1

Multiple Equilibrium

2

3

4

Weak AcidsEquilibrium in Solution

HA + H2O H3O+ + A-

2 H2O H3O+ + OH-

If we know the analytical concentration (CHA) of the acid, can we calculate the concentration of all other species in solution ? (ie. [H3O+], [OH-], [A-], [HA]: 4 unknowns)

Equations?1) acid equilibrium

[ ][ ][HA]

AOHK 3a

−+

=

2) water equilibrium Kw = [H3O+][OH-]3) mass balance CHA = [A-] + [HA]4) charge balance [H3O+] = [A-] + [OH-]

4 eqns, 4 unknowns ... we can solve it in principle, try isolating [H3O+] in 1).5) [A-] = [H3O+] - Kw/[H3O+] from 2) and 4)6) [HA] = CHA - [H3O+] + Kw/[H3O+] from 3) and 5)

Substitute 5) and 6) into 1) to give :

[H3O+]3 + Ka[H3O+]2 - (Kw+KaCHA)[H3O+]- KaKw = 0 A cubic equation! @#$%

Weak acids-cont'dIn principle, the previous cubic equation could be solved. In practice, we can simplify. Look at equation 4. If we have a weak acid, then[OH-] << 1x10-7 M AND 1x10-7 M << [A-]. So, [OH-]<<[A-]Look at equation 4. If [OH-]<<[A-], then [H3O+] ≈ [A-]. Substituting into 1...

7) 1)in3and4(from]O[H-C

]O[HK3HA

23

a +

+

=

8) [H3O+]2 + Ka[H3O+] - CHAKa = 0 a = 1 b = Ka c = - CHAKa

2a4acbbx

2 −±−=9)

2C4KKK

]O[H HAa2

aa3

+±−=+

We can further simplify the chemistry by assuming [H3O+]<<CHA, into 7 gives...

HAa3 CK]O[H =+10)

Our first approximation is almost always true, therefore we can use the quadratic. The second approximation is not always true. It is true for high concentrations of weak acids. If we use 10) and find that our approximation:[H3O+] <<CHA , is not true, use the quadratic formula or solve quickly using an

iterative method appoximations (calculate [H3O+], put it into the right side of 11 and recalculate). Then put next next estimate of [H3O+] into 11 and solve again. keep on going until answer converges.

11). ])O[H(CK]O[H 3HAa3++ −=

5

Weak acid-example 1 (in class)Example 1: Calculate [H3O+] for a 1% (w/v) solution of acetic acid.

Calculate pH and pOH.

...

Answer: [H3O+] = 1.7x10-3M pH = 2.77, pOH=11.23

Weak acid - example 2 (in class)Example 2: Calculate the pH of a 0.01M solution of HF; pKa = 3.17

Use both the succesive approximation solution and the quadratic solution if necessary.

...

Answer: [H3O+] = 2.290x10-3M pH = 2.64,

6

Weak Bases

BbCK][OH =−

B + H2O OH- + HB+

2 H2O H3O+ + OH- Kw = [H3O+][OH-]

Similar to the case for acids, we can derive a set of 4 equations.With the approximation [OH-] << CB, we get…

If this is not true, then use succesive approximations…

or use quadratic:

)][OH(CK][OH nBb1n−

+− −=

[B]]HB][OH[Kb

+−

=

2C4KKK

][OH Bb2

bb +±−=−

Note that for these are the identical equations derived for an acid but with the transformations… Ka Kb AND [H3O+] [OH-]

Weak base - example 1 (in class)Example 1: 0.01 M solution of NH3.

Calculate pH, pOHCalculate the concentration of all species in solution.Calculate the fraction composition of NH3; ie [NH3]/([NH3]+[NH4

+])

Answers: [H3O+] = 2.44x10-11M [OH-] = 4.10x10-4M [NH3]=9.6x10-3M [NH4+]=4.10x10-4M

pH = 10.6, NH3 fraction = 0.96 (96%); NH4+ fraction = 4%.

7

BuffersMixtures of acid/conjugate base pair that maintains the pH at a relatively constant value. This is desirable for many chemical and biological systems. Lets review how to derive the Henderson-Hasselback equation…

HA + H2O H3O+ + A-

( )

][A[HA]logpKapH

][A[HA]loglogK]Olog[H

][A[HA]Klog]O[Hlog

][A[HA]K]O[H

[HA]]][AO[HK

a3

a3

a3

3a

−

−+

−+

−+

−+

−=

−−=−

⎟⎟⎠

⎞⎜⎜⎝

⎛−=−

=

=

If we mix an amount of acid, HA, and the Na salt of the conjugate base, NaA and approximate that [HA] = CHA; [A-] = CNaA, the we get the HH eqn…

HA

NaA

NaA

HA

CClogpKapHOR

CClogpKapH +=−=

More exact buffer derivation

Mixture of HA and NaA

Mass Balance: CNaA + CHA = [HA] + [A-] (3)CNaA = [Na+] (4)

Charge Balance: [H3O+] + [Na+] = [A-] + [OH-] (5)From (5) and (4) [A-] = CNaA + [H3O+] - [OH-] (6)From (3) and (6) [HA] = CHA - [H3O+] + [OH-] (7)

Substitute 6 and 7 into 1 rearranged to give H3O+

(2)]][OHO[HK(1)[HA]

]][AO[HK 3w3

a−+

−+

==

][OH]O[HC][OH]O[HCK]O[H -

3NaA

-3HA

a3 −++−

= +

++

A more exact solution for buffer solutions...use suitable approximations in acidic ([H3O+] >> [OH-]) and basic ([H3O+] << [OH-]) buffer solutions.

][OH]O[HC][OH]O[HClogpKapH -

3HA

-3NaA

+−−+

+= +

+

8

9

from weak base example

10

11

12

Other Notes

Acid Base Titrations Applications

Volumetric acid base titrations are performed to determine the unknown acid or base content of a sample. Our analytical signal is the endpoint. ideally the endpoint should be the same as the equivalence point. The equivalence point is the point where we have added equal equivalents of base as are present in our unkown acid sample; or visa versa.

What is an equivalent? In an acid base titration, an equivalent is that amount of sample or standard that produces or consumes 1 mole of H+. Normality, is the concentration of equivalents expressed as equivalents per liter.

Normality (N) = Concentration (M) x equivalents / mole

For monoprotic acid or base, normality is equal to the concentration since there is only one mole of acid produced or consumed per mole of sample or standard. (ie- NaOH, HCl, CH3COOH, CH3CH2NH2, NH3). For polyprotic acids and bases, the equivalents per mole depend on our application. (oxalic acid –HOOC-COOH has two equivalents per mole if we titrate both ionizable H's

2 NaOH + HOOC-COOH 2H2O + 2 Na+ + 2 C2O42-

At the equivalence point NacidVacid = NbaseVbase

13

Titration Curves (Theory)

Strong Acid-Strong Baseeg. NaOH(aq) + HCl(aq) H2O + NaCl(aq)

This is really just a titration of H3O+ with OH-

H3O+(aq) + OH-

(aq) 2H2O

The calculation of pH for this type of titration is defined by three regions:i) before equivalenceii) equivalence pointiii) after equivalence

Region I [H3O+] is given by the unreacted H3O+. This is much greater than the [H3O+] from water autoprotolysis.

NaOHHCl

NaOHNaOHlHClHCl3

total

addedoriginal33

VVVCVC]O[H

V][OH]O[H

]O[H

+−

=

−=

+

−++

from buret

strong acid-strong base cont'd

Region IIat the equivalence point, the pH is given by that of the neutralized solution, in this case, the pH of a solution of NaCl(aq) in H2OIn this case, the solution is trivial. Since Na+ and Cl- are not conjugate acids or bases, [H3O+] is given by that resulting from autoprotolysis of water,

[H3O+] = 1×10-7M pH = 7.0

Region IIIafter the equivalence point, all H3O+ has been titrated (producing H2O), and we only are left with have an excess of OH-. The amount of excess OH- is easily calculated…

][][H

VVVCVC][OH

V]O[H ][OH

][OH

3

NaOHHCl

HClHClNaOHNaOHl

total

original3added

−+

−

+−−

=

+−

=

−=

OHKO w

14

Example Spreadsheet Calculation of Titration

0

2

4

6

8

10

12

14pH

0 10 20 30 40 50NaOH (mL)

0.1M

0.01M

0.001M

0.0001M

Strong Acid - Strong Base

Titration of 25 ml HCl by NaOHCHCl and CNaOH

0.01

0.1

1

10

100

dpH

/ dV

0 10 20 30 40 50NaOH (mL)

0.1M

0.01M

0.001M

0.0001M

Strong Acid - Strong Base

Titration of 25 ml HCl by NaOH

dpH/dV

CHCl and CNaOH

15

d2pH/dV

-100-80-60-40-20

020406080

100

d(dp

H/ d

V)/d

V

22 23 24 25 26 27 28NaOH (mL)

0.1M

0.01M

0.001M

0.0001M

Strong Acid - Strong Base

Titration of 25 ml HCl by NaOHCHCl and CNaOH

Titration - weak acid sample with strong base

The calculation of pH for this type of titration is defined by four regions:

I) initial weak acid sample solutionII) buffer region before E.P.III) equivalence pointIV) beyond equivalence (excess OH-)

region I

region II

region III

region IV

Region I)Initial pHThe pH of our initial sample solution depends on CHA and pKa and is calculated as we have shown previously:

2C4KKK

]O[H HAa2

aa3

+±−=+

])O[H(CK]O[H 3HAa3++ −=

16

weak acid/weak base-cont'dii) Region II Buffer region. Remenber our neutralization reactionHA(aq) + OH-

(aq) H2O + A-(aq)

In this region, we need to calculate the concentration of HA and A-. We use a form of the HH equation derived earlier.

][OH]O[HC'][OH]O[HC'logpKpH

][HA][AlogpKpH -

3HA

-3-A

a

-

a +−−+

+=⇒+= +

+

But

NaOHHA

NaOHNaOH

total

added--A

NaOHHA

NaOHNaOHHAHA

total

addedoriginalHA

VVVC

V][OH][AC'

VVVCVC

V][OH[HA]

[HA]C'

+==≈

+−

=−

=≈

−

−

NaOHNaOHHA

NaOHNaOHa

NAOHHA

NaOHNaOHHA

NAOHHA

NaOHNaOH

a VCCVClogpKpH

VVVCC

VVVC

logpKpH−

+≈⇒

+−+

+≈HAHA VV

Weak Acid/Strong base titration - cont'd

ii) Region III - Equivalence pointAt the equivalence point, the acid has been neutralized and exists as the conjugate base. There is no excess OH-.

But, what is CB at the equivalence point. Number of moles of conjugate base = number of moles of acid in original mixture, but the concentration is less due to dilution as the titration proceeds.

ii) Region IV- Post Equivalence regionAfter the equivalence point, we have an excess of OH-, just as`we did for a strong acid/strong base. The pH will largely be determined by this concentration of excess (ie subtract out amount required to neutralize the acid. This will determine the pH.

2C4KKK

][OH Bb2

bb +±−=−

NaOHHA

HAHAB VV

VCC+

=

]log[HpH][OH

K]O[H VV

VCVC][OH w3

NaOHHCl

HAHANaOHNaOH +−

+− −==+−

=

17

Example – spreadsheet (in class)

Titration of CClH2COOH with NaOH

0.0

2.0

4.0

6.0

8.0

10.0

12.0

0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0

Volume of Titrant (mL)

pH

Polyprotic Acid and Base TitrationsAmphiprotic SaltsIn our titration of polyfunctional acids and bases, we will come across solutions of these salts at intermediate endpoints. These salts are capable of acting both as an acid and a base. How do we calculate the pH?examples - NaHCO3 or amino acid HLIn a titration of H2CO3 with NaOH, our first endpoint contains a solution of only NaHCO3(aq). How would you calculate the pH at the first equivalence point?

example: for the titration of a 25 mL aliquot of 0.01M H2CO3 with 0.015M NaOH, at the first endpoint we have:

VNaOH = VH2CO3*MH2CO3/MNaOH= 25mL*0.01/0.015 = 16.67mL

Vtotal = VH2CO3 + VNaOH

CNaHCO3 = moles NaHCO3/Vtotal

= moles NaOH/Vtotal = .025*0.01M/(.02500 + .01667) = 0.006M

What is the pH of a 0.006M solution of NaHCO3? ANSWER : pH = 8.33

18

Amphiprotic Salt – pH predictionReactions

HCO3- + H2O H3O+ + CO3

2-

HCO3- + H2O OH- + H2CO3

EquationsKa2 = [H3O+][CO3

2-]/[HCO3-] (1)

Kb = [OH-][H2CO3]/[HCO3-] = Kw/Ka1 (2)

Mass Balance Csalt = [H2CO3] + [HCO3-] + [CO3

2-] = [Na+] (3)

Charge Balance [Na+] + [H3O+] = [OH-] + [HCO3-] + 2[CO3

2-] (4)

Kw = [H3O+][OH-] (5)

5 equations, 5 unknowns. Try the derivation on your own. For a general amphiprotic salt of concentration Csalt, we have...

salta

wasaltaa

a

salt

wsalta

CKKKCKK

KC

KCKH++

=+

+=+

1

121

1

2

1][

19

Titration of Mixtures of AcidsMixtures of Strong AcidsIt is not possible to separate mixtures of strong acids by titration with strong base.

example: 0.05 M HCl and 0.05 M HNO3

In solution this gives 0.10M [H3O+], 0.05M [Cl-], 0.05M [NO3-]. The titration

curve would look identical to that for a titration of 0.10 M HCl alone. Water is too strong a base and does not differentiate between the stregths of these acids. This is termed the levelling effect.

Levelling EffectThe strongest acid in water is H3O+ and the strongest base is OH-. If a stronger acid than H3O+ is added to water, it protonates water to produce H3O+. If a base stronger than OH- is added, it strips a proton from water to produce OH-. Strong acids and strong bases appear to have the same acid and bas strength in water because of this.

Non- Aqueous TitrationsNote that the definition of strong is dependent on the solvent. In another solvent, separation might be possible. The way to separate strong acids in a titration is to use a solvent that is less basic than water (less able to accept H+) than water. ie- acetic acid, methyisobutylketone, ...

Mixtures of strong and weak acidsConsider a mixture of a strong acid and weak acid.ie - HCl (strong) and benzoic acid (pKa = 4.2).

HA + H2O H3O+ + A- Ka = 10-4.2

HCl (aq) H3O+ + Cl-

titration OH- + H3O+ 2H2O

While the strong acid is in solution, it supresses the dissociation of the weak acid, by Le Chatelier's pronciple. If we titrate with base, the OH-

will preferentially neutralize the H3O+ from the strong acid first. The dissociation of the weak acid remains supressed until [H3O+] becomes comparable to what might be expected from the weak acid alone. As this point occurs, [H3O+] decreses rapidly (pH increses) giving a distinct endpoint, before HA starts to be neutralized. An endpoint will occur when all the strong acid has been consumed. Clarity of the endpoint depends on the pKa of the weak acid, ie. pKa > 4 for a reasonable endpoint.

see autotitrator for example of a titration of mixture of strong an weak acid.

20

Autotitrator

pH electrode

sample flask with stir bar

standardized titrant

microprocessor with optional display screen(pH vs. Vtitrant)

magneticstirrer

Autotitrator - endpoint detectionmixture of 2 acids

pH vs volume - endpoint occurs at points with steepest slope, also points of inflection as seen in 2nd derivative plot…2 endpoints occur in the titration to the right at V1 and V2. Concentrations determined by:

C1∆V1 = CNaOHVNaOH

C2∆V2 = CNaOHVNaOH

dpH/dV - endpoints are detected at maximum points, computer algorithm searches for peaks in the 1st derivative plot.

d2pH/d2V - endpoints occur at points of zero crossing. Algorithm searches for points of zero crossing after a peak.

1st endpoint at V1

2nd endpoint at V2

∆V1

∆V2

21

Endpoint detection – cont'd

2nd derivative endpoints

22

Acid Base Titrations in PracticeEndpoint Detection

1) indicators

2) pH measurement (autotitrations)1st derivative plots - ∆pH/∆V vs Vtitrant

endpoint at maximum (slope of pH vs volume is maximum)

2nd derivative plots∆(∆pH/∆V)/∆V vs. Vtitrantendpoint at zero crossings (inflection points)

3) Gran plots- sometimes used due to slow electrode response in the region of the endpoint- doesn’t require data in the endpoint region, just data before it.

Acid Base IndicatorsUsually weak organic acids and bases in which the colour of the conjugate acid and conjugate base are different. (ie- different electronic transitions).

HInd + H2O H3O+ + Ind-

The human eye is not very sensitive to changes in the intensity of colour of one particular form. For visible changes, the human eye is most sensitive when the relative concentrations of the the two indicatior forms switch:ie: [HInd]/[Ind-] must change from ~ 10 (mostly [HInd])

to ~0.1. (mostly [Ind-]).In these cases, we have:Acid form of Indicator:

Base form of indicator:

Therefore, indicator changes color (endpoint) in range, pH = pKa +/- 1. The most appropriate indicator depends on the pH at the end-point!!

[HInd]]][IndO[HK 3

a

−+

=

1)10log(]O[Hlog

10][Ind

[HInd]K]O[H

3

a3

−=×−=−=

×==

+

−+

aa

a

pKKpH

K

1)10.0log(]O[Hlog

10.0][Ind

[HInd]K]O[H

3

a3

+=×−=−=

×==

+

−+

aa

a

pKKpH

K

23

Indicators

Indicators – cont'd

24

Using suitable indicatorsUse of incorrect indicators can lead cto either positive OR negative errors in a volumetric determination (classroom discussion) . The error CAN be calculated. Think about it!

Gran PlotsGran Plots- a pH meter will respond to the activity of hydronium ion, not [H3O+]. For the equilibrium during titration:

HA + H2O W H3O+ + A-

Ka = (H+(A- [H3O+][A-]/(HA[HA]During the titration in the buffer region we can approximate:Ka = (H+ (A- [H3O+][A-]/(HA[HA][A-] = VbCb/(Va + Vb)[HA]= (VaCa - VbCb)/(Va + Vb)

if we substitute these into the equilibrium expression above and rearrange, we get :Vb (H+[H3O+] = (HAKa/(A- (VaCa - VbCb)/Cband then ......Gran Plot equation

We can plot Vb*10-pH vs Vb (volume of base added).The slope of the curve gives us the Ka. The x intercept gives us the equivalence point volume, without actually having to use the pH meter data at the equivalence point.

( )baseeequivalencaA

HApHb VVK

γγ10V −=×

−

−

25

Gran Plot example

The x intercept can be determined by least squares, and gives us Veq(=CaVa/Cb), without actually having to use the pH meter data at the equivalence point. Useful for reactions with slow completion times

Complexation Titrations

Metal ChelatesMetal ions tend to be Lewis acids, accepting electron pairs from electron donators (Lewis acids).Ligands that can bind to metals are known as chelates. The order of the chelate refers to the number of atoms on the chelate molecule available for electron donation to the metal ion.

Monodentate : cyanide, CN-

Mn+ :C≡N: Others : :F-, :OH-, H2O:

Bidentate: ethylendiamine, H2NCH2CH2NH2

H2N NH2

Mn+

26

Metal ChelatesATP ADP ∆G =-34.3 kJ/mol

Tetradendate: Adenosine Triphosphate (ATP)

See handout given in class for all the notes for Complexation and Complexation Titrations. Come to

my office if you did not get these.