Acid-Base Equilibria(Buffers )

Green & Damji Chapter 8, Section 18.2

Chang Chapter 16

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A buffer solution is a solution of:

1. A weak acid or a weak base and

2. The salt of the weak acid or weak base

Both must be present!

A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.

16.3

16.3

Add strong acid: H+ ions react with conj base… H+ ions are therefore removed from solution, pH increases back to near its original level.

H+ (aq) + CH3COO- (aq) CH3COOH (aq)

Add strong base: OH- (aq) reacts with undissociated acid… OH- (aq) are therefore removed from solution, pH decreases back to near its original level.

OH- (aq) + CH3COOH (aq) CH3COO- (aq) + H2O (l)

Consider an equal molar mixture of

CH3COOH and CH3COONa

CH3COOH (aq) CH3COO- (aq) + H+ (aq)

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Add strong acid: H+ ions react with base… H+ ions are therefore removed from solution,

pH increases back to near its original level.

NH3(aq) + H+(aq) NH4+(aq)

OH- (aq) + NH4+(aq) H2O (l) + NH3(aq)

Consider an equal molar mixture of

NH3 and NH4+ (from ammonium chloride)

NH3(aq) + H2O (l) NH4+(aq) + OH- (aq)

Add strong base: OH- (aq) reacts with ammonium ions… OH- (aq) are therefore removed from solution, pH decreases back to near its original level.

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Which of the following are buffer systems?

(a) KF/HF

(b) KBr/HBr,

(c) Na2CO3/NaHCO3

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Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na2CO3/NaHCO3

(a) HF is a weak acid and F- is its conjugate base

buffer solution

(b) HBr is a strong acid

not a buffer solution

(c) CO32- is a weak base and HCO3

- is its conjugate acid

buffer solution

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Chemistry In Action: Maintaining the pH of Blood

16.3

CO2(g) + H2O (g) H+ (aq) + HCO3- (aq)

Maintaining the pH of blood (7.4) is critical because

enzymes only function effectively over a limited pH range.The buffer system in action here is carbonic acid and sodium bicarbonate…

H2CO3 (g) H+ (aq) + HCO3- (aq)

The buffer system is typically written as:

Chemistry In Action: Maintaining the pH of Blood

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To make a buffer – Method 1

• Mix a soln of a weak acid (HA) and a salt of the weak acid (NaA)– Ex: ethanoic acid and sodium ethanoate

• Mix a soln of a weak base and a salt of the weak base– Ex: ammonia and ammonium chloride

To make a buffer – Method 2

• Mix a small amount of a strong base to an excess of weak acid– Ex: sodium hydroxide with excess ethanoic

acid

• Mix a small amount of a strong acid to an excess of weak base– Ex: hydrochloric acid to excess ammonia

Calculations involving buffers:For buffers systems involving weak acids and their conjugate base, we can use the formula for the acid dissociation constant:

We can assume that the [HA] is the concentration of the acid and the [A-] is the concentration of the conjugate base… and rearrange to this form…

Ka =[H+][A-][HA]

[H+] =Ka [HA]

[A-]

Calculations involving buffers:Or this form:

The pH of a buffer system depends on…• Ka of the weak acid• ratio of conj base to acid

– pH of a buffer does not change with dilution… but is less effective when buffer components decrease

pH = pKa + log [A-][HA]

Henderson-Hasselbach equation

Or…. pH = pKa - log [HA-][A-]

Calculations involving buffers:For buffer systems involving a weak base and its conjugate acid…

we can use the formula for the base dissociation constant:

We can assume that the [B] is the concentration of the base and the [BH+] is the concentration of the conjugate base… and rearrange to this form…

Kb =[BH+][OH-]

[B]

[OH-] =Kb [B]

[BH+]

Calculations involving buffers:

Or this form:

pOH = pKb + log [BH+][B]

Henderson-Hasselbach equation

Or…. pH = pKa - log[B]

[BH+]

Effective buffers• Concentration of the weak acid & its salt or

weak base & its salt must be >> than the strong acid/base added

• Conc of weak acid = conc of conj base• pH = pKa• Effective buffer range of any weak

acid/base is in the range of pKa + 1

Solid sodium ethanoate is added to 0.200 M ethanoic acid until it reaches a concentration of 0.0500 M.

Calculate the pH of the buffer solution formed. [Ka for ethanoic acid is 1.74 x 10-5 M.]

Assume no volume change on dissolving the solid.

Try this….

Solid sodium ethanoate is added to 0.200 M ethanoic acid until it reaches a concentration of 0.0500 M.

Calculate the pH of the buffer solution formed. [Ka for ethanoic acid is 1.74 x 10-5 M.]

Assume no volume change on dissolving the solid.

Method 1:

[H+] = 1.74 x 10-5 x (0.200/0.0500)

= 6.95 x 10-5 M

pH = - log [H+] = -log (6.95 x 10-5 M)

= 4.16

[H+] =Ka [HA]

[A-]

Solid sodium ethanoate is added to 0.200 M ethanoic acid until it reaches a concentration of 0.0500 M.

Calculate the pH of the buffer solution formed. [Ka for ethanoic acid is 1.74 x 10-5 M.]

Assume no volume change on dissolving the solid.

Method 2:

pH = log (1.74 x 10-5 ) – log (0.200/0.0500)

= 4.76 – 0.6

= 4.16

pH = pKa - log[HA]

[A-]

Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system.

NH4+ (aq) H+ (aq) + NH3 (aq)

pH = pKa + log[NH3][NH4

+]pKa = 9.25 (you’d have to look this up)

pH of the buffer = 9.25 + log[0.30][0.36]

= 9.17

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Textwork: 18.2

Read Section 18.2 on pp. 221-223

Do Ex 18.2 # 1-6

Due: _______________________