chapter 13 applications of aqueous equilibria
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Chapter 13 Applications of Aqueous Equilibria. 13.1 Solutions of Acids or Bases Containing a Common Ion 13.2 Buffered Solutions 13.3 Exact Treatment of Buffered Solutions (skip) 13.4 Buffer Capacity 13.5 Titrations and pH Curves 13.6 Acid-Base Indicators - PowerPoint PPT PresentationTRANSCRIPT
04/22/23 Zumdahl Chapter 8 1
Chapter 13Applications of Aqueous Equilibria
13.1 Solutions of Acids or Bases Containing a Common Ion
13.2 Buffered Solutions 13.3 Exact Treatment of Buffered Solutions (skip)13.4 Buffer Capacity 13.5 Titrations and pH Curves 13.6 Acid-Base Indicators 13.7 Titration of Polyprotic Acids (skip)13.8 Solubility Equilibria and the Solubility Product 13.9 Precipitation and Qualitative Analysis (skip)13.10 Complex Ion Equilibria (skip)
The Common Ion Effect (1)
HF (aq) H+ (aq) + F- (aq) Ka = 7.2 x 10-4
When F- is added (from NaF), then [H+] must decrease (Le Chatelier’s principle). The pH increases. See example 13.1 in the text to see how to quantitatively determine this effect, with an ICE box)
This applies to weak acids, weak bases and solubility of salts when a common ion is added to the equilibrium reaction. You can change the pH, but adding salt.
NaF (s) → Na+ (aq) + F– (aq)
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Ka =[H3O
+][F−]
[HF]
The Common Ion Effect (2)
If a solution and a salt to be dissolved in it have an ion in common, then the solubility of the salt is depressed relative to pure.
General equation
AB (s) A+ (aq) + B- (aq)
If you add BH(aq), which dissociates into B- and H+, then the [A+]
decreases, and AB is driven out of solution.
Buffers: resist change in pH when acid or base is added.
Buffer Solutions: contain a common ion and are important in biochemical and physiological processes
Organisms (and humans) have built-in buffers to protect them against changes in pH.
Applications of Aqueous Equilibrium
Buffered Solutions
Human blood is a buffered solution
The Common Ion Effect on Buffering
Blood: (pH 7.4)Death = 7.0 <pH > 7.8 = Death
Human blood is maintained by a combination of CO3-2, PO4
-3 and protein buffers.
A solution that is buffered by acetic acid/acetate
Unbuffered solution
How Do Buffers Work?
• Ka = [H+][A-]/[HA] => [H+] = Ka[HA]/[A-]
•If Ka is small (weak acid) then [H+] does not change much when [HA] and [A-] change.
If [HA] and [A-] are large, and [HA]/[A-] ≈ 1,then small additions of acid ([H+]) or base ([OH-]) don’t change the ratio much.
HA H+ + A –
HA = generic acid
Example: A solution of 0.5 M acetic acid plus 0.5 M acetate Ka = 1.8x10-5 pKa = 4.7
HA H+ + A-
Ka = [H+][A-]/[HA] pH = pKa + log[A-]/[HA]
Use an ICE box to calculate the pH[AH]ini = 0.5, [A-]ini = 0.5, [H+]ini = 0
=> => pH = 4.74 i.e., pH=pKa
Example: A solution of 0.5 M acetic acid plus 0.5 M acetate Ka = 1.8x10-5 pKa = 4.74
HA H+ + A-
Ka = [H+][A-]/[HA] pH = pKa + log[A-]/[HA]
Use an ICE box to calculate the pH[AH]ini = 0.5, [A-]ini = 0.5, [H+]ini = 0
=> => pH = 4.74 i.e., pH=pKa
Now add NaOH to 0.01 M(in an unbuffered solution this would give a pOH of 2 and pH of 12)
Use HA + OH- → H2O + A- Redo the ICE Box[AH]ini = 0.49, [A-]ini = 0.51
=> => pH = 4.76
Buffer Calculation: Add Acid (#1) to a Buffered Solution Acetic Acid: Ka = 1.8x10-5 pKa = 4.74HA H+ + A-
Ka = [H+][A-]/[HA] => -pKa = -pH + log[A-]/[HA] => pH = pKa + log[A-]/[HA]
Case 1) [CH3COOH]tot = [CH3COOH] + [CH3COO-] = 1.0M pH = pKa => [CH3COOH] = [CH3COO-] = 0.5
Now add 0.01 M HCl (strong acid)
[CH3COOH] = 0.51 [CH3COO-] = 0.49
pH = pKa + log[A-]/[HA] = 4.74 + log(0.49/0.51) = 4.74 – 0.02 = 4.72
ΔpH = -0.02
Buffer Calculation: Add Acid (#2) to a dilute Buffered Solution
Acetic Acid: Ka = 1.8x10-5 pKa = 4.74HA H+ + A-
Ka = [H+][A-]/[HA] => -pKa = -pH + log[A-]/[HA] => pH = pKa + log[A-]/[HA]
Case 2) [CH3COOH]tot = [CH3COOH] + [CH3COO-] = 0.10 M pH = pKa => [CH3COOH] = [CH3COO-] = 0.05
Now add 0.01 M HCl (strong acid)
[CH3COOH] = 0.06 [CH3COO-] = 0.04
pH = pKa + log[A-]/[HA] = 4.74 + log(0.04/0.06) = 4.74 – 1.5 = 3.54
ΔpH = -1.5
Adding Base to a Buffered Solution:OH- ions do not accumulate but are replaced
by A- ions.
OH– + HA A⇌ – + H2O
When the OH- is added, the concentrations of HA and A- change, but only by small amounts. Under these conditions the
[HA]/[A-] ratio and thus the [H+] stay virtually constant.
When protons are added to a buffered solution, the conjugate base (A–) reacts
H+ + A– ⇌ HA
OH– + HA A⇌ – + H2O
Characteristics of Buffered solutions
• Contain relatively large concentrations of a weak acid and its conjugate base.
• When acid is added, it reacts with the conjugate base
• When base is added, it reacts with the acid.
• pH is determined by the ratio of the base and acid.
⎟⎟⎠
⎞⎜⎜⎝
⎛+=
−
[HA]
][Alog pKpH a
Buffer Capacity
• Buffer capacity is the amount of protons or hydroxide ions that can be absorbed without a significant change in pH.
• pH is determined by the ratio of [A–]/[HA] and pKa
• Capacity is determined by the magnitudes of [HA] and [A–].
€
pH = pKa + log[A−]
[HA]
⎛
⎝ ⎜ ⎞
⎠ ⎟
Equivalence Point
04/22/23 Zumdahl Chapter 8 16
Buffer Summary
Buffer Design: Add known amount of HA (weak acid) and salt of HA (its conjugate base, A─)
[H+] or pH depends on Ka and the ratio of acid to salt or [A─].
Thus if both conc. HA and A- are large then small additions of acid or base don’t change the ratio much
How to Actually Make a Buffer (a buffered solution in a lab)
HA H+ + A –
Zumdahl Chapter 8
Acid-Base Titrations
A controlled addition of measured volumes of a solution of known concentration (the Titrant) from a buret to a second solution of unknown concentration under conditions in which the solutes react cleanly (without side reactions), completely, and rapidly.
Titration:
A titration is complete when the second solute is fully consumedCompletion is signaled by a change in some physical property, such as the color of the reacting mixture or the color of an indicator that has been added to it.
[NaOH]
“X”
Indicator phenolphthalein
04/22/23 Zumdahl Chapter 8 20
Titrations and pH Curves
Zumdahl Chapter 8 21
Strong acid
Titration of a weak acid with a strong base.
The base reacts with the acid.
.
OH– + HA A⇌ – + H2O
04/22/23 Zumdahl Chapter 8 22
Equivalence point determined defined by the Stoichiometry, not by the pKa.
nbase = nacid
number of moles of acid = number of moles of base
For a variety of weak acids:
The Equivalence point occurs at the same stoichiometric amount of base added
The weaker the acid, the greater the pH value for the equivalent point.
04/22/23 Zumdahl Chapter 8 24
Similar for titration of weak bases with strong acids
The indicator phenolphthalein is pink in basic solution and colorless in acidic solution.
Acid-Base Indicators
04/22/23 Zumdahl Chapter 8 26
Indicators (Weak Acid Equilibria)
pH = -log10[H3O+]
04/22/23 Zumdahl Chapter 8 27
IndicatorsA soluble compound, generally an organic dye, that changes its color noticeably over a fairly short range of pH.
Typically, Indicators are a weak organic acid that has a different color than its conjugate base.
Acid: HIn (aq)
HIn(aq) + H2O(l) ↔ H3O+(aq) + In–(aq)
Phenolphtalein
Indicator denoted by In
Conjugate base: In– (aq)
28
Acid: HIn (aq) Conjugate base: In- (aq)
HIn(aq) + H2O(l) ↔ H3O+(aq) + In-(aq)
pH = -log10[H3O+]
[H3O+][OH-] = Kw
04/22/23 Zumdahl Chapter 8 29
Methyl Red
Bromothymol blue
Phenolphtalein
HIn(aq) + H2O(l) ↔ H3O+(aq) + In-(aq)
04/22/23 Zumdahl Chapter 8 30
The useful pH ranges for several common indicators
HIn(aq) + H2O(l) ↔ H3O+(aq) + In–(aq)
04/22/23 Zumdahl Chapter 8 31
Could use either indicator Methyl red changes color to early
Weak acid
Indicator Selection
• Want indicator color change and titration equivalence point to be as close as possible
• Easier with a large pH change at the equivalence pointStrong acid
04/22/23 Zumdahl Chapter 8 32
Solubility Product Ksp
Describes a chemical equilibrium in which an excess solid salt is in equilibrium with a saturated aqueous solution of its separated ions.
General equation
AB (s) ↔ A+ (aq) + B- (aq)
Ksp =
Ksp =The solubility expression controls the amount of solid that will dissolve
04/22/23 Zumdahl Chapter 8 33
Ksp Values at 25°C for Common Ionic Solids
04/22/23 Zumdahl Chapter 8 34
The Solubility of Ionic Solids
The Solubility Product
AgCl(s) ↔Ag+(aq) + Cl-(aq)
= 1.6 10-10 at 25oC
Ksp =
Ksp
The solid AgCl, which is in excess, is understood to have a concentration of 1 mole per liter.
excess
04/22/23 Zumdahl Chapter 8 35
The Solubility of Ionic SolidsThe Solubility Product
Ag2SO4(s) ↔2Ag+(aq) + SO42-(aq)
Ksp =
Fe(OH)3(s) ↔Fe+3(aq) + 3OH-1(aq)
Ksp =
excess
excess
04/22/23 Zumdahl Chapter 8 36
Solubility and Ksp
Determine the mass of lead(II) iodate dissolved in 2.50 L of a saturated aqueous solution of Pb(IO3)2 at 25oC. The Ksp of Pb(IO3)2 is 2.6 10-13.
Pb(IO3)2(s) ↔ Pb2+(aq) + 2 IO3-(aq)
Let “y” = molar solubility in mol/L
04/22/23 Zumdahl Chapter 8 37
Determine the mass of lead(II) iodate dissolved in 2.50 L of a saturated aqueous solution of Pb(IO3)2 at 25oC. The Ksp of Pb(IO3)2 is 2.6 10-13.
Pb(IO3)2(s) ↔ Pb2+(aq) + 2 IO3-(aq)
[Pb2+][IO3-]2 = Ksp
[Pb2+][IO3-]2 =
y = 4.0 10-5 [Pb(IO3)2] = [Pb2+] = y = 4.0 10-5 mol L-1
[IO3-] = 2y = 8.0 10-5 mol L-1
= (4.0 10-5 mol L-1) (557 g mol-1)
= 0.0223 g L-1 2.50 LMolar Mass of lead (II) iodate
Pb(IO3)2 = 557g per mole
Gram solubility of
Lead (II) iodate
“y” = molar solubility
04/22/23 Zumdahl Chapter 8 38
The Solubility of SaltsSolubility and KspExercise 9-3
Compute the Ksp of silver sulfate (Ag2SO4) at 25oC if its mass solubility is 8.3 g L-1.
1 Ag2SO4 (s) ↔ 2 Ag+(aq) + 1 SO42-(aq)
04/22/23 Zumdahl Chapter 8 39
Compute the Ksp of silver sulfate (Ag2SO4) at 25oC if its mass solubility is 8.3 g L-1.
[y] = (8.3 g Ag2SO4 L-1)
[Ag+]2[SO42-] = Ksp
Ksp =
(1 mol Ag2SO4/311.8 g)
[y] = 2.66 10-2 mol Ag2SO4 L-1
1 Ag2SO4 (s) ↔ 2 Ag+(aq) + 1 SO42-(aq)
04/22/23 Zumdahl Chapter 8 40
The Nature of Solubility Equilibria
Dissolution and precipitation are reverse of each other.
General reaction
X3Y2 (s) ↔ 3X+2 (aq) + 2Y-3 (aq)
Ksp =
Dissolution (Solubility)
[s]
s = molar solubility
expressed in moles per liter
04/22/23 Zumdahl Chapter 8 41
A salt’s Ksp value gives us information about its solubility.
Salt ↔ Cation + Anion
If the salts being compared produce the same number of ions, eg., AgI, CuI, CaSO4
s = [cation]s = [anion]
Ksp = [cation] [anion] = s2
Salt molar solubility = s = (Ksp)1/2
Salt Ksp Solubility
AgI 1.5 x 10-16 1.2 x 10-8
CuI 5.0 x 10-12 2.2 x 10-6
CaSO4 6.1 x 10-5 7.8 x 10-3
Solubility CaSO4 > CuI > AgI
Relative Solubilities
04/22/23 Zumdahl Chapter 8 42
The Effects of pH on SolubilitySolubility of Hydroxides
Zn(OH)2(s) ↔Zn2+(aq) + 2 OH-(aq)
[Zn2+][OH-]2 = Ksp = 4.5 10-17
Make more acidic:
[Zn2+][OH-]2 = Ksp[Zn2+][OH-]2 = Ksp[Zn2+][OH-]2 = Ksp
Zinc hydroxide is more soluble in acidic solution than in pure water.
Many solids dissolve more readily in more acidic solutions
If pH decreases (or made more acidic), the [OH-] decreases. In order to maintain Ksp the [Zn2+] must increase and consequently more solid Zn(OH)2 dissolves.
04/22/23 Zumdahl Chapter 8 43
The Effects of pH on SolubilityEstimate the molar solubility of Fe(OH)3 in a solution that is buffered to a pH of 2.9. Lookup Ksp = 1.1x10-36
In pure water:
[OH-] = 3y = 1.3 10-9 mol L-1
pOH = 8.87 (and pH = 5.13)
(pH = 2.9 and) pOH = 11.1 [OH-] = 7.9 10-12 mol L-1
[Fe3+] = Ksp/[OH-]3 = 1.1 10-36 / (7.9 10-12)3 [Fe3+] = [Fe(OH)3] = 2.2 10-3 mol L-1 answer
[Fe3+] = y [OH-] = 3y
y(3y)3 = 27y4 = Ksp = 1.1 10-36
y = 4.5 10-10 mol L-1 = [Fe3+] = [Fe(OH)3]=
[Fe3+][OH-]3 = Ksp
In pure water, Fe(OH)3 is 5 x 10 6 less soluble than at pH = 2.9
Fe(OH)3(s) ↔Fe3+(aq) + 3 OH-(aq)
04/22/23 Zumdahl Chapter 8 44
The Common Ion Effect
The Ksp of thallium(I) iodate (TlO3) is 3.1 10-6 at 25oC. Determine the molar solubility of TlIO3 in 0.050 mol L-1 KIO3 at 25oC.
TlIO3(s) ↔ Tl+(aq) + IO3-(aq)
[Tl+] (mol L-1) [IO3-] (mol L-1)
Initial concentration
Equilibrium concentration Change in concentration
[Tl+][IO3-] = Ksp
s = [TlIO3] = 6.2 × 10-5 mol L-1 = molar solubility
04/22/23 Zumdahl Chapter 8 45
The Common Ion Effect
The Ksp of thallium(I) iodate (TlO3) is 3.1 10-6 at 25oC. Determine the molar solubility of TlIO3 in 0.050 mol L-1 KIO3 at 25oC.
TlIO3(s) ↔ Tl+(aq) + IO3-(aq)
With common ion (from previous calculation)
s = [TlIO3] = 6.2 × 10-5 mol L-1 = molar solubility
With common ion, s = [TlIO3]= 6.2 × 10-5 mol L-1
[s] [s] [s]
What if no common ion is added? i.e., dissolve thallium iodate in pure water
s= 1.76x10-3 mol L-1 = molar solubility
[Tl+][IO3-] = Ksp
04/22/23 Zumdahl Chapter 8 46
Chapter 8Applications of Aqueous Equilibria
8.1 Solutions of Acids or Bases Containing a Common Ion
8.2 Buffered Solutions 8.3 Exact Treatment of Buffered Solutions (skip)8.4 Buffer Capacity 8.5 Titrations and pH Curves 8.6 Acid-Base Indicators 8.7 Titration of Polyprotic Acids (skip)8.8 Solubility Equilibria and the Solubility Product 8.9 Precipitation and Qualitative Analysis (skip)8.10 Complex Ion Equilibria (skip)