common ion effect revisited. ph changes due to common ion effect

Common Ion Effect Revisited

• Recall that the addition of a common ion causes an equilibrium to shift– Earlier we related this to the solubility of a salt– The idea is just an application of Le Chatelier’s

Principle

pH changes due to Common Ion Effect

• Consider adding sodium acetate (NaCH3COO) to a solution of acetic acid:

• The addition of a common ion here (CH3COO-) will increase the pH– By consuming H+ ions

HaqCOOCHaqCOOHCH )()( 33

Henderson-Hasselbalch

• Consider:

• Rearranging Ka for [H+]we get:

)()()( aqAaqHaqHA

][

][][

A

HAKH a

][

]][[

HA

AHKa

Henderson-Hasselbalch

• Take negative log of both sides:

or:

][

][loglog]log[

HA

AKH a

][

][log

Acid

baseConjugatepKpH a

][

][log

Base

acidConjugatepKpOH b

Finding pH with common ion present

• Calculate the pH of a solution containing both 0.20 M CH3COOH and 0.30 M CH3COONa? The Ka of CH3COOH is 1.8x10-

5.– Sodium acetate fully dissociates in solution

MCOOCH 30.0][ 3

• Can use I.C.E. table, or Henderson-Hasselbach

Finding pH with common ion present

]20.0[

]30.0[log)]108.1log([ 5 xpH

][

][log

3

3

COOHCH

COOCHpKpH a

92.4pH

Effect of Common Ion on pH

• Consider calculating the pH for a 0.20 M acetic acid solution– pH = 2.72

• From our last example, its obvious the pH has increased due to the common ion

Buffer Solutions

• Buffers are a solution of (1) weak acid or weak base and (2) its salt

• Buffers resist changes in pH upon addition of acid or base

Buffer Solutions

• Consider a solution of acetic acid and sodium acetate– Upon addition of an acid, H+ is consumed:

– Upon addition of a base, OH- is consumed:)()()( 33 aqCOOHCHaqHaqCOOCH

)()()()( 233 lOHaqCOOCHaqOHaqCOOHCH

Buffer Animation

• http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/buffer12.swf

Buffer Problem

• (a) Calculate the pH of a buffer system containing 1.0M CH3COOH and 1.0 M CH3COONa. (b) What is the pH of the buffer system after the addition of 0.10 mole of gaseous HCl to 1.0 L of the solution? Assume volume of sol’n does not change.

Buffer Problem

• To calculate pH of buffer, I.C.E. or H.H.

• With addition of HCl, we are adding 0.10 M H+, which will react completely with acetate ion

]0.1[

]0.1[log)]108.1log([ 5 xpH

74.4 apKpH

Buffer Problem

Now acetic acid will still dissociate and amount of H+ formed is the pH of sol’n

)()()( 33 aqCOOHCHaqHaqCOOCH

Initial (mol) 1.0 0.10 1.0

Change (mol) -0.10 -0.10 +0.10

Equilibrium (mol) 0.90 0 1.10

Buffer Problem

• Using H.H. (or I.C.E.)

HaqCOOCHaqCOOHCH )()( 33

]10.1[

]90.0[log)]108.1log([ 5 xpH

66.4pH

Finding Buffers of Specific pH

• If concentrations of both species are equal this means:

• Using H.H., to find specific pH, search for pKa pH

0][

][log

HA

A


• Describe how you would prepare a “phosphate buffer” with a pH of about 7.40

4243 POHHPOH

2442 HPOHPOH

34

24 POHHPO

12.2;105.7 3 aa pKxK

21.7;102.6 8 aa pKxK

32.12;108.4 13 aa pKxK


• Using the HPO42-/H3PO4

- buffer:

][

][log21.740.7

42

24

POH

HPO

19.0][

][log

42

24

POH

HPO

5.110][

][ 19.0

42

24

POH

HPO


• To obtain this solution, disodium hydrogen phosphate (Na2HPO4) and sodium dihydrogen phosphate (NaH2PO4) is in 1.5:1.0 ratio– Meaning it is 1.5 M Na2HPO4 and 1.0 M NaH2PO4

Acid-Base Titrations

• Three situations will be considered:– Strong Acid/ Strong Base– Weak Acid/ Strong Base– Strong Acid/ Weak Base

• Titrations of weak acid/base are complicated by hydrolysis

Strong Acid-Strong Base Titration

• Reacting HCl and NaOH, the net ionic equation would be:

• For our example, 0.100 M HCl is being titrated by 0.100 M NaOH

• Calculating pH changes depends on the stage of the titration

)()()( 2 lOHaqOHaqH


• Scenario #1: After addition of 10.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl.– Total volume = 35.0 mL


• Scenario #1

HClmolxmL

HClmolxmL 31050.2

1000

100.00.25

NaOHmolxmL

NaOHmolxmL 31000.1

1000

100.00.10

leftHClmolxmolxmolx 333 105.11000.11050.2

HClML

mLx

mL

molx0429.0

1

1000

0.35

1050.1 3

37.1 pH


• Scenario #2: After addition of 25.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl (aka equivalence point)– Because equivalent moles of acid/base


• Scenario #2

–Since no hydrolysis occurs (both are strong) equivalence point of all strong acid/base titrations has a pH = 7• Keep in mind, equivalence point means

[OH-]=[H+]


• Scenario #3: After addition of 35.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl– Total volume = 60.0 mL


• Scenario #3

NaOHmolxmL

NaOHmolxmL 31050.3

1000

100.00.35

NaOHmolxHClmolxNaOHmolx 333 100.11050.21050.3

22.1278.100.1400.14 pOHpH

Weak Acid-Strong Base Titration

• Consider the neutralization between acetic acid and sodium hydroxide:

• The net ionic is:

• Let’s calculate the pH for this reaction at different stages

)()()()( 233 lOHaqCOONaCHaqNaOHaqCOOHCH



• Scenario #1: 25.0 mL of 0.100 M acetic acid is titrated with 10.0 mL of 0.100 M NaOH– Total volume = 35.0 mL


• Key things to note:–Work in moles, not molarity in I.C.F. table• If volume is the same for all species, ratio of

moles is equal to ratio of molar concentration– This means for Ka, no need to convert back to

molarity

– Equivalance point is not 7• Hydrolysis of salt causes this shift


• Scenario #1– Buffer system exists here

(CH3COONa/CH3COOH)

• Calculate moles of each

COOHCHmolxmL

molxmL 3

31050.21000

100.00.25

NaOHmolxmL

NaOHmolxmL 31000.1

1000

100.00.10



Initial (mol)2.5x10-3 1.00x10-3 0

Change (mol) -1.00x10-3 -1.00x10-3 +1.00x10-3

Equilibrium (mol) 1.50x10-3 0 1.00x10-3

Buffer System Wooo!!


]1050.1[

]1000.1[log)]108.1log([

3

35

x

xxpH

57.4pH


• Scenario #2: Adding 25.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M acetic acid– Equivalence point is not at 7!


NaOHmolxmL

NaOHmolxmL 31050.2

1000

100.00.25

)()()()( 233 lOHaqCOONaCHaqNaOHaqCOOHCH

Initial (mol)2.5x10-3 2.50x10-3 0

Change (mol) -2.50x10-3 -2.50x10-3 +2.50x10-3

Equilibrium (mol) 0 0 2.50x10-3


• Acid/Base concentration being zero at the equivalence point, pH is determined by hydrolysis of salt

)()()()( 323 aqOHaqCOOHCHlOHaqCOOCH

ML

molxCOOCH 05.0

05.0

1050.2][

3

3


• I.C.E. it! Equation becomes:

x

xx

05.0106.5

210

10

3

3 106.5][

]][[

xCOOCH

OHCOOHCHKb

72.8,103.5][ 6 pHMxOHx


• Scenario #3: Adding 35.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M acetic acid


NaOHmolxmL

NaOHmolxmL 31050.3

1000

100.00.35


Initial (mol)2.5x10-3 3.50x10-3 0

Change (mol) -2.50x10-3 -2.50x10-3 +2.50x10-3

Equilibrium (mol) 0 1.00x10-3 2.50x10-3


• At this point, both OH- and hydrolysis of CH3COO- affects pH

• Since OH- is much stronger, we can ignore the impact of the hydrolysis

ML

molxOH 0167.0

060.0

1000.1][

3

22.12 pH

Strong Acid-Weak Base Titration

• Consider the titration of HCl with NH3:

• Or

• Let’s calculate the pH for this titration at different stages

)()()( 43 aqClNHaqNHaqHCl

)()()( 43 aqNHaqNHaqH


• Key main differences from Weak Acid/ Strong Base:– Equivalence point is less then 7 due to salt

hydrolysis– Calculation of pH after equivalence point is

based on remaining [H+] only• Even though salt still exists, effect on pH minimal

– First part of titration is the same (buffer!)


• Calculate the pH at the equivalence point when 25.0 mL of 0.100 M NH3 is titrated by a 0.100 M HCl solution


• At equivalence point, moles of acid equal moles of base

333 1050.2

1000

100.00.25 NHmolx

mL

NHmolxmL


)()()( 43 aqNHaqNHaqH Initial (mol)

2.5x10-3 2.50x10-3 0

Change (mol) -2.50x10-3 -2.50x10-3 +2.50x10-3

Equilibrium (mol) 0 0 2.50x10-3


ML

molxNH 05.0

05.0

1050.2][

3

4

)()()( 34 aqNHaqHaqNH

Initial (mol)0.0500 0 0

Change (mol) -x +x +x

Equilibrium (mol) 0.0500 - x x x


• Approx. good (0.01%)

][

]][[

4

3

NH

HNHKa

0500.00500.0106.5

2210 x

x

xx

Mxx 6103.5

28.5)103.5log( 6 xpH

common ion effect revisited. ph changes due to common ion effect

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