1
Buffer Solutions
pH of solution adding 0.10 M HCl to 100 mL water
HCl added pH
0 mL 7.00
2 mL 2.71
5 mL 2.32
10 mL 2.04
20 mL 1.78
50 mL 1.48
mL of 0.10 M HCl added
0 10 20 30 40 501
2
3
4
5
6
7
Buffer Solutions
A buffer helps a solution maintain its pH whenacid or base is added
A buffer must contain two components to work
a weak acid that reacts with added base
a weak base that reacts with added acid
Buffers usually contain approximately equalamounts of a weak acid and its conjugate base
2
Buffer Solutions
Solution that is 0.100 M CH3COOH (acetic acid)and 0.100 M NaCH3COO (sodium acetate)
Find pH of buffer solution:CH3COOH(aq) + H2O CH3COO-(aq) + H3O
+(aq)
[CH3COOH] [CH3COO-] [H3O+]
initial 0.100 0.100 0
-x x x
equil 0.100 – x 0.100 + x x
Buffer Solutions
Find pH of buffer solution:CH3COOH(aq) + H2O CH3COO-(aq) + H3O
+(aq)
Ka = [CH3COO-][H3O
+]
[CH3COOH] =
(.100 + x)x
(.100 - x) = 1.8 x 10-5
x = 1.80 x 10-5 M
pH = 4.74
assume x is negligiblecompared to .100 M
3
Buffer Solutions
Add 5 mL .10 M HCl
Find pH of resulting solution
Assume all acid added reacts with acetate ion to formacetic acid (remember that acids react with bases)
CH3COOH(aq) + H2O CH3COO-(aq) + H3O+(aq)
[H3O+] added = (5 mL)(.10 M)/(105 mL) = 4.76x10-3 M
[CH3COOH] = (0.100)(100/105) + 0.005 = 0.100 M
[CH3COO-] = (0.100)(100/105) - 0.005 = 0.090 M
(100/105)=dilution factor for addition of HCl
Buffer Solutions
Now let solution come to equilibrium
[CH3COOH] [CH3COO-] [H3O+]
initial .100 .090 0
-x x x
equil .100 – x .090 + x x
Ka = [CH3COO-][H3O
+]
[CH3COOH] =
(.090 + x)x
(.100 - x) = 1.8 x 10-5
x = 2.00 x 10-5 M
pH = 4.70
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Buffer Solutions
pH of buffered solution adding 0.10 M HCl to 100mL soln
HCl added pH
0 mL 4.74
5 mL 4.70
10 mL 4.66
15 mL 4.61
25 mL 4.52
50 mL 4.28mL of 0.10 M HCl added
0 10 20 30 40 501
2
3
4
5
6
7
Buffered solution
Buffer Solutions
Henderson-Hasselbach Equation
Allows calculation of pH of a buffer ifconcentrations of conjugate acid and conjugatebase are known
HA(aq) + H2O H3O+(aq) + A-(aq)
Ka = [H3O
+][A-][HA]
[H3O+] =
Ka[HA]
[A-]
5
Buffer Solutions
Take -log of both sides
log[H3O+] = - log
Ka[HA]
[A-]
= - log Ka( ) - log
[HA]
[A-]
-log(Ka) = pKa log[HA]
[A ]
= log
[A-]
[HA]
pH = pKa + log[A-]
[HA]
Henderson-Hasselbach Eqn
Buffer Solutions
Using the Henderson-Hasselbach Eqn, we can:
Determine pH of a solution
Determine ratio of conjugate base toconjugate acid to achieve specific pH
pH = pKa + log[A-]
[HA]
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Buffer Solutions
Let’s go back to problem of adding HCl to buffersolution:
We can use H-H eqn. to make the calculationsmuch easier
[CH3COOH] = 0.100 + [HCl]added
[CH3COO-] = 0.100 – [HCl]added
pH = pKa + log[A-]
[HA]
Buffer Solutions
VHCl [HCl] [CH3COOH] [CH3COO-] pH
5mL (.1)(5mL)/105mL .095+.005 .095-.005
= .00476 M =.100 M =.090 M 4.70
10mL (.1)(10)/110 .091+.009 .091-.009
= .00909 M =.100 M =.082 M 4.66
25mL (.1)(25)/125 .080+.020 .080-.020
= .0200 M =.100 M =.060 M 4.52
50mL (.1)(50)/150 .067+.033 .067-.033
= .0333 M =.100 M =.034 M 4.28
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Buffer Solutions
Buffer Capacity—the amount of acid or base thatcan be added to a buffer without the pHsignificantly changing
Suppose we acid to a buffer solution:
The acid will react with the conjugate base until it isdepleted
Past this point, the solution behaves as if no bufferwere present
Acid-Base Titrations
A titration is a method used to determine theconcentration of an unknown species
Add a measured amount of a known reactant
Determine when the reaction has gone tocompletion
[unknown] + [known] products
At the equivalence point
moles unknown = moles known added
CunknownVunknown = CknownVknown
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Acid-Base Titrations
Titrate 50.00mL unknownHCl soln.with 0.2137M NaOH
mL NaOH added
0 5 10 15 20 25 30 35 40
1
2
3
4
5
6
7
8
9
10
equivalence point
Acid-Base Titrations
At equivalence point, VNaOH
= 23.96 mL
mol(NaOH) =
(.2137 M)(.02396 L)
= 5.120 x 10-3 mol
mol(HCl) = 5.120 x 10-3 mol
(mol known = mol unknown)
[HCl] =
(5.120x10-3 mol)/(.05000 L)
= 0.1024 M
mL NaOH added
0 5 10 15 20 25 30 35 40
1
2
3
4
5
6
7
8
9
10
equivalence point
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Indicators
An indicator is a chemical species that changescolor depending on the pH of the solution
An indicator is a conjugate acid-conjugate basepair in which the acid and base forms of thecompound have different colors
HIn(aq) + H2O In-(aq) + H3O+(aq)
color 1 color 2
Indicators are used to determine the endpointof a titration
Indicators
The pKa of the indicator determines the pHrange over which the color changes
[HIn]/[In-] 10 acid color
[HIn]/[In-] 0.1 base color
[HIn]/[In-] 1 intermediate color
Remember: pH = pKa + log{[In-]/[HIn]}
If [HIn]/[In-] = 1, log{[HIn]/[In-]} = 0
pH = pKa at point when indicator is
changing color
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Indicators
Indicator pKa pH range color change
Methyl orange 3.7 3.1 – 4.4 red to yellow
Bromophenol blue 4.0 3.0 – 4.6 yellow to blue
Methyl red 5.1 4.2 – 6.3 red to yellow
Bromothymol blue 7.0 6.0 – 7.6 yellow to blue
Phenol red 7.9 6.8 – 8.4 yellow to red
Phenolphthalein 9.3 8.2 – 10.0 clear to pink
Indicators
Figure 17.5: pH curve fortitration of 0.100 M HCl with0.100 M NaOH
change before endpoint
change after endpoint
changearound
endpoint
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Indicators
Titration of weak acid with strong base
HA(aq) + OH-(aq) A-(aq) + H2O
At equivalence point
A-(aq) + H2O HA(aq) + OH-(aq)
the solution is basic because conjugate base ofweak acid reacts with water to form OH-(aq)
Indicators
Titrate 25.00 mL 0.100 M formic acid (HCOOH) with0.100 M NaOH
Ka = 1.8 x 10-4
Find pH at equivalence point and select appropriateindicator
At equivalence point, mol(HCOOH) = mol(OH-)
mol(fa) = (0.100 M fa)(0.02500 L) = 2.5 x 10-3 mol
fa = formic acid
VNaOH added = (2.5 x 10-3 mol)/(0.100 M) = 25.0 mL
Vtotal = 50.0 mL
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Indicators
Assume HCOOH + OH- reaction goes to completion:
[HCOO-] = (2.5 x 10-3 mol)/(0.0500 L) = 0.0500 M
Determine Keq for reaction of formate ion:
HCOO-(aq) + H2O HCOOH(aq) + OH-(aq)
Keq = [HCOOH][OH-]
[HCOO ] =
Kw
Ka
= 6.67 x 10-11
Indicators
[HCOO-] [HCOOH] [OH-]
initial .0500 0 0
-x x x
equil .0500 – x x x
Keq = [HCOOH][OH-]
[HCOO ] =
Kw
Ka
= 6.67 x 10-11
Keq = [HCOOH][OH-]
[HCOO-] =
x2
0.0500 x = 6.67 x 10-11
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Indicators
x = 1.83 x 10-6 M = [OH-]
pOH = -log(1.83 x 10-6) =5.74
pH = 14.00 – 5.74 = 8.26
Phenol red (6.8 – 8.4) or phenolphthalein (8.2 – 10.0)would be appropriate indicators
Keq = [HCOOH][OH-]
[HCOO-] =
x2
0.0500 x = 6.67 x 10-11
Indicators
Titration of weak base with strong acid
B(aq) + H3O+(aq) BH+(aq) + H2O
At equivalence point
BH+(aq) + H2O B(aq) + H3O+(aq)
the solution is acidic because conjugate acid ofweak base reacts with water to form H3O
+(aq)
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Indicators
Figure 17.8: titration of 0.100 MNH3 with 0.100 M HCl
Acid Rain
Carbon dioxide in the air is in equilibrium withH2O in atmospheric water droplets (clouds &fog):
CO2(aq) + H2O H2CO3(aq)
carbonic acid Ka = 4.2 x 10-7
H2CO3(aq) + H2O H3O+(aq) + HCO3
-(aq)
Natural rain water has pH = 5.6
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Acid Rain
Emitted pollutants can form additional acidsources in clouds:
NO2:
2 NO2(aq) + H2O HNO3(aq) + HNO2(aq)
nitric acid nitrous acid
strong Ka = 4.5 x 10-4
Acid Rain
Emitted pollutants can form additional acidsources in clouds:
SO2:
SO2(aq) + H2O H2SO3(aq)
sulfurous acid Ka = 1.2 x 10-2
2 SO2(g) + O2(g) 2 SO3(g)
SO3(aq) + H2O H2SO4(aq)
sulfuric acid
strong
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Acid Rain
Solubility Products
Many salts are only slightly soluble
The solubility product is a measure of theconcentration of ions in a solution saturated withthe saltMA(s) M+(aq) + A-(aq) Ksp = [M+][A-]
Examples
AgCl(s) Ag+(aq) + Cl-(aq) Ksp=[Ag+][Cl-]=1.8x10-10
PbCl2(s) Pb2+(aq) + 2 Cl-(aq) Ksp=[Pb2+][Cl-]2=1.7x10-5
AuBr3(s) Au3+(aq) + 3 Br-(aq) Ksp=[Au3+][Br-]3=4.0x10-36
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Solubility Products
Knowing the Ksp, we can calculate theconcentration of ions in solution
Examples
AgCl(s) Ag+(aq) + Cl-(aq) Ksp=[Ag+][Cl-]=1.8x10-10
-x x x
x2 = 1.8 x 10-10 x = 1.3 x 10-5 M = [Ag+] = [Cl-]
Solubility Products
Examples
PbCl2(s) Pb2+(aq) + 2 Cl-(aq) Ksp=[Pb2+][Cl-]2=1.7x10-5
-x x 2x
x(2x)2 = 1.7 x 10-5
4x3 = 1.7 x 10-5 x = 1.6 x 10-2 M
[Pb2+] = 1.6 x 10-2 M
[Cl-] = 3.2 x 10-2 M
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Solubility Products
Examples
AuBr3(s) Au3+(aq) + 3 Br-(aq) Ksp=[Au3+][Br-]3=4.0x10-36
-x x 3x
x(3x)3 = 4.0 x 10-36
27x4 = 4.0 x 10-36 x = 6.2 x 10-10 M
[Au3+] = 6.2 x 10-10 M
[Br-] = 1.9 x 10-9 M
Solubility Products
Examples—Common ion effect
How much PbI2 will dissolve in a 0.0100 M solution of NaI?
PbI2(s) Pb2+(aq) + 2 I-(aq) Ksp = 8.7 x 10-9
-x x 2x + .0100
x(2x+.0100)2 = 8.7 x 10-9
x(4x2 + 0.0200x + 1.0x10-4) = 8.7 x 10-9
4x3 + .0200x2 + 1.0x10-4x – 8.7x10-9 = 0
x = 8.6 x 10-5 M
vs 1.3 x 10-3 M if no I-(aq) were present initially
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Factors Affecting Solubility
Salts that are slightly soluble in water can bemuch more soluble in acid if one or both of itsions are moderately basic:CaCO3(s) Ca2+(aq) + CO3
2-(aq) Ksp = 8.7 x 10-9
But CO32-(aq) is the conjugate base of HCO3
-(aq)CO3
2-(aq) + H3O+(aq) HCO3
-(aq) + H2O Kb = 2.1x10-4
HCO3-(aq) + H3O
+(aq) H2CO3(aq) + H2O Kb = 2.4x10-8
H2CO3(aq) H2O + CO2(g) Keq 105
Works for carbonates, some sulfides, phosphates,
etc. (species that behave as bases [no too weak])
Precipitation
Define ion quotient, Q, as:
MxAy(s) x My+(aq) + y Ax-(aq)
Q = [My+]x[Ax-]y
A precipitate will form only when Q exceeds Ksp
Q < Ksp: solution is unsaturated—no precipitate
Q > Ksp: solution is saturated—precipitate forms
Q = Ksp: solution at saturation point