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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 added0 10 20 30 40 50

pH

1

2

3

4

5

6

7

Buffer Solutions

n  A buffer helps a solution maintain its pH when acid or base is added

n  A buffer must contain two components to work n  a weak acid that reacts with added base n  a weak base that reacts with added acid

n  Buffers usually contain approximately equal amounts of a weak acid and its conjugate base

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 Δ -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)

5-

3

3-

3a 10 x 1.8

x)-(.100x)x(.100

]COOHCH[]O][HCOO[CH K =

+==

+

x = 1.80 x 10-5 M

pH = 4.74

assume x is negligible compared to .100 M

Buffer Solutions

Add 5 mL .10 M HCl Find pH of resulting solution

Assume all acid added reacts with acetate ion to form acetic 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 + 0.005 = 0.105 M [CH3COO-] = 0.100 - 0.005 = 0.095 M

Buffer Solutions

Now let solution come to equilibrium [CH3COOH] [CH3COO-] [H3O+]

initial .105 .095 ≈0 Δ -x x x equil .105 – x .095 + x x

5-

3

3-

3a 10 x 1.8

x)-(.105x)x(.095

]COOHCH[]O][HCOO[CH K =

+==

+

x = 1.99 x 10-5 M

pH = 4.71

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Buffer Solutions

pH of buffered solution adding 0.10 M HCl to 100 mL soln

HCl added pH 0 mL 4.74 5 mL 4.71 10 mL 4.66 15 mL 4.58 25 mL 4.57 50 mL 4.45

mL of 0.10 M HCl added0 10 20 30 40 50

pH

1

2

3

4

5

6

7

Buffered solution

Buffer Solutions

n  Henderson-Hasselbach Equation n  Allows calculation of pH of a buffer if

concentrations of conjugate acid and conjugate base are known HA(aq) + H2O ↔ H3O+(aq) + A-(aq)

][A[HA]K ]OH[

[HA]]][AO[H K

-a

3

-3

a

=

=

+

+

Buffer Solutions

n  Take -log of both sides

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎭⎬⎫

⎩⎨⎧

=− +

][A[HA]log- Klog-

][A[HA]Klog- ]OHlog[ -a-

a3

-log(Ka) = pKa ⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛−

− [HA]][Alog

]A[]HA[log

-

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

[HA]][Alog pK pH

-a Henderson-Hasselbach Eqn

Buffer Solutions

n  Using the Henderson-Hasselbach Eqn, we can: n  Determine pH of a solution n  Determine ratio of conjugate base to

conjugate acid to achieve specific pH

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

[HA]][Alog pK pH

-a

Buffer Solutions

n  Let’s go back to problem of adding HCl to buffer solution:

n  We can use H-H eqn. to make the calculations much easier

[CH3COOH] = 0.100 + [HCl]added [CH3COO-] = 0.100 – [HCl]added

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

[HA]][Alog pK pH

-a

Buffer Solutions

VHCl [HCl] [CH3COOH] [CH3COO-] pH 5mL (.1)(5mL)/105mL .100+.005 .100-.005

= .00476 M =.105 M =.095 M 4.70 10mL (.1)(10)/110 .100+.009 .100-.009

= .00909 M =.109 M =.091 M 4.66 25mL (.1)(25)/125 .100+.020 .100-.020

= .0200 M =.120 M =.080 M 4.56 50mL (.1)(50)/150 .100+.033 .100-.033

= .0333 M =.133 M =.067 M 4.44

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Buffer Solutions

n  Buffer Capacity—the amount of acid or base that can be added to a buffer without the pH significantly changing

n  Suppose we acid to a buffer solution: n  The acid will react with the conjugate base until it is

depleted n  Past this point, the solution behaves as if no buffer

were present

Acid-Base Titrations

n  A titration is a method used to determine the concentration of an unknown species

n  Add a measured amount of a known reactant n  Determine when the reaction has gone to

completion [unknown] + [known] → products

Acid-Base Titrations

n  For the generic titration reaction

n  At the equivalence point moles unknown = moles known added CunknownVunknown = CknownVknown

Acid-Base Titrations

n  For the generic titration reaction αHA + βB → products At the equivalence point

BHA V]B[1

V ]HA[1

base moles1

HA moles1

β=

α

β=

α

Acid-Base Titrations

Titrate 50.00 mL unknown HCl soln. with 0.2137 M NaOH

The titration requires 23.96 mL of NaOH to reach the equivalence point

mL NaOH added0 5 10 15 20 25 30 35 40

pH

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 added0 5 10 15 20 25 30 35 40

pH

1

2

3

4

5

6

7

8

9

10

equivalence point

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Indicators

n  An indicator is a chemical species that changes color depending on the pH of the solution

n  An indicator is a conjugate acid-conjugate base pair in which the acid and base forms of the compound have different colors HIn(aq) + H2O → In-(aq) + H3O+(aq) color 1 color 2

n  Indicators are used to determine the endpoint of a titration

Indicators

n  The pKa of the indicator determines the pH range over which the color changes [HIn]/[In-] ≥ 10 acid color [HIn]/[In-] ≤ 0.1 base color [HIn]/[In-] ≈ 1 intermediate color

n  Remember: pH = pKa + log{[In-]/[HIn]} If [HIn]/[In-] = 1, log{[HIn]/[In-]} = 0 ∴ pH = pKa at point when indicator is

changing color

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 for titration of 0.100 M HCl with 0.100 M NaOH

change before endpoint

change after endpoint

change around endpoint

Indicators

n  Titration of weak acid with strong base HA(aq) + OH-(aq) → A-(aq) + H2O

n  At equivalence point A-(aq) + H2O ↔ HA(aq) + OH-(aq) the solution is basic because conjugate base of weak acid reacts with water to form OH-(aq)

Indicators

n  Titrate 25.00 mL 0.100 M formic acid (HCOOH) with 0.100 M NaOH Ka = 1.8 x 10-4

n  Find pH at equivalence point and select appropriate indicator

n  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

n  Assume HCOOH + OH- reaction goes to completion: [HCOO-] = (2.5 x 10-3 mol)/(0.0500 L) = 0.0500 M

n  Determine Keq for reaction of formate ion: HCOO-(aq) + H2O ↔ HCOOH(aq) + OH-(aq)

11-

aw

-eq 10 x 6.67

KK

]HCOO[][HCOOH][OH K ===

−

Indicators

[HCOO-] [HCOOH] [OH-] initial .0500 0 0 Δ -x x x equil .0500 – x x x

11-

aw

-eq 10 x 6.67

KK

]HCOO[][HCOOH][OH K ===

−

11-2

-

-eq 10 x 6.67

x0500.0x

][HCOO][HCOOH][OH K =

−==

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

11-2

-

-eq 10 x 6.67

x0500.0x

][HCOO][HCOOH][OH K =

−==

Indicators

n  Titration of weak base with strong acid B(aq) + H3O+(aq) → BH+(aq) + H2O

n  At equivalence point BH+(aq) + H2O ↔ B(aq) + H3O+(aq) the solution is acidic because conjugate acid of weak base reacts with water to form H3O+(aq)

Indicators

Figure 17.8: titration of 0.100 M NH3 with 0.100 M HCl

Acid Rain

n  Carbon dioxide in the air is in equilibrium with H2O 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)

n  Natural rain water has pH = 5.6

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Acid Rain

n  Emitted pollutants can form additional acid sources in clouds: NO2: 2 NO2(aq) + H2O ↔ HNO3(aq) + HNO2(aq) nitric acid nitrous acid strong Ka = 4.5 x 10-4

Acid Rain

n  Emitted pollutants can form additional acid sources 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

Acid Rain

http://www.sourcewatch.org/images/8/8d/US_electricity_prod_by_coal.PNG

Acid Rain Sulfur content of coal by percentage in the northeastern U.S. http://pubs.usgs.gov/circ/circ1204/images/knrb_fig011.gif

Acid Rain Acid Rain

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Acid Rain Acid Rain Effects

n  10,000 lakes in Sweden have pH below 6.0—5000 lakes in Sweden have pH below 5.0

n  In the Adirondak Mountain (NY) between 1929 – 1937, 4% of lakes contained no fish. By 1970, half of lakes had pH below 5.0, and of those, 90% contained no fish

n  Lake acidification has been measured in remote areas such as the Uintah Mountains in Utah

Solubility Products

n  Many salts are only slightly soluble n  The solubility product is a measure of the

concentration of ions in a solution saturated with the salt MA(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

Solubility Products

n  Knowing the Ksp, we can calculate the concentration 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

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

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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

Factors Affecting Solubility

n  Salts that are slightly soluble in water can be much more soluble in acid if one or both of its ions are moderately basic: CaCO3(s) ↔ Ca2+(aq) + CO3

2-(aq) Ksp = 8.7 x 10-9 n  But CO3

2-(aq) is the conjugate base of HCO3-(aq)

CO32-(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

n  Works for carbonates, some sulfides, phosphates, etc. (species that behave as bases [no too weak])

Precipitation

n  Define ion quotient, Q, as: MxAy(s) ↔ x My+(aq) + y Ax-(aq) Q = [My+]x[Ax-]y

n  A precipitate will form only when Q exceeds Ksp n  Q < Ksp: solution is unsaturated—no precipitate n  Q > Ksp: solution is saturated—precipitate forms n  Q = Ksp: solution at saturation point