lect w9 buffers_exercises
TRANSCRIPT
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Key to ProblemsKey to Problems
Week 9
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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
0.7000.700 0.6000.600 00
-x-x +x+x +x+x
0.700 - x0.700 - x 0.600 + x0.600 + x xx
Assuming that Assuming that x << 0.700 and 0.600x << 0.700 and 0.600, we , we can use the Henderson-Hasselbalch can use the Henderson-Hasselbalch EquationEquation
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pH = pKpH = pKaa + log + log[Base][Base] [Acid][Acid]
pKpKaa = -logK = -logKaa pKpKaa = -log(1.8 x 10 = -log(1.8 x 10--
55))
pKpKaa = 4.74 = 4.74[Base] = 0.600 M; [Acid] = 0.700 M[Base] = 0.600 M; [Acid] = 0.700 M
pH = 4.74 + log[0.600]pH = 4.74 + log[0.600] [0.700][0.700]
pH = 4.67pH = 4.67
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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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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 pKpKaa
HSOHSO44- - / SO/ SO44
2-2- 1.2 x 101.2 x 10-2-21.921.92
CHCH33COOH/CHCOOH/CH33COOCOO-- 1.8 x 101.8 x 10-5-54.744.74
HCN / CNHCN / CN-- 4.0 x 104.0 x 10-10-10 9.409.40
The acetate buffer would be chosen due The acetate buffer would be chosen due to the proximity of the pKa to the to the proximity of the pKa to the desired buffer pH.desired buffer pH.
Preparing a BufferPreparing a Buffer
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You want to buffer a solution at You want to buffer a solution at pH = pH = 4.304.30 If you have a 0.100 M
concentration of the base, what concentration of the acid would you
need?
pH = pKpH = pKaa + log + log[Base][Base] [Acid][Acid]
pH = 4.30; pKpH = 4.30; pKaa = 4.74 = 4.74[Base] = 0.100 M; [Acid] = ?[Base] = 0.100 M; [Acid] = ?
4.30 = 4.74 + log4.30 = 4.74 + log[0.100][0.100] [HA][HA]
-0.44 = log-0.44 = log[0.100][0.100] 10 10-0.44-0.44 = = [0.100][0.100] [HA][HA] [HA] [HA]
[HA] = 0.28 M[HA] = 0.28 M
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The 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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H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-(aq)
Ka= 4.2 x 10-7
Use this information to determine the concentration ratio [HCO3
-]/[H2CO3] in blood (pH = 7.40).
pH = pKpH = pKaa + log + log[Base][Base] [Acid][Acid]
pH = 7.40; pKpH = 7.40; pKaa = 6.38 = 6.38
[Base]/[Acid] = ?[Base]/[Acid] = ?
7.40 = 6.38 + log7.40 = 6.38 + log[Base][Base][Acid][Acid]
1.02 = log1.02 = log[[HCO3-]] 10 101.021.02 = = [[HCO3
-]] [[H2CO3]] [ [H2CO3]]
[[HCO3-] = 10.5] = 10.5
[[H2CO3]]
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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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@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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C
CH2
O
OHH3N
@pH = 1.0@pH = 1.0
: :
::
+C
CH2
O
OH3N
: :
::
+C
CH2
O
OH2N
: :
::
@pH = 7.0@pH = 7.0 @pH = 12.0@pH = 12.0
pKpKaa(carb. acid) = 2.0(carb. acid) = 2.0
pKpKaa(prot. amine) = 10.5(prot. amine) = 10.5
Key hints:Key hints:
-at low pH’s, the overall molecules will be + or -at low pH’s, the overall molecules will be + or neutralneutral
-at high pH’s, the overall charge on the -at high pH’s, the overall charge on the molecules will be neutral or -molecules will be neutral or -
:-
:-
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C
CH2
O
OHH3N
@pH = 1.0@pH = 1.0
: :
::
+C
CH2
O
OH3N
: :
::
+C
CH2
O
OH2N
: :
::
@pH = 7.0@pH = 7.0 @pH = 12.0@pH = 12.0
:-
:-
pH = 1.0pH = 1.0 pH = 7.0pH = 7.0 pH = 12.0pH = 12.0
pKpKaa = 2.0 = 2.0
-COOH -COO-
pKpKaa = 10.2 = 10.2
-NH3+ -NH2
-COO-
-NH3+
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Solubility of Salts
Consider: AgCl(s) Ag+(aq) + Cl-(aq)
In a saturated solution [Ag+] =1.34 10-5 M.
What is the concentration of Cl¯ ions in the system?
[Cl-] = [Ag+] = 1.34 10-5 M
What is the value of Ksp for this salt?
Ksp =[Ag+][Cl-] = (1.34 10-5)2 = 1.80 x 10-
10
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Estimate the solubility of the following salt:
PbSO4 (Ksp= 1.8 10-8)
PbSO4 Pb2+ + SO42-
I solid 0 0
C -x +x +x
E solid x x
Ksp= [Pb2+][SO42-]
1.8 10-8 = [x][x]
solubility = x = 1.3 x 10-4 M
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Determine the molar solubility of PbCl2? Ksp = 1.7 x 10-5
PbCl2 Pb2+ + 2 Cl- I solid 0 0
C -x +x +2x
E solid x 2x
Ksp= [Pb2+][Cl-]2
1.7 10-5 = [x][2x]2
solubility = x = 0.016 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 [Pb if [Pb2+2+ ] = 0.00130 ] = 0.00130
M in a saturated solution.M in a saturated solution.
PbI2 Pb2+ + 2 I- I solid 0 0
C -x +x +2x
E solid x 2x
Ksp= [Pb2+][I-]2
Ksp = [0.00130][2(0.00130)]2
Ksp = 8.79 10-9
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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)
solubility will because CO32- is basic: the
acid will react with the base, thereby removing a product, shifting the equilibrium to the right
b) Hydrochloric acid (HCl)
solubility will because CO32- is basic: the
acid will react with the base, thereby removing a product, shifting the equilibrium to the right
c) Calcium chloride (CaCl2)
solubility will because you are adding a common ion (Ca2+) that shifts the equilibrium to the left
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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)
increase; SO42- interacts with Ag+ to form an
insoluble compound, removing a product and shifting equilibrium to the right
b) Hydrochloric acid (HCl)
decrease; because Cl- is a common ion and shifts the equilibrium to the left
c) Calcium Chloride (CaCl2)
decrease; again, Cl- is a common ion and shifts the equilibrium to the left, thus reducing the solubility
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Consider 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
Ksp = [Ag+][Cl-]
First determine actual concentrations. For both:
M1V1=M2V2 M2 = (1.5 10-5 mole/L)(0.500 L)
(1.000 L)
Q = (7.5 10-6)(7.5 10-6) = 5.6 10-11
Q < K, therefore AgCl will not precipitate
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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??
At the KAt the Kspsp, there is no precipitate, but is the , there is no precipitate, but is the
threshold for when solid will form.threshold for when solid will form.
KKspsp = [Hg = [Hg22+2+][Cl][Cl--]]22
1.1 x 101.1 x 10-18-18 = [0.010][Cl = [0.010][Cl--]]2 2
[Cl[Cl--] = 1.0 x 10] = 1.0 x 10-8 M
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Consider a solution containing 0.020 M ClConsider a solution containing 0.020 M Cl-- and and 0.010 M CrO0.010 M CrO44
2-2- ions in which Ag ions in which Ag++ ions are added ions are added slowly. slowly.
Which precipitates first, AgCl or AgWhich precipitates first, AgCl or Ag22CrOCrO44??
KKspsp for Ag for Ag22CrOCrO44 = 9.0 x 10 = 9.0 x 10-12-12
KKsp sp for AgClfor AgCl = 1.8 x 10= 1.8 x 10-10-10
Solve for [AgSolve for [Ag++] ] the one with the lower the one with the lower concentration of Agconcentration of Ag++ will be the one that will be the one that
precipitates first.precipitates first.Ksp = [AgAg++]2[CrOCrO442-2-]
9.0 10-12 = [AgAg++]2[0.0100.010][Ag+] = 3.0 10-5 M
Ksp = [AgAg++][ClCl--]1.8 10-10 = [AgAg++][0.0200.020]
[Ag+] = 9.0 10-9 M
AgCl will precipitate first, because a lower concentration of silver is required to create a
precipitate with Cl-
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0.020 M Cl0.020 M Cl-- and 0.010 M CrO and 0.010 M CrO442-2- ions in which ions in which
AgAg++ ions are added ions are added
KKspsp for Ag for Ag22CrOCrO44 = 9.0 x 10 = 9.0 x 10-12-12
KKsp sp for AgClfor AgCl = 1.8 x 10= 1.8 x 10-10-10
Calculate the [AgCalculate the [Ag++] when the maximum AgCl is ] when the maximum AgCl is precipitated, but none of the Agprecipitated, but none of the Ag22CrOCrO44 has has precipitated. What percent of Clprecipitated. What percent of Cl-- remains in remains in solution?solution?
[Ag+] = 3.0 10-5 M when Ag2CrO4 will be nearly precipitated. At this point, AgCl will have precipitated, but the Ag2CrO4 will not.
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What percent of ClWhat percent of Cl-- remains in solution? remains in solution?To calculate the [Cl-] remaining, take the [Ag+] (at the point where Ag2CrO4 is nearly precipitated) and plug it into the Ksp for AgCl and solve for [Cl-].[Cl-] consumed by Ag+:Ksp = [Ag+][Cl-] 1.8 x 101.8 x 10-10-10 = [3.0 10-5][Cl-][Cl-] = 6.0 10-6 M
We know the initial [Cl-] = 0.020 M[Cl-] remaining0.020 M Cl- (initially) 6.0 10-6 M Cl-
(consumed by Ag+)
= 0.01999 moles Cl- remaining0.01999 100% = 99.95% Cl- remaining 0.020
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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?[]/Ka = 0.150/0.17 ≈ 1
the approximation doesn’t work!!HIO3 + H2O IO3
- + H3O+
I 0.150 0.350C -x +x +xE 0.150-x 0.350+x xKa = [IO3
-][H3O+] 0.17 = [0.350+x][x] [HIO3] [0.150-x]
Quadratic x2 + 0.52x – 0.255 = 0x = 0.308 = [H3O+]pH = 0.511
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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 MW = 78.08 g/mol
CaF2 Ca2+ + 2 F- Ksp = [Ca2+][F-]2
I 0 0C +x +2xE x 2x1.46 10-10 = (x)(2x)2
x = 3.32 10-4 = molar solubility of CaF2
3.32 10-4 mol CaF2 0.500 L 78.08 g CaF2 = L mol CaF2
0.0259 g CaF2 in 500 mL
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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 CaF2 Ca2+ + 2 F- Ksp = [Ca2+][F-]2
I 0 0.10C +x +2xE x 0.10+2x1.46 10-10 = (x)(0.10+2x)2
1.46 10-10 = (x)(0.10)2 (2x will be small)x = 1.46 10-8 = molar solubility
Even though the soln has F-, some CaF2 is expected to dissolve