mol H C2H3O2 and 0.300 M NaC2H3O2 to enough water to make 1 liter
solution. The pH of the buffer is 4.74. Calculate the pH of this solution (a) after 0.0200 mol of NaOH is
added (neglect any volume changes)
Solution (a)
Step 1: Stoichiometric calculation: The OH- provided by NaOH reacts with HC2H3O2 . The following
table summarizes the concentrations before and after neutralization:
Before rxn
Change
After Rxn
HC2H3O2(aq)
0.300 M
-0.0200 M
0.280M
+
OH®
0.020 M
-0.020M
0.0M
H2O (l)
+ C2H3O2-(aq)
--------------0.300 M
--------------+0.020M
-------------0.320 M
Step 2: Equilibrium calculation: the solution after neutralization contains
H C2H3O2 - C2H3O2 conjugate acid-base pair. We next consider the proton-transfer equilibrium in order
to determine the pH of solution.
HC2H3O2(aq)
+ H2O
H3O+ (l)
+ C2H3O2-(aq)
0.280 M
0.020 M
0
0.300 M
Before rxn
Change
-xM
-----------+xM
+xM
After Rxn
(0.280 - x) M
----------xM
(0.320 + x) M
Ka = [H3O+][ C2H3O2-] / [H C2H3O2] = (x) (0.320 + x)/(0.280 -x) = (x) (0.320)/0.280 =
1.8 X 10-5
+
-5
-5
* x = [H3O ] = {(0.280)(1.8 x 10 )}/0.320 = 1.6 x 10 M
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pH = -log (1.6 x 10-5) = 4.80
*In step 2, you could also use the Henderson-Hasselbalch equation:
pH = pKa + log [base]/[acid]
pKa = -log (1.8 x 10-5) = 4.74
4.74 + log [0.320]/[0.280] = 4.80
IV. Precipitation and Separation of ions
Consider
BaSO4 X Ba2+ + SO42At any instant in time
Q = [Ba2+][SO42-]
1. If Q<Ksp precipitation occurs until Q = Ksp
2. If Q = Ksp, equilibrium exists
3. If Q > Ksp, solid dissolves until Q=Ksp
Based on solubilities, ions can be selectively removed from solutions.
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Consider a mixture of Zn2+9aq) and Cu2+(aq). CuS (Ksp = 6 x 10-37) is less soluble than ZnS (Ksp
= 2 x 10-25). CuS will be removed from solution before ZnS.
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As H2S is added to the green solution, black CuS forms in a colorless solution of Zn2+. When
more H2S is added, a second precipitate of white ZnS forms
Selective precipitation of Ions
Ions can be separated from each other based on their salt solubilities. Example: if Hcl is added to a
solution containing Cu2+ and Ag+, the silver precipitates (Ksp = 1.8 x 10-10) while the Cu2+ remains in
solution.>>>>>Selective Precipitation
V. Qualitative analysis
remember old lab: review how you detect certain ions
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Qualitative analysis :designed to detect the presence of metal ions
Quantitative analysis: determine how much metal is present.
VI. Acid-Base titrations
Titration curve: a plot of pH vs volume during a titration (we will be doing this in lab)
A. Strong acid-strong base
At equivalence point, pH = 7
Indicator: Phenolphthalein changes color between 8.3 and 10.0
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In acid colorless, in base pink
Equivalence point : the point at which the acid and base are present in stoichiometric mole relationship
Endpoint: the observed end of the titration (color change)
Titration error: the difference between the observed and equivalence point.
B. Strong base, weak acid
acetic acid and sodium hydroxide leaves water and acetate ion which is basic. Therefore the titration ends
with a basic pH greater than 7.
C. Weak acid- Weak Base titration
At the equivalence point, there is a steep change in pH and then a leveling off due to buffer effects.
For very weak acids it is impossible to detect the equivalence point.
D. Polyprotic acids
In a titration there are n equivalence points based on the number of ionizable H