Transcript Acid-Base Equilibria and Solubility Equilibria

Acid-Base Equilibria and
Solubility Equilibria
Chapter 16
Dr. Ali Bumajdad
Chapter 16 Topics
Acid –Base Equilibria
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Homogeneous Versus Heterogeneous Solution Equilibria
The Common Ion Effect
Buffer Solutions
Acid-Base Titrations and Indicators
Solubility Equilibria
Relation between solubility and Ksp
The Common Ion Effect and Solubility
pH and Solubility
Dr. Ali Bumajdad
Homogeneous Equilibria
H2O (l)
H+ (aq) + OH- (aq)
Heterogeneous Equilibria
AgNO3(aq) +NaCl(aq)
AgCl(s) + NaNO3 (aq)
The Common Ion Effect
•Common ion effect is the shift in equilibrium to the left
(reactant) caused by the addition of a compound (strong
electrolyte) having an ion in common with the dissolved
substance (weak electrolyte). This is because of the Le
Chatelier Principle.
The presence of a common ion decreases
the ionization of a weak acid or a weak base.
Consider mixture of CH3COONa (strong electrolyte) and
CH3COOH (weak acid).
CH3COONa (s)
Na+ (aq) + CH3COO- (aq)
CH3COOH (aq)
H+ (aq) + CH3COO- (aq)
pH will increase
common
ion
Consider mixture of salt NaA and weak acid HA.
NaA (s)
Na+ (aq) + A- (aq)
HA (aq)
H+ (aq) + A- (aq)
Ka [HA]
+
[H ] =
[A-]
[H+][A-]
Ka =
[HA]
(1)
-log [H+] = -log Ka - log
[HA]
[A-]
-]
[A
-log [H+] = -log Ka + log
[HA]
[A-]
pH = pKa + log
[HA] where
pKa = -log Ka
[conjugate base]
pH = pKa + log
[acid]
(2)
Henderson-Hasselbalch
equation
Q) What is the pH of a solution containing 0.30 M HCOOH
and 0.52 M HCOOK? (Ka of HCOOH = 1.710-4 )
Mixture of weak acid and conjugate base!
HCOOH (aq)
Initial (M)
Change (M)
Equilibrium (M)
Common ion effect
0.30 – x  0.30
0.52 + x  0.52
HCOOH pKa = 3.77
H+ (aq) + HCOO- (aq)
0.30
0.00
0.52
-x
+x
+x
0.30 - x
x
0.52 + x
[HCOO-]
pH = pKa + log
[HCOOH]
[0.52]
= 4.01
pH = 3.77 + log
[0.30]
- Na is spectator ion and do not affect pH
Another solution:
- Find [H+] then find pH
[H+]
Ka [HCOOH]
=
[HCOO-]
pH=-log [H+]
(1)
Buffer solution
• Buffer solution is a solution that resists change in pH
because it contain both acids to neutralize OH- and base to
neutralize H+ ions
•
Buffer composition:
1. A weak acid + its salt (that is salt of weak conjugate base)
2. A weak base + its salt (that is salt of weak conjugate acid)
Consider an equal molar mixture of CH3COOH and CH3COONa
Add strong acid
H+ (aq) + CH3COO- (aq)
Add strong base
OH- (aq) + CH3COOH (aq)
CH3COOH (aq)
CH3COO- (aq) + H2O (l)
HCl
HCl + CH3COO-
H+ + ClCH3COOH + Cl-
Q) Which of the following are buffer systems? (a) KF/HF
(b) KBr/HBr, (c) Na2CO3/NaHCO3
(a) HF is a weak acid and F- is its conjugate base
buffer solution
(b) HBr is a strong acid
not a buffer solution
(c) CO32- is a weak base and HCO3- is its conjugate acid
buffer solution
Buffer capacity:
The greater the amount of the conjugate acid-base pair the
greater the buffer capacity (that is the grater the resistance
of pH change). Some buffer solution have the same pH but
their capacity are different
Q) Which of the following buffer solutions has greater pH? And
which has greater buffer capacity?
1) 1M HC2H3O2 + 1 M NaC2H3O2
2) 0.1M HC2H3O2 + 0.1M NaC2H3O2
[conjugate base]
pH = pKa + log
[acid]
(2)
Q) Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer
system. What is the pH after the addition of 20.0 mL of 0.050 M
NaOH to 80.0 mL of the buffer solution? (Ka NH4+ =5.610-10; Kb
NH3 = 1.810-5)
NH4+ (aq)
H+ (aq) + NH3 (aq)
[NH3] (2)
[0.30]
pH = pKa + log
pKa = 9.25 pH = 9.25 + log
= 9.17
+
[NH4 ]
[0.36]
start (moles)
(better to use mole
then convert to M)
end (moles)
0.029
0.001
NH4+ (aq) + OH- (aq)
0.028
0.0
0.024
H2O (l) + NH3 (aq)
0.025
final volume = 80.0 mL + 20.0 mL = 100 mL
[NH4
+]
0.028
0.025
=
[NH3] =
0.10
0.10
[0.25]
pH = 9.25 + log
= 9.20
[0.28]
Sa. Ex. How many mole of NH4Cl must be added to 2.0 L of
0.10 M NH3 to form Buffer solution with pH =9.00? (assume adding
NH4Cl do not affect the total volume (Kb = 1.810-5)
pKa+pKb = pKw=14 For base or acid and its salt
[conjugate base]
pH = pKa + log
[acid]
or
pOH = 14 - pH
[OH-]= 10-pOH
Kb= [NH4+ ][OH-] / [NH3]
(2)
(Ka=1.810-5)
(a) pH = pKa + log [conjugate base]
[acid]
(2)
(b) The added strong acid (HCl) will be completely
consumed by the reaction with the buffer and that’s
way it will not affect the pH much
CH3COO- + H+
CH3COOH
Start molarity:1.0
0.1
1.0
Change
: -0.1
-0.1
+0.1
End molarity: 0.9
0
1.1
Use equation 2
Sa. Ex. Buffer made by adding 0.300 mol HC2H3O2 and 0.300 mol
NaC2H3O2 to enough water to make 1.00 L solution. The pH of the
buffer is 4.74.
(a) Calculate the pH of this solution after 0.20 mol of NaOH is
added (neglect any volume change)
(b) calculate the pH when 0.020 mol of NaOH is added to 1.00L
of water
(a) OH- will react with the acid and the no. of mole of the acid
and the conjugate base will change
HC2H3O2 + OH-
H2O + C2H3O2-
Before reaction
Change
After reaction
H+ + C2H3O2(2)
[conjugate base]
pH = pKa + log
[acid]
HC2H3O2
Titrations
In a titration a solution of accurately known concentration is
added gradually added to another solution of unknown
concentration until the chemical reaction between the two
solutions is complete.
• Equivalence point : the point at which the reaction is complete
• Indicator : substance that changes color at (or near) the
equivalence point
Slowly add base
to unknown acid
UNTIL
The indicator
changes color
(pink)
Strong Acid-Strong Base Titrations
NaOH (aq) + HCl (aq)
OH- (aq) + H+ (aq)
H2O (l) + NaCl (aq)
H2O (l)
Weak Acid-Strong Base Titrations
CH3COOH (aq) + NaOH (aq)
CH3COOH (aq) + OH- (aq)
CH3COONa (aq) + H2O (l)
CH3COO- (aq) + H2O (l)
At equivalence point (pH > 7) because of the hydrolysis:
CH3COO- (aq) + H2O (l)
OH- (aq) + CH3COOH (aq)
Strong Acid-Weak Base Titrations
HCl (aq) + NH3 (aq)
H+ (aq) + NH3 (aq)
NH4Cl (aq)
NH4Cl (aq)
At equivalence point (pH < 7) because of the hydrolysis:
NH4+ (aq) + H2O (l)
NH3 (aq) + H+ (aq)
Q) Exactly 100 mL of 0.10 M HNO2 are titrated with 100 ml of 0.10
M NaOH solution. What is the pH at the equivalence point ?
start (moles)
end (moles)
0.01
0.01
HNO2 (aq) + OH- (aq)
0.0
0.0
Final volume = 200 mL
NO2- (aq) + H2O (l)
0.01
NO2- (aq) + H2O (l)
Initial (M)
Change (M)
0.01
= 0.05 M
0.200
OH- (aq) + HNO2 (aq)
[NO2-] =
0.05
0.00
0.00
-x
+x
+x
x
x
Equilibrium (M) 0.05 - x
[OH-][HNO2]
x2
-11
=
2.2
x
10
Kb =
=
[NO2-]
0.05-x
pOH = 5.98
0.05 – x  0.05 x  1.05 x 10-6 = [OH-]
pH = 14 – pOH = 8.02
Acid-Base Indicators
• Indiators are weak organic acid or base
•End point: the point when the indicator change color
HIn (aq)
H+ (aq) + In- (aq)
[HIn]
 10 Color of acid (HIn) predominates
[In ]
[HIn]
-) predominates
Color
of
conjugate
base
(In

10
[In-]
pH
The titration curve of a strong acid with a strong base.
Methyl red and ph.ph are good indicators but Thymol blue is not
Q) Which indicator(s) would you use for a titration of HNO2 with
KOH ?
Weak acid titrated with strong base.
At equivalence point, will have conjugate base of weak acid.
At equivalence point, pH > 7
Use cresol red or phenolphthalein
Solubility Equilibria
Ag+ (aq) + Cl- (aq) Heterogeneous equilibrium
AgCl (s)
Ksp = [Ag+][Cl-]
Ksp is the solubility product constant
Ksp is the product of the concentrations of the ions involve in the
equilibrium all rise to a power = the coefficient
MgF2 (s)
Ag2CO3 (s)
Ca3(PO4)2 (s)
Mg2+ (aq) + 2F- (aq)
2Ag+ (aq) + CO32- (aq)
3Ca2+ (aq) + 2PO43- (aq)
Ksp = [Mg2+][F-]2
Ksp = [Ag+]2[CO32-]
Ksp = [Ca2+]3[PO43-]2
Relation between Molar solubility and Ksp
•Molar solubility, s, (mol/L) is the number of moles of solute
dissolved in 1 L to form a saturated solution.
•Solubility (g/L) is the number of grams of solute dissolved in
1 L to form a saturated solution.
Molar solubility = the highest concentration of solution
Molar solubility change upon adding another solutes
Molar solubility change upon changing temperature
Molar solubility  M.m. = solubility
(3)
Ag+ (aq) + Cl- (aq)
But [Ag+] = solubility of AgCl
Ksp= [Ag+] [Cl-]
and [Cl-] = solubility of AgCl
(5)
 Ksp= s s = s2 (4)  s = (Ksp)1/2
AgCl (s)
Solubility Product versus Molar Solubility
Salt type
AB
(e.g. NaCl , KIO3)
KSP
KSP =S2
A2B or AB2
KSP =4 S3
A3B or AB3
KSP = 27 S4
A2B3 or A3B2
KSP = 108 S5
(e.g. CaF2 , Ag2CrO4)
(e.g. Ce(NO3)3 , AlCl3)
(e.g. Bi2S3)
S
S = K SP
 K SP 

4


S=
 K SP 

27


S =
 K SP 

108


S =
1
3
1
4
1
5
/ M.m.
 M.m.
Q) What is the solubility of silver chloride in g/L ? Ksp = 1.6 x 10-10
AgCl (s)
Initial (M)
Change (M)
Equilibrium (M)
[Ag+] = 1.3 x 10-5 M
Ag+ (aq) + Cl- (aq)
0.00
0.00
+s
+s
s
s
[Cl-] = 1.3 x 10-5 M
Ksp = 1.6 x 10-10
Ksp = [Ag+][Cl-]
Ksp = s2
s = Ksp
s = 1.3 x 10-5
1.3 x 10-5 mol AgCl 143.35 g AgCl
Solubility of AgCl =
x
= 1.9 x 10-3 g/L
1 L soln
1 mol AgCl
Ksp of Cu(OH)2 = 2.210-20
Dissolution of an ionic solid in aqueous solution:
Q < Ksp
Unsaturated solution
Q = Ksp
Saturated solution
Q > Ksp
Supersaturated solution
No precipitate
Precipitate will form
Ion concentration at equilibrium
Ion concentration not at equilibrium
Q) If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M
CaCl2, will a precipitate form? Ksp of Ca(OH)2 = 8.0 x 10-6
The ions present in solution are Na+, OH-, Ca2+, Cl-.
Only possible precipitate is Ca(OH)2 (solubility rules).
Is Q > Ksp for Ca(OH)2?
[Ca2+]0 = 0.100 M
[OH-]0 = 4.0 x 10-4 M
Q = [Ca2+]0[OH-]02 = 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8
Ksp = [Ca2+][OH-]2 = 8.0 x 10-6
Q < Ksp
No precipitate will form
Ksp of BaSO4 = 1.110-10
Q) What concentration of Ag is required to precipitate ONLY
AgBr in a solution that contains both Br- and Cl- at a
concentration of 0.02 M?
AgBr (s)
Ag+ (aq) + Br- (aq)
Ksp = 7.7 x 10-13
Ksp = [Ag+][Br-]
-13
K
7.7
x
10
sp
-11 M
=
=
3.9
x
10
[Ag+] =
0.020
[Br-]
AgCl (s)
[Ag+]
Ag+ (aq) + Cl- (aq)
Ksp = 1.6 x 10-10
Ksp = [Ag+][Cl-]
Ksp
1.6 x 10-10
-9 M
=
=
8.0
x
10
=
0.020
[Cl-]
3.9 x 10-11 M < [Ag+] < 8.0 x 10-9 M
The Common Ion Effect and Solubility
The presence of a common ion decreases the solubility
of the salt. This due to LeChatelier’s principle
Q) What is the molar solubility of AgBr in (a) pure water and (b)
0.0010 M NaBr?
AgBr (s)
Ag+ (aq) + Br- (aq)
Ksp = 7.7 x 10-13
s2 = Ksp
s = 8.8 x 10-7
NaBr (s)
Na+ (aq) + Br- (aq)
[Br-] = 0.0010 M
AgBr (s)
Ag+ (aq) + Br- (aq)
[Ag+] = s
[Br-] = 0.0010 + s  0.0010
Ksp = 0.0010 x s
s = 7.7 x 10-10
Solubility not molar solubility
pH and Solubility
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The presence of a common ion decreases the solubility.
Insoluble bases dissolve in acidic solutions
Insoluble acids dissolve in basic solutions
remove
add
Mg(OH)2 (s)
Mg2+ (aq) + 2OH- (aq)
Ksp = [Mg2+][OH-]2 = 1.2 x 10-11
Ksp = (s)(2s)2 = 4s3
4s3 = 1.2 x 10-11
s = 1.4 x 10-4 M
[OH-] = 2s = 2.8 x 10-4 M
pOH = 3.55 pH = 10.45
At pH less than 10.45
Lower [OH-]
OH- (aq) + H+ (aq)
H2O (l)
Increase solubility of Mg(OH)2
At pH greater than 10.45
Raise [OH-]
Decrease solubility of Mg(OH)2
S-2 is a conjugate base of a weak acid
Cl- is a conjugate base of strong acid
SO42- is a conjugate base of a weak acid
Weak base
Extremely weak base
Weak base