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ANALYTICAL CHEMISTRY - CLUTCH 1E

CH.9 - POLYPROTIC ACID-BASE EQUILIBRIA

CONCEPT: DIPROTIC ACIDS

Diprotic acids and bases are compounds that can donate or accept _______ H+ ion.

For diprotic acids ________ their equations can be illustrated by:

For diprotic bases ________ their equations can be illustrated by:

Based on these equations the relationship between the different forms of diprotic species are:

As a result of these equations for diprotic acids and bases the relationship between Ka and Kb will be:

Ka1 ⋅Kb2 =Kw

Ka2 ⋅Kb1 =Kw

When dealing with diprotic acids:

1) H2A can be treated as a monoprotic acid and we use ________ can be used to find pH.

2) HA – represents the intermediate form and we use ________ can be used to find pH.

3) A2– represents the basic form and we use ________ can be used to find pH.

H2A (aq) + H2O (l) HA– (aq) + H3O+ (aq) Ka1 =ProductsReac tan ts

=

HA– (aq) + H2O (l) A2– (aq) + H3O+ (aq) Ka2 =ProductsReac tan ts

=

A2– (aq) + H2O (aq) HA – (aq) + OH – (aq) Kb1 =ProductsReac tan ts

=

HA – (aq) + H2O (aq) H2A (aq) + OH – (aq) Kb2 =ProductsReac tan ts

=



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PRACTICE: DIPROTIC ACID CALCULATIONS 1

EXAMPLE 1: Sulfurous acid, H2SO3, represents a diprotic acid with a Ka1 = 1.6 x 10-2 and Ka2 = 4.6 x 10-5. Calculate the pH and concentrations of H2SO3, HSO3– and SO32– when given 0.200 M H2SO3.

EXAMPLE 2: Determine the pH of 0.080 M Na2S. Hydrosulfuric acid, H2S, contains Ka1 = 1.0 x 10-7 and Ka2 = 9.1 x 10-8.



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PRACTICE: DIPROTIC ACID CALCULATIONS 2

EXAMPLE: If Ka1 = 4.46 x 10-7 and Ka2 = 4.69 x 10-11 for H2CO3 what is the pH for a 0.15 M solution of NaHCO3?

PRACTICE: An unknown diprotic acid has an initial concentration of 0.025 M. What is the pH of the solution if pka1 is 3.25 and pKa2 is 6.82?



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CONCEPT: POLYPROTIC ACIDS

Our understanding of diprotic acids and bases can be used to understand polyprotic acids and bases.

For polyprotic acids ________ their equations can be illustrated by:

For polyprotic bases ________ their equations can be illustrated by:

As a result of these equations for polyprotic acids and bases the relationship between Ka and Kb will be:

Ka1 ⋅Kb3 =Kw

Ka2 ⋅Kb2 =Kw

Ka3 ⋅Kb1 =Kw

When dealing with polyprotic acids:

• H3A can be treated as a monoprotic acid and we use ________ can be used to find pH.

• A3– represents the basic form and we use ________ can be used to find pH.

H2A−

[H+ ] ≈ Ka1Ka2[ ]0 +Ka1Kw

Ka1 + [ ]0 HA2−

[H+ ] ≈ Ka2Ka3[ ]0 +Ka2Kw

Ka2 + [ ]0

H3A (aq) + H2O (l) H2A– (aq) + H3O+ (aq) Ka1 =ProductsReac tan ts

=

H2A– (aq) + H2O (l) HA2– (aq) + H3O+ (aq) Ka2 =ProductsReac tan ts

=

HA2– (aq) + H2O (l) A3– (aq) + H3O+ (aq) Ka3 =ProductsReac tan ts

=

A3– (aq) + H2O (l) HA2– (aq) + OH – (aq) Kb1 =ProductsReac tan ts

=

HA2– (aq) + H2O (l) H2A – (aq) + OH – (aq) Kb2 =

ProductsReac tan ts

=

H2A – (aq) + H2O (l) H3A (aq) + OH – (aq) Kb3 =

ProductsReac tan ts

=



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PRACTICE: POLYPROTIC ACID CALCULATIONS 1

EXAMPLE 1: Calculate the equilibrium concentrations of H3PO4, H2PO4–, HPO42-, PO43-, and H3O+ for 0.35 M H3PO4.

Ka1 = 7.2 x 10-3, Ka2 = 6.3 x 10-8, and Ka3 = 4.2 x 10-13.



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PRACTICE: POLYPROTIC ACID CALCULATIONS 2

EXAMPLE 1: Determine the pH of 0.250 M sodium hydrogen phosphate, Na2HPO4. Phosphoric acid, H3PO4, contains Ka1 = 7.5 x 10-3, Ka2 = 6.2 x 10-8 and Ka3 = 4.2 x 10-13.

EXAMPLE 2: Determine the pH of 0.150 M citric acid, H3C6H5O7. It possesses Ka1 = 7.4 x 10-4, Ka2 = 1.7 x 10-5 and Ka3 =

4.0 x 10-7.



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CONCEPT: MONOPROTIC & DIPROTIC BUFFERS

A diprotic buffer can be approached in a way similar to monoprotic buffers. The key difference is that a diprotic acid has 2 pKa values.

For Monoprotic Buffers

HA A–

Acid Form Base Form

Ka

pH = pKa + log (conjugate base)(weak acid)

For Diprotic Buffers

H2A HA– A2–

Acid Form Intermediate Form Base Form

Ka1 Ka2

pH = pKa1 + logHA−

H2A⎛

⎝⎜

⎞

⎠⎟ pH = pKa2 + log

A2−

HA−

⎛

⎝⎜

⎞

⎠⎟

0.20 M H2SO3 & 0.25M NaHSO3 Ka1 =1.6 x10−2

30.0 mL of 0.10 M Na2SO3 Ka2 = 6.4 x10−8

20.0 mL of 0.20 M NaHSO3

HClO + NaClO (aq)Weak oxyacid Conjugate base

(Has 1 less Hydrogen)



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PRACTICE: MONOPROTIC & DIPROTIC BUFFERS CALCULATIONS 1

EXAMPLE 1: What is the pH of a solution consisting of 2.5 M potassium dihydrogen phosphite (KH2PO3) and 2.75 M phosphorus acid (H3PO3)? Ka1 = 3.0 x 10-2 and Ka2 = 1.66 x 10-7.

EXAMPLE 2: Sulfurous acid, H2SO3, is a major component in the creation of commercial fertilizers. What is the buffer component concentration ratio of a buffer that has a pH of 1.15? Ka1 = 1.39 x 10-2 and Ka2 = 6.73 x 10-8.

PRACTICE: Calculate the pH of a solution made by mixing 8.627 g of sodium butanoate in enough 0.452 M butanoic acid, HC4H7O2, to make 250.0 mL of solution. Ka = 1.5 x 10-5.



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CONCEPT: POLYPROTIC BUFFERS

A triprotic buffer can be approached in a way similar to diprotic buffers. The key difference is that a triprotic acid has 3 pKa values.

For Polyprotic Buffers

H2A– HA2 – A3 –H3A

Acid Form Intermediate 1 Base FormIntermediate 2

Ka1 Ka2 Ka3

pH = pKa1 + logH2A

–

H3A⎛

⎝⎜

⎞


HA2–

H2A–

⎛

⎝⎜

⎞


A3–

HA2–

⎛

⎝⎜

⎞

⎠⎟

0.10 M H3C6H5O7 Ka1 = 7.4×10−4

0.15M NaH2C6H5O7

0.25 moles Na2HC6H5O7 Ka2 =1.7×10−5

0.17 moles NaH2C6H5O7

50.0 mL of 0.32 M Na3C6H5O7 Ka3 = 4.0×10−7

60.0 mL of 0.25M NaHC6H5O7



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CONCEPT: PRINCIPAL SPECIES

The acid constant, Ka, of an acid tells us the numerical value that an acidic hydrogen can be removed.

For a monoprotic acid: When the pH < pka then [HA] is ________________ than [A-].

When the pH > pka then [HA] is ________________ than [A-].

• The relationship between pH and pKa can be furthered applied to diprotic and polyprotic acids.

EXAMPLE 1: What is the predominant form of the diprotic acid, methionine, at a pH equal to 4.18? Ka1 = 6.6 x 10-3 and Ka2

= 8.3 x 10-10.

EXAMPLE 2: What is the predominant form of histidine at a pH equal to 8.00? pKa1 = 1.6, pKa2 = 5.97 and pKa3 = 9.28.

EXAMPLE 3: What is the second most predominant form in the previous question?



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CONCEPT: ISOELECTRIC AND ISOIONIC POINT Isoelectric or isoionic points represent the pH where a polyprotic acid doesn’t migrate to an electric field because it’s neutral.

Below Isoelectric Point At Isoelectric Point Above Isoelectric Point

R

H

COOHH3N

R

H

COOH3N

R

H

COOH2N

At the isoionic point the polyprotic acid exists as an intermediate and so we can utilize past equations to determine [H+].

Diprotic Acid Polyprotic Acid

[H+ ]=Ka1Ka2F+Ka1

Kw

Ka1+F

H2A−

[H+ ] ≈ Ka1Ka2[ ]0 +Ka1Kw

Ka1 + [ ]0

HA2−

[H+ ] ≈ Ka2Ka3[ ]0 +Ka2Kw

Ka2 + [ ]0

The isoelectric point is the pH where [H2A] = [A–] and therefore the average charge is equal to zero.

Diprotic Acid Polyprotic Acid

pH =12(pK a1

+pKa2) pH =

12(pK a1

+pKa2) pH =

12(pK a2

+pKa3)



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PRACTICE: ISOELECTRIC AND ISOIONIC POINT CALCULATIONS 1 EXAMPLE 1: Calculate the isoelectric and isoionic pH of 0.025 M glutamine. pKa1 = 2.19, pKa2 = 9.00.

EXAMPLE 2: Draw the structures and charge of aspartic acid at pH = 9.82. PRACTICE 1: Calculate the pI value for histidine. pKa1 (carboxyl group) = 1.60, pKa2 (ammonium group) = 9.28, pKa3 (R-group) = 5.97.

PRACTICE 2: Calculate the pI value for gluatamic acid. pKa1 (carboxyl group) = 2.16, pKa2 (ammonium group) = 9.96, pKa3 (R-group) = 4.30.



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ANALYTICAL CHEMISTRY - CLUTCH 1E CH.9 ...lightcat-files.s3.amazonaws.com/packets/admin_analytical...A diprotic buffer can be approached in a way similar to monoprotic buffers. The