Prepared by
Prof. Odyssa Natividad R.M. Molo
 Consider a solution that contains not only a weak acid
(HC2H3O2) but also a soluble salt (NaC2H3O2) of that
acid.
NaC2H3O2(aq)  Na+(aq) + C2H3O2-(aq)
HC2H3O2(aq)  H+(aq) + C2H3O2-(aq)
 What happens when salt is added? (Remember Le
Chatelier’s Principle)
 Shifts equilibrium
 pH of solution increases
 [H+] decreases
The Common-Ion Effect
 “The dissociation of a weak electrolyte is decreased by
adding to the solution a strong electrolyte that has an
ion common with the weak electrolyte”
 Plays an important role in determining the pH of the
solution & the solubility of a slightly soluble salt
 Despite distinctive name, this is simply a special case of
Le Chatelier’s principle
Sample
Problem
1) What is the
 STEPS:
pH of a soln
made by
adding a 0.30
mol acetic
acid & 0.30
mol sodium
acetate to
enough
water to
make 1.0L of
soln?
1) Identify the major species in
the soln & consider their
acidity & basicity
2) Identify the important eqlbm
rxn
3) Calculate the initial & eqlbm
conc of each species that
participate in the eqlbm
4) Calculate the pH from the
eqlbm conc
Practice Exercise
1) Calculate the pH of a solution containing 0.085M
nitrous acid (Ka = 4.5 x 10-4) & 0.10M potassium
nitrite. What would be its pH if no salt were present.
2) Calculate the pH of a solution containing 0.20 M
acetic acid & 0.30M sodium acetate. What would be
the pH if no salt were present.
Buffered Solution/Buffers
 Solution that resist a drastic change in pH upon
addition of small amounts of acid or base
 Contain weak conjugate acid-base pairs
 Example:
 Human blood: bicarbonate-carbonic acid buffer




Normal pH: 7.35 – 7.45
Acidosis: condition when pH falls below 7.35
Alkalosis: when pH rises above 7.45
Death may result if blood pH < 6.8 or > 7.8
Composition of Buffers
 A buffer resist in pH because it contains both an acidic
specie to neutralize OH- ions & a basic one to
neutralize H+ ions. The acidic & basic species that
make up the buffer, however, must not consume each
other through a neutralization rxn. These
requirements are fulfilled by a weak acid-base
conjugate pair.
Composition cont….
 Buffers are often prepared by mixing a WA or a WB
with a salt of that acid or base.
 Example:


HC2H3O2 - C2H3O2- buffer (NaC2H3O2 & HC2H3O2)
NH4+ - NH3 buffer (NH4Cl & NH3)
 By choosing appropriate components & adjusting their
relative concentration, one can buffer a solution at
virtually any pH
How Buffer Works
 Ex: Buffer composed of weak acid (HX) & its salt
(MX)
HX(aq)  H+(aq) + X-(aq)
 If OH- ions are added, it reacts with the acid
component of the buffer to produce water & its
base component (X-)
OH-(aq) + HX(aq)  H2O(l) + X-(aq)
 Result: [HX] dec & [X-] inc
How Buffer Works cont…
 If H+ ions are added, it reacts with the base
component of the buffer to produce water & its
acid component (HX)
H+(aq) + X-(aq)  HX(aq)
OR
H3O+(aq) + X-(aq)  H2O(l) + HX(aq)
 Result: [HX] inc & [X-] dec
Buffer pH
HX(aq)  H+(aq) + X-(aq)
so
 pH is determined by 2 factors:
 (1) value of Ka
 (2) ratio of conjugate acid-base pair
 If [HX] = [X-], [H+] = Ka, pH = pKa
 Result: select a buffer whose acid form has a pKa close to
desired pH
2 important characteristics of
buffer:
 (1) its capacity & (2) its pH
 Buffer capacity
 Is the amount of acid or base the buffer can neutralize
before the pH begins to change to an appreciable degree
 Depends on the amount of Acid & Base from which the
buffer is made
 pH of buffer depends on
 The Ka for the acid & on the relative concentration of A
& B that comprise the buffer
To solve buffer pH
 Use the same procedure to treat the common-ion effect OR
the Henderson-Hasselbalch equation
 Sample Problem: Calculate the pH of a buffer that is 0.12M
lactic acid (Ka = 1.4 x 10-4) & 0.10M sodium lactate.
 Practice: Calculate the pH of a buffer composed of 0.12M
benzoic acid, HC7H5O2, (Ka = 6.3 x 10-5) & 0.20M sodium
benzoate
Addition of SA/SB to Buffers
 When a SA is added, the H+ is consumed by X- to
produce HX;
 Result:[HX] inc & [X-] dec
 When a SB is added, the OH- is consumed by HX
to produce X Result: [HX] dec & [X-] inc
Steps to calculate pH
1) Identify the neutralization rxn (Strong Acid &
Weak Base or Strong Base & Weak Acid)
2) Analyze and set-up condition before & after
neutralization
3) Calculate pH based on what is left during
equilibrium condition.
Sample Exercise
 A buffer is made by adding 0.300 mol acetic acid &
0.300 mol sodium acetate to enough water to make
1.00L soln. Calculate its pH (a) at the start; (b) after
0.020 mol KOH is added; (c) after 0.010 mol HCl is
added. For comparison, what is the the pH of (d) 0.020
mol NaOH and (e) 0.010 mol HCl added in 1.00L pure
water.
Practice Exercise
What are the effects on the pH of adding (0.0060 mol
HCl and (b) 0.0060 mol NaOH to 0.300L of a buffer
solution that is 0.250M acetic acid (Ka = 1.8 x 10-5) &
0.560 M sodium acetate. For comparison, what is the pH
of (c) 0.0060 mol HCl & (d) 0.0060 mol NaOH in 0.300L
water.
2) A 1.00 L volume of buffer is made with concentrations of
0.350 M NaCHO2 (sodium formate) and 0.550M HCHO2
(Ka = 1.8 x 10-4). (a) What is the initial pH? (b) What is
the pH after the addition of 0.0050 mol HCl? (assume
that the volume remains 1.00L) (c) What would be the
pH after the addition of 0.0050 mol NaOH to the original
buffer?
1)