lect w9 152 - buffers and ksp_alg
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General Chemistry IIGeneral Chemistry IICHEM 152 Unit 2CHEM 152 Unit 2
Week 9
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Week 9 Reading Assignment
Chapter 16 – Sections 16.2 (buffers), 16.3 (buffers), 16.4 (titrations), 16.5 (Ksp)
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Equilibrium in Equilibrium in Aqueous Solutions Aqueous Solutions
Acids and BasesAcids and Bases
We have explored the properties of acids and bases as we add them
independently to water. But what
happens when you mix them?
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TitrationsTitrations
You have covered the topic of mixing acids and bases for
titrations in lab. We will assume that
you understand this topic and cover
some other topics of mixing acids and
bases.
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What is the effect on the pH of adding What is the effect on the pH of adding NHNH44Cl to 0.25 M NHCl to 0.25 M NH33(aq)?(aq)?
NHNH33(aq) + H(aq) + H22O O NH NH44++(aq) + OH(aq) + OH--(aq)(aq)
Here we are adding an ion Here we are adding an ion COMMONCOMMON to the to the equilibrium (equilibrium (COMMON ION EFFECTCOMMON ION EFFECT))
Let us first calculate the pH of a Let us first calculate the pH of a 0.25 M NH0.25 M NH33 solution: solution:
[NH[NH33]] [NH[NH44++]] [OH[OH--]]
initialinitial 0.250.25 0 0 00changechange -x-x +x +x +x +xequilibequilib 0.25 - x0.25 - x x x xx
Mixing Conjugate PairsMixing Conjugate Pairs
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NHNH33(aq) + H(aq) + H22O O NH NH44++(aq) + OH(aq) + OH--(aq)(aq)
Kb 1.8 x 10-5 = [NH4
+][OH- ][NH3 ]
= x2
0.25 - xKb 1.8 x 10-5 =
[NH4+][OH- ]
[NH3 ] =
x2
0.25 - x
pH of Aqueous NHpH of Aqueous NH33
Assuming x is << 0.25, we haveAssuming x is << 0.25, we have
[OH[OH--] = x = [K] = x = [Kbb(0.25)](0.25)]1/21/2 = 0.0021 M = 0.0021 M
This gives pOH = 2.67This gives pOH = 2.67
and so pH = 14.00 - 2.67and so pH = 14.00 - 2.67
= = 11.3311.33 for 0.25 M NH for 0.25 M NH33
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What is the pH of a solution with 0.25 M What is the pH of a solution with 0.25 M NHNH44Cl and 0.25 M NHCl and 0.25 M NH33(aq)?(aq)?
NHNH33(aq) + H(aq) + H22O O NHNH44++(aq) + OH(aq) + OH--(aq)(aq)
We expect that the pH will decline on We expect that the pH will decline on adding NHadding NH44Cl. Let’s test that!Cl. Let’s test that!
pH of NHpH of NH33/NH/NH44++ Mixture Mixture
[NH[NH33]] [NH[NH44++]] [OH[OH--]]
initialinitial 0.250.25 0.250.25 00
changechange -x-x +x+x +x+x
equilibequilib 0.25 - x0.25 - x 0.25 + x 0.25 + x xx
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Kb 1.8 x 10-5 = [NH4
+][OH- ][NH3 ]
= x(0.25 + x)
0.25 - x
pH of NHpH of NH33/NH/NH44++ Mixture Mixture
[[OHOH--] = x = (0.25 / 0.25) K] = x = (0.25 / 0.25) Kbb = 1.8 x 10 = 1.8 x 10-5-5
MM
This gives pOH = 4.74 and pH = 9.26This gives pOH = 4.74 and pH = 9.26
pH drops from 11.33 to 9.26 on pH drops from 11.33 to 9.26 on
adding a common ion.adding a common ion.
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pH of NHpH of NH33/NH/NH44++ Mixture Mixture
What is the pH when 1.00 mL of 1.00 M HCl is What is the pH when 1.00 mL of 1.00 M HCl is added to 1.00 L of a mixture that has [NHadded to 1.00 L of a mixture that has [NH33]= ]=
0.25 M and [NH0.25 M and [NH44++]= 0.25 M ]= 0.25 M
(pH = 9.26)(pH = 9.26)
HH33OO++ + NH + NH33 NH NH44++ + H + H22OO
The reaction occurs completely because K is The reaction occurs completely because K is large.large.
HH33OO++ + NH + NH33 NH NH44++ + H + H22OO
Before rxnBefore rxn 0.001 0.250.001 0.25 0.250.25
Change -0.001 –0.001 +0.001Change -0.001 –0.001 +0.001
After rxn ~0 0.249 +0.251After rxn ~0 0.249 +0.251
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pH of NHpH of NH33/NH/NH44++ Mixture Mixture
Equilibrium will be re-established:
Ka=5.6 x 10-10 NH4+ + H2O H3O+ + NH3
Equilibrium 0.251-x x 0.249+x
[H3O ] [NH4+]
[NH3]
• Ka 0.2510.249
• (5.6 x 10-10 )
[H[H33OO++] = 5.64 x 10] = 5.64 x 10-10-10 M M pH = 9.25pH = 9.25
The pH has not changed much (pH = 9.26The pH has not changed much (pH = 9.26 to 9.25) on adding HCl to the mixture! on adding HCl to the mixture!
Why?Why? What would happen if we add NaOH?What would happen if we add NaOH?
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Formation of a BufferFormation of a Buffer
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The The NHNH33/NH/NH44++ mixture is a typical example mixture is a typical example
of a of a BUFFERBUFFER. A buffer is a solution that . A buffer is a solution that resists changes in the pH when limited resists changes in the pH when limited
amounts of base or acid are added to it.amounts of base or acid are added to it.
Buffer CompositionBuffer CompositionWeak AcidWeak Acid + Conjugate Base+ Conjugate Base
CHCH33COOHCOOH ++ CH CH33COOCOO
HH22POPO44 ++ HPO HPO44
22
Weak BaseWeak Base ++ Conjugate AcidConjugate Acid
NHNH33 ++ NH NH44++
Buffer SolutionsBuffer Solutions
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H+? OH-?
Action of a BufferAction of a Buffer
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[CH[CH33COOH] + HCOOH] + H22O O [CH [CH33COOCOO--]+ [H]+ [H33OO++]]initialinitialchangechangeEquilibEquilib
What is the pH of a buffer that has [CHWhat is the pH of a buffer that has [CH33COOH] COOH] = 0.700 M and [CH= 0.700 M and [CH33COOCOO--] = 0.600 M?] = 0.600 M?
CHCH33COOH + HCOOH + H22O O CH CH33COOCOO-- + H + H33OO++
KKaa = 1.8 x 10 = 1.8 x 10-5-5
Buffer SolutionsBuffer Solutions
0.7000.700 0.6000.600 00
-x-x +x+x +x+x
0.700 - x0.700 - x 0.600 + x0.600 + x xx
Ka 1.8 x 10-5 = [H3O+](0.600)
0.700 Ka 1.8 x 10-5 =
[H3O+](0.600)0.700
[H[H33OO++] = 2.1 x 10] = 2.1 x 10--
55 pH = 4.68pH = 4.68
Assuming that Assuming that x << 0.700 and 0.600x << 0.700 and 0.600, we , we havehave
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Notice that the expression for calculating Notice that the expression for calculating the Hthe H33OO++ concentration of the buffer is concentration of the buffer is
Buffer SolutionsBuffer Solutions
[H3O] [Acid]
[Conj. base]•Ka[H3O]
[Acid][Conj. base]
•Ka
Take the Take the negative lognegative log of both sides of of both sides of this equationthis equation
pH pKa + log[Conj. base]
[Acid]pH pKa + log
[Conj. base][Acid]
pH pKa - log[Acid]
[Conj. base]
Henderson-Henderson-Hasselbalch Hasselbalch
EquationEquation
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Preparing a BufferPreparing a Buffer
Imagine that you want to prepare a Imagine that you want to prepare a buffer solution with a pH = 4.30. buffer solution with a pH = 4.30.
It is best to choose an acid such that It is best to choose an acid such that [H[H33OO++] is about equal to K] is about equal to Kaa (or pH ≈ (or pH ≈
pKpKaa).).
Then you get the exact pH by adjusting Then you get the exact pH by adjusting the ratio of acid to conjugate base.the ratio of acid to conjugate base.
pH pKa + log[Conj. base]
[Acid]pH pKa + log
[Conj. base][Acid]
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You want to buffer a solution at You want to buffer a solution at pH = pH = 4.304.30 and have access to the following and have access to the following
acids:acids:
POSSIBLE ACIDSPOSSIBLE ACIDS KKaa
HSOHSO44- - / SO/ SO44
2-2- 1.2 x 101.2 x 10-2-2
CHCH33COOH/CHCOOH/CH33COOCOO--1.8 x 101.8 x 10-5-5
HCN / CNHCN / CN-- 4.0 x 104.0 x 10-10-10
Preparing a BufferPreparing a Buffer
Which one would you choose? If you have a 0.100 M
concentration of the base, what concentration of the acid would
you need?
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If you have a 0.100 M concentration of the base, what concentration of the acid would you need?
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Your TurnThe blood of mammals is an aqueous solution that maintains a constant pH. The normal pH
of human blood is 7.40. Carbon dioxide provides the most important blood buffer. In
solution, CO2, reacts with water to form H2CO3, which ionizes to produce H3O+ and
HCO3- ions:
CO2(g) + H2O(l) H2CO3(aq)
H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-(aq)
Ka= 4.2 x 10-7Use this information to determine the concentration ratio [HCO3
-]/[H2CO3] in blood.
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Determine the concentration ratio [HCO3
-]/[H2CO3] in blood
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If the pH of human blood decreases below 7.35, a condition known as acidosis occurs;
increasing the pH above 7.45 causes alkalosis. Both of these conditions can be
life-threatening.
CO2(g) + H2O(l) H2CO3(aq)
H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-(aq)
Ka= 4.2 x 10-7
What is an easy way to treat alkalosis?
(Apply Le Chatelier’s Principle)
Your Turn
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More with Henderson-More with Henderson-HasselbalchHasselbalch
CH2
H2N C
O
OH
pKpKaa(carb. acid) = 2.0(carb. acid) = 2.0
pKpKaa(prot. amine) = (prot. amine) = 10.510.5
Draw the structure of the amino acid at Draw the structure of the amino acid at
a)a)pH = 1.0, b) pH = 7.0, and c) pH = 12.0pH = 1.0, b) pH = 7.0, and c) pH = 12.0
Start by using the Henderson-Hasselbalch Start by using the Henderson-Hasselbalch equation!equation!
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Functional group structuresFunctional group structuresAcidic formsAcidic forms Basic formsBasic forms
C
O
OH
C
O
O
N
H
H
N
H
H
H
Which version of each functional group should Which version of each functional group should exist at a particular pH?exist at a particular pH?
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@pH = 1.0@pH = 1.0
The acid form (-COOH) is more prevalent. The acid form (-COOH) is more prevalent.
For the carboxylic acid group:For the carboxylic acid group:
[acid][base]
log2.01.0 [acid][base]
0.10
The acid form (-NHThe acid form (-NH33++) is more prevalent. ) is more prevalent.
For the amine group:For the amine group:
[acid][base]
log10.51.0 [acid][base]
103.2 10
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@pH = 1.0@pH = 1.0
CH2
H3N C
O
OH
Now – in your groups – Draw the prevalent Now – in your groups – Draw the prevalent structures of the amino acid at pH=7 and at structures of the amino acid at pH=7 and at
pH = 12.pH = 12.
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What ishappenin
ghere?
solution
solution
Solubility
Equilibria
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Solubility of Salts Solubility of Salts
Not all salts are completely soluble in water. Many of them only dissolve to a
small extent.
When equilibrium has been established, no
more AgCl dissolves and the solution is
SATURATED.
When an insoluble salt is placed in water, chemical
equilibrium can be established:
AgCl(s) Ag+(aq) + Cl-(aq)
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Solubility of Salts
Consider: AgCl(s) Ag+(aq) + Cl-(aq)
Write the expression of the equilibrium constant Ksp (solubility product)
associated with this process.
Ksp = [Ag+][Cl-]/ [AgCl(s)]
Ksp = [Ag+][Cl-]
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Solubility of Salts
Consider: AgCl(s) Ag+(aq) + Cl-(aq)
The concentration of [Ag+] in a saturated solution of AgCl is
1.34x10-5 M. This number represents the SOLUBILITY of the salt.
What is the concentration of Cl¯ ions in the system?
What is the value of Ksp for this salt?
Ksp =[Ag+][Cl-] = (1.34 x10-5)2 = 1.80 x 10-10
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Estimate the solubility of the following salts:
PbSO4
Your Turn
Ksp= 1.8 x10-8
Ksp= (x)(x)
x = 1.3 x 10-4 M
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Determine the molar solubility of PbCl2? Ksp = 1.7 x 10-5
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Estimate the solubility of the following salt:
Your Turn
PbCl2
Ksp= 1.7 x 10-5
Ksp= (x)(2x)2
x =1.6 x 10-2 M
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PbIPbI22(s) (s) Pb Pb2+2+(aq) + 2I(aq) + 2I--(aq)(aq)
Calculate KCalculate Kspsp for PbI for PbI22 if if
[Pb[Pb2+2+ ] = 0.00130 M in a saturated ] = 0.00130 M in a saturated solution.solution.
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Consider PbIConsider PbI22 dissolving dissolving
in water in water
PbIPbI22(s) (s) Pb Pb2+2+(aq) + 2I(aq) + 2I--
(aq)(aq)
Calculate KCalculate Kspsp for PbI for PbI22 if if
[Pb[Pb2+2+ ] = 0.00130 M ] = 0.00130 M
PbIPbI22(s) (s) Pb Pb2+2+(aq) + 2I(aq) + 2I--
(aq)(aq)
KKspsp =[0.00130][0.00260] =[0.00130][0.00260]2 2 = =
8.79 x108.79 x10-9-9
0.00130 0.00260
Your Turn
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Solubility of Salts Consider the following insoluble salt:
CaCO3(s)
Will the solubility of this salt increase, decrease, or remain the same after
adding:
a)Sulfuric acid (H2SO4)
b)Hydrochloric acid (HCl)
c) Calcium Chloride (CaCl2)
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Solubility of Salts
Consider the following insoluble salt:
AgCl(s)
Will the solubility of these salts increase, decrease, or remain the
same after adding:
a)Sulfuric acid (H2SO4)
b)Hydrochloric acid (HCl)
c) Calcium Chloride (CaCl2)
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The Common Ion EffectThe Common Ion EffectAdding an ion “common” to an equilibrium causes the
equilibrium to shift back to reactant.
AgCH3CO2(s) Ag+(aq) + CH3CO2-(aq)
AgNO3(aq) Ag+(aq) + NO3-(aq)
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Kidney StonesKidney stones are normally
composed of: calcium oxalate (CaC2O4), calcium
phosphate (Ca3(PO4)2), magnesium ammonium phosphate (MgNH4PO4).Precipitate formed from calcium ions
from food rich in calcium, dairy products, and oxalate ions from chocolate, spinach,
celery, black tea:Ca+2 + C2O4
2- CaC2O4 How would you try to dissolve the
CaC2O4 stones?
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Precipitating an Insoluble SaltPrecipitating an Insoluble SaltConsider the following problem:
Imagine you mix 500.0 mL of a solution of AgNO3 1.5 x 10-5 M with 500.0 mL of a solution of NaCl 1.5 x 10-5 M. Will AgCl
precipitate?AgCl(s) Ag+(aq) + Cl-(aq) Ksp= 1.8 x 10-10
Reactant Quotient, Q=[Ag+]o[Cl-]o
Q < Ksp - no precipitate forms
unsaturated solution
Q > Ksp - precipitate may form supersaturated solution
Q = Ksp - no precipitate forms saturated solution
DID YOU CONSIDER DILUTION?
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Precipitating an Insoluble SaltPrecipitating an Insoluble Salt
Consider the case: Consider the case:
HgHg22ClCl22(s) (s) Hg Hg222+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)
KKspsp = 1.1 x 10 = 1.1 x 10-18-18 = [Hg = [Hg222+2+] [Cl] [Cl--]]22
If [HgIf [Hg222+2+]=0.010 M, what [Cl]=0.010 M, what [Cl--] is required to ] is required to
just begin the precipitation of Hgjust begin the precipitation of Hg22ClCl22??
That is, what is the maximum [ClThat is, what is the maximum [Cl--] that can be ] that can be
in solution with 0.010 M Hgin solution with 0.010 M Hg222+2+ without without
forming Hgforming Hg22ClCl22??KKspsp = 1.1 x 10 = 1.1 x 10-18-18 = [Hg = [Hg22+2+] [Cl] [Cl--]]22 = [0.010][Cl = [0.010][Cl--]]2 2
[Cl[Cl--]] = 1.0 x 10 = 1.0 x 10-8 M
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Separating Salts by Differences in Separating Salts by Differences in KKspsp
If two or more substances can precipitate If two or more substances can precipitate in a solution, the substance whose in a solution, the substance whose KKspsp is is
first exceeded precipitates first. first exceeded precipitates first.
Consider a solution containing 0.020 Consider a solution containing 0.020 M ClM Cl-- and 0.010 M CrO and 0.010 M CrO44
2-2- ions in which ions in which Ag+ ions are added slowly. Ag+ ions are added slowly.
Which precipitates first, AgCl or Which precipitates first, AgCl or AgAg22CrOCrO44??
KKspsp for Ag for Ag22CrOCrO44 = 9.0 x 10 = 9.0 x 10-12-12
KKspsp for AgClfor AgCl = 1.8 x 10= 1.8 x 10-10-10
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Separating Salts by Differences in Separating Salts by Differences in KKspsp
Consider a solution containing 0.020 Consider a solution containing 0.020 M ClM Cl-- and 0.010 M CrO and 0.010 M CrO44
2-2- ions in which ions in which Ag+ ions are added slowly. Ag+ ions are added slowly.
KKspsp for Ag for Ag22CrOCrO44 = 9.0 x 10 = 9.0 x 10-12-12
KKspsp for AgClfor AgCl = 1.8 x 10= 1.8 x 10-10-10
What is [AgWhat is [Ag++] when the maximum AgCl ] when the maximum AgCl is precipitated, but none of the is precipitated, but none of the
AgAg22CrOCrO44 has precipitated? has precipitated?
What percent of ClWhat percent of Cl-- remains remains
in solution?in solution?
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Separating Salts by Differences in Separating Salts by Differences in KKspsp
If
PbCl2(s) Pb2+ (aq) + 2Cl-(aq) Ksp=1.7 x 10-5
PbCrO4(s) Pb2+(aq) + CrO42-(aq)
Ksp=1.8 x 10-4
What is the equilibrium constant for the reaction:
PbCl2 + CrO42- PbCrO4 + 2Cl-?
What does this tell you?
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1. What is the pH of a solution of 0.150 M HIO3 (Ka = 0.17)
and 0.350 M NaIO3?
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2. What is the maximum mass of calcium fluoride that will
dissolve in 500. mL of aqueous solution?Ksp = 1.46 x 10-10 M = 78.08 g/mol
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3. What is the molar solubility of CaF2 in a solution of 0.10
M NaF?Ksp = 1.46 x 10-10