Acid Base Hydrolysis Worked Examples

8/3/2019 Acid Base Hydrolysis Worked Examples

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Acid Base Hydrolysis

Worked Examples


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pH

Determine the pH of a solution which contains 4

x 10-5

M (moles/litre) H+

The decimal part is always positive

Log 4.0 = 0.602

log (4 x 10-5) = -5 + 0.602 = -4.398

pH = -log[H+] = -(-4.398) = 4.398

Or log 0.00004 = -4.398ve log = 4.398

pHHHpH 101loglog 1010


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pH

Find the hydrogen ion concentration

corresponding to pH 5.643 pH = -log[H+] = 5.643

log[H+] = -5.643 = (0.357) + (-6.000)

Antilog of 0.357 = 2.28 x 10-6

[H+] = 2.28 x 10-6

pHHHpH

101loglog 1010


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Hydrolysis

Salt of a weak acid and strong base

Calculate:

1) the hydrolysis constant,

2) the degree of hydrolysis and

3) the hydrogen concentration for 0.01 M solution ofsodium acetate

Na+ + Ac - + H2O Na+ + HAc + OH-

10

5

14

105.51082.1

100.1

a

wh

K

KK


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Hydrolysis

The degree of dissociation is given by

Substituting for Kh and V = 1/c we obtain

Solving this quadratic equation, x = 0.000235 or0.0235%

Vxx

Kh

1

2

x

x

1

01.0105.5

210


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Hydrolysis

If the solution was completely hydrolysed, theconcentration of acetic acid produced would be 0.01 M.But the degree of hydrolysis is 0.0235%. Therefore theconcentration of acetic acid is 2.35 x 10-6 M

This is also equal to the hydroxide ion concentration

[OH-] = 2.35 x 10-6

M pOH = 5.63

pH = pKw pOH i.e. pH = 14 5.63 = 8.37

The pH may also be determined using pH = pKw + pKa + log c

= 7.0 + 2.37 + (-2)

pH = 8.37


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Hydrolysis

Salt of a strong acid and weak base

Determine the pH of a 0.2 M NH4Cl solution

NH4- + Cl- + H2O NH4OH + Cl

- + H+

Since [NH4OH] and [H+] are equal

V

KKBaseAcid

NHOHNHHK

b

wh

x-1

x

SaltedUnhydrolys

2

4

4


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Hydrolysis

Ammonia in water Kb = 1.85 x 10-5 and

pKb = 4.74

pH = pKw - pKb - log c

cpKpK

K

KcH

KK

cH

NHOHNHHK

bw

b

w

b

wh

logpH21

21

21

2

4

4


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Hydrolysis

pH = (14) - (4.74) - (-0.6989)

pH = 7.0 2.37 + 0.3495

pH = 4.98


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Hydrolysis

Salt of a weak acid and weak base

NH4+ + Ac- + H2O NH4OH + HAc

2

4

4

4

4

(1)..................

4

4

4

4

SaltHydrolysed

AcidBaseK

AcNH

HAcOHNHK

ffff

AcNHHAcOHNH

aaaaK

h

h

AcNH

HAcOHNH

AcNH

HAcOHNHh


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Hydrolysis

If x is the degree of hydrolysis of 1g mol. of saltdissolved in V litres of solution, then

[MOH] = [HA] = x/V; [M+] = [A-] = (1 x)/V

Substituting the values in the equation 1 given in theprevious slide

The degree of hydrolysis and consequently pH isindependent of the concentration of the solution.

22

111 x

x

VxVx

VxVxKh


12/24

Hydrolysis

Remembering the following equations must hold

simultaneously

Also, it can be shown that

OHNH OHNHKHAcAcHK

OHHK

ba

w

4

4and

bawh

bawh

pKpKpKpK

KKKK


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Hydrolysis

The hydrogen ion concentration of the hydrolysedsolution is calculated in the following manner.

h

aaa

Kxx

x

xK

Vx

VxK

A

HAKH

1but

11

baw

bawha

pKpKpKpH

KKKKKH

21

21

21or


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Hydrolysis

If the ionisation constant of the acid and base are equal,

i.e. Ka= Kb, pH = pKw=7 and the solution is neutral,although the hydrolysis may be considerable.

Ka> Kb, pH < 7

Kb> Kc, pH > 7

The pH of a solution of ammonium acetate is given by:

pH = 7.0 + 2.37 2.37 = 7.0 and the solution isapproximately neutral

On the other hand, for a solution of ammonium formate

pH = 7.0 +1.88 2.37 = 6.51

Formic acid; Ka = 1.77 x 10-4; pKa = 3.75 and the solution

reacts slightly acid.


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pH Buffers

A solution of 0.0001 M HCl should have a pH of 4, butthis solution is extremly sensitive to traces of alkali fromthe glass container and ammonia in the air.

Likewise a solution of 0.0001 M NaOH should have a pH

of 10 but this solution is sensitive to carbon dioxide in theair.

An aqueous solution of KCl pH = 7

Likewise, an aqueous solution of NH4Ac pH = 7

The addition of 1 mL of 1M HCl to 1 litre of KCl changesthe pH to 3

The addition of 1 mL of 1M HCl to 1 litre of NH4Ac doesnot change the pH much at all.


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pH Buffers

This is because the H+ ions added are mopped up by theacetate ions: H+ + Ac- HAc(the equilibrium lies very

much to the right hand side of this equation).

The ammonium acetate it is behaving as a buffer andresisting changes in pH. A buffer possesses reservedacidity and reserved alkalinity.


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pH Buffers For a buffer formed from a weak acid (HA) and the salt

of a weak acid (MA). HA H+ + A-

Making an approximation of activities concentrations

If the concentration of the acid = ca

The concentration of the salt = cs

Then the concentration of the un-dissociated portion ofthe acid = ca [H

+]

The solution must be electrically neutral [A-] = cs + [H+]

a

KA

HAH


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pH Buffers

The quadratic equation can be solved for [H+]

It can also be simplified by considering that in a mixtureof weak acid with its salt, the dissociation constant of theacid is repressed by the common ion effect. Thehydrogen ion concentration is negligibly small incomparison with ca and cs

Equation reduces to:

as

a KHc

HcH

aas

a KHKc

cH

Salt

Acid


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pH Buffers

Or

If the concentrations of an acid and salt are equal (i.e.half neutralised), then pH = pKa

For acetic acid, Ka = 1.82 x 10-5, pKa = 4.74

At half titre point a 0.1 M HAc solution would have a pHof 4.74. This would also be true for higher and lowerconcentrations of HAc, e.g 1.0 M and 0.05 M HAc

BaseSaltlogorSaltBaseOH

Similarly

AcidSaltlog

-

bb

a

pKpOHK

pKpH


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pH Buffers

If we add a small concentration of H+ ions they would

combine with acetate ions to form acetic acid.

H+ + CH3COO- CH3COOH

Similarly a small concentration of OH- added willcombine with H+ ions from dissociation of HAc and form

water. More acetic will dissociate to replace the H+ ionsdepleted in this manner.

Example: Calculate the pH of a solution produced byadding 10 mL of 1M HCl to 1 liter of a solution which is

0.1M in acetic acid and 0.1M in sodium acetate. The pH of the acetic acid sodium acetate buffer is

given by:


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pH Buffers

Neglecting the volume change from 1000 to 1010 mL. The HCl reacts with acetate ion forming pratically

undissociated acetic acid.

H+ + CH3COO- CH3COOH

Therefore [CH3COO-] = 0.1- 0.01 = 0.09 And [CH3COOH] = 0.1 + 0.01 = 0.11

pH = 4.74 + log 0.09/0.11 = 4.74 0.09 = 4.65

74.400.074.4Acid

Saltlog apKpH


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pH Buffers The addition of the strong acid to the buffer change the

pH by 4.74 - 4.65 = 0.09 pH Adding 10 mL of 1M HCl to 1 litre of pure water (pH 7),

the pH would have changed from 7 tolog(0.01) = 2, by5 pH unites.

A solution that contains equal concentrations of acid andsalt, or a half neutralized solution, has the maximumbuffer capacity. Other mixtures will also haveconsiderable buffer capacity, but the pH will differ slightly

from the half-neutralized acid. E.g. a quarter neutralized solution of acid, [Acid] = 3[Salt]

pH = pKa + log

= pKa 0.477


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pH Buffers E.g. a three quarter neutralized acid solution, 3[Acid] =

[Salt] pH = pKa + log 3

= pKa + 0.477

Generally speaking the buffering capacity is maintained

for mixtures within the range:

1 acid : 10 salt and 10 acid : 1 salt

pH = pKa 1

The concentration of the acid is usually of the order 0.05-0.20 M.

Similar remarks apply to buffers of weak bases(NH3/NH4Cl)


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pH Buffers

HAc (x mL) NaAc (y mL) pH

9.5 0.5 3.42

9.0 1.0 3.72

8.0 2.0 4.057.0 3.0 4.27

6.0 4.0 4.45

5.0 5.0 4.63

4.0 6.0 4.80

3.0 7.0 4.99

2.0 8.0 5.23

1 0 9 0 5 57

pH of 0.2 M HAc, 0.2 M NaAc Buffer Mixtures 10 mL Vol.

Date post:	06-Apr-2018
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Acid Base Hydrolysis Worked Examples

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