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Acid-Base Eqm (4): Buffer Solutions. Acid-Base Eqm (4): Buffer Solutions. p.01. 0.001 mol HCl. 0.001. = 0.010 M. [H 3 O + ] =. 100/1000. 100 cm 3 H 2 O.  pH would decrease from 7 to 2 after adding 0.001 mol HCl at 298K!.

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C. Y. Yeung (CHW, 2009)

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Acid-Base Eqm (4):Buffer Solutions

Acid-Base Eqm (4):Buffer Solutions

p.01

0.001 mol HCl

0.001

= 0.010 M

[H3O+] =

100/1000

100 cm3 H2O

 pH would decrease from 7 to 2 after adding 0.001 mol HCl at 298K!

Q.:What is the new pH after addition of 0.001 mol HCl at 298K? (assume no volume change)

 pH = - log(0.010) = 2

C. Y. Yeung (CHW, 2009)

p.02

x2

0.001 mol HCl

= 1.7610-5

0.1 – x

x = 1.3310-3

pH = 2.88

0.001

= 0.01M

extra [H3O+] =

100/1000

y(0.01+y)

= 1.7610-5

0.1 – y

 pH would decrease from 2.88 to 1.99 after adding 0.001 mol HCl at 298K!

Q.:What is the new pH after addition of 0.001 mol HCl at 298K? (assume no volume change)

100 cm3 0.1M CH3COOH Ka = 1.7610-5M

y = 1.7610-4

pH = -log (0.01+y) = 1.99

p.03

x (0.1+x)

= 1.7610-5

0.1 – x

0.001 mol HCl

x = 1.7610-5

pH = 4.75

(0.09+y) y

= 1.7610-5

0.11 – y

 pH would decrease from 4.75 to 4.67 after adding 0.001 mol HCl at 298K!

Q.:What is the new pH after addition of 0.001 mol HCl at 298K? (assume no volume change)

100 cm3 0.1M CH3COOH and 0.1M CH3COO- Na+

After addition of HCl: extra [H3O+] = 0.01M

[CH3COO-] remained = 0.1 – 0.01 = 0.09M

y = 2.1510-5

pH = -log (y) = 4.67

p.04

tends to resist the changes in pH when small amounts of acid is added

i.e. it is a “BUFFER” solution!

Note the difference …

It tends to resist changes in pH when small amounts of acid or base are added, and their pH is not affected by dilution.

p.05

CH3COO-+ H3O+

CH3COOH + H2O

When acid is added, the additional H3O+ would combine with the conjugate base so that the eqm shifts BW.

When base is added, some H3O+ ions are reacted. Some acid molecules dissociate to produce H3O+ so that the eqm shifts FW.

It is an ACIDIC BUFFER solution! (i.e. weak acid + conjugate base)

100 cm3 0.1M CH3COOH and 0.1M CH3COO- Na+

i.e. [H3O+] does not change much, pH remains almost unchanged.

i.e. [H3O+] does not change much, pH remains almost unchanged.

p.06

NH4++ OH-

NH3 + H2O

When acid is added, some OH- are neutralized, some NH3 molecules would dissociate so that the eqm shifts FW.

When base is added, additional OH- ions would combine with NH4+ so that the eqm shifts BW.

It is an BASIC BUFFER solution! (i.e. weak base + conjugate acid)

100 cm3 0.1M NH3 and 0.1M NH4Cl

i.e. [OH-] does not change much, pH remains almost unchanged.

i.e. [OH-] does not change much, pH remains almost unchanged.

pH of buffer is NOT affected by dilution.

p.07

conjugate base

constant at constant temp.

acid

pH of the buffer could be adjusted by varying the ratio of HA and A-.

[A-(aq)]

- log

pH

pKa =

[HA(aq)]

 WHY?

During dilution, both [A-] and [HA] are diluted to the same extent. Thus the ratio [A-]/[HA] is unchanged.

Therefore, pH of buffer remains unchanged in dilution!

p.08

[CH3CH(OH)COO-]

log

pKa =

pH –

[CH3CH(OH)COOH]

[CH3CH(OH)COO-]

2.14 104 =

[CH3CH(OH)COOH]

Q.1:Lactic acid (乳酸), CH3CH(OH)COOH, is produced in muscle tissues during physical exertion.

At body temperature, human blood pH is 7.40.

Whether the lactic acid produced exists mainly in form of “undissociated molecule” or “dissociated lactate ions” in blood at body temperature?

(Given: Ka of lactic acid at body temperature = 8.5010-4 mol dm-3)

With a given Ka, the ratio of [salt] and [acid] could be found from the pH value.

 Lactic acid formed is mainly in form of dissociated lactic ions.

p.09

After mixing,

new [CH3COOH] = 0.10M

new [CH3COONa] = 0.20M

x(0.20+x)

pH of acidic buffer could be adjusted by [acid] and [salt].

1.7510-5 =

0.10 – x

Q.2:Calculate the pH value of the resultant solution for 10cm3 of 0.20M CH3COOH are mixed with 10cm3 of 0.40M CH3COONa. (Ka = 1.7510-5 mol dm-3)

x = 8.75 10-6

 pH = 5.06

An acid-salt mixture (acid and salt have comparable conc.)  ACIDIC BUFFER!

p.10

After mixing,

new [NH3] = 0.50M

new [NH4+] = 0.50M

x(0.50+x)

pH of basic buffer could be adjusted by [base] and [salt].

1.7810-5 =

0.50 – x

Q.3:Calculate the pH value of the resultant solution for 10cm3 of 1.00M NH3 are mixed with 10cm3 of 1.00M NH4Cl. (Kb = 1.7810-5 mol dm-3)

x = 1.78 10-5

 pOH = 4.75

 pH = 9.25

A base-salt mixture (base and salt have comparable conc.)  BASIC BUFFER!

p.11

Acid-Base Equilibria & pH Regulation in Blood Plasma

p.12

Next ….

Acid-Base Indicators (Book 2 p. 164 – 168)

Assignment

Study the examples in p. 158 - 163,

p.163 Check point 18.2 ,

p.179 Q.1 - 6

[due date: 23/4(Thur)]