Acids and bases, pH and buffers
Dr. Mamoun Ahram
Lecture 2
ACIDS AND BASES
Acids versus bases
• Acid: a substance that produces H+ when dissolved
in water (e.g., HCl, H2SO4)
• Base: a substance that produces OH- when dissolved
in water (NaOH, KOH)
• What about ammonia (NH3)?
Brønsted-Lowry acids and bases
• The Brønsted-Lowry acid: any substance able to give
a hydrogen ion (H+-a proton) to another molecule
– Monoprotic acid: HCl, HNO3, CH3COOH
– Diprotic acid: H2SO4
– Triprotic acid: H3PO3
• Brønsted-Lowry base: any substance that accepts a
proton (H+) from an acid
– NaOH, NH3, KOH
Acid-base reactions
• A proton is transferred from one substance (acid) to
another molecule
Ammonia (NH3) + acid (HA)  ammonium ion (NH4+) + A-
–
–
–
–
Ammonia is base
HA is acid
Ammonium ion (NH4+) is conjuagte acid
A- is conjugate base
Water: acid or base?
• Both
• Products: hydronium ion (H3O+) and hydroxide
Amphoteric substances
• Example: water
NH3 (g) + H2O(l) ↔ NH4+(aq) + OH–(aq)
HCl(g) + H2O(l) → H3O+(aq) + Cl-(aq)
Acid-base reactions
Acid + base  salt + H2O
• Exceptions:
• Carbonic acid (H2CO3)-Bicarbobate ion (HCO3-)
• Ammonia (NH3)-
Acid/base
strength
Rule
• The stronger the acid, the weaker the conjugate base
HCl(aq)
NaOH(aq)
→H
→ Na
+
+
Cl
(aq)
(aq)
+
(aq)
+ OH-(aq)
HC2H3O2 (aq) ↔ H+(aq) + C2H3O2-(aq)
NH3 (aq) + H2O(l) ↔ NH4+(aq) + OH-(aq)
Equilibrium constant
HA <--> H+ + A-
Ka: >1 vs. <1
Expression
• Molarity (M)
• Normality (N)
• Equivalence (N)
Molarity of solutions
moles = grams / MW
M = moles / volume (L)
grams = M x vol (L) x MW
Exercise
• How many grams do you need to make 5M NaCl
solution in 100 ml (MW 58.4)?
grams = 58.4 x 5 moles x 0.1 liter = 29.29 g
Normal solutions
N= n x M (where n is an integer)
• n =the number of donated H+
Remember!
The normality of a solution is NEVER less than the
molarity
Equivalents
• The amount of molar mass (g) of hydrogen ions that
an acid will donate
– or a base will accept
• 1M HCl = 1M [H+] = 1 equivalent
• 1M H2SO4 = 2M [H+] = 2 equivalents
Exercise
• What is the normality of H2SO3 solution made by
dissolving 6.5 g into 200 mL? (MW = 98)?
Example
• One equivalent of Na+ = 23.1 g
• One equivalent of Cl- - 35.5 g
• One equivalent of Mg+2 = (24.3)/2 = 12.15 g
• Howework:
Calculate milligrams of Ca+2 in blood if total
concentration of Ca+2 is 5 mEq/L.
Titration
• The concentration of acids and bases can be
determined by titration
Excercise
• A 25 ml solution of 0.5 M NaOH is titrated until
neutralized into a 50 ml sample of HCl. What was the
concentration of the HCl?
• Step 1 - Determine [OH-]
• Step 2 - Determine the number of moles of OH• Step 3 - Determine the number of moles of H+
• Step 4 - Determine concentration of HCl
A 25 ml solution of 0.5 M NaOH is titrated
until neutralized into a 50 ml sample of HCl
• Moles of base = Molarity x Volume
• Moles base = moles of acid
• Molarity of acid= moles/volume
Another method
MacidVacid = MbaseVbase
Note
• What if one mole of acid produces two moles of H+
MacidVacid = 2MbaseVbase
Homework
• If 19.1 mL of 0.118 M HCl is required to neutralize
25.00 mL of a sodium hydroxide solution, what is the
molarity of the sodium hydroxide?
• If 12.0 mL of 1.34 M NaOH is required to neutralize
25.00 mL of a sulfuric acid, H2SO4, solution, what is
the molarity of the sulfuric acid?
Equivalence point
Ionization of water
H3O+ = H+
Equilibrium constant
Keq = 1.8 x 10-16 M
Kw
• Kw is called the ion product for water
PH
What is pH?
Acid dissociation constant
•
•
•
•
Strong acid
Strong bases
Weak acid
Weak bases
pKa
What is pKa?
HENDERSON-HASSELBALCH
EQUATION
The equation
pKa is the pH where 50% of acid is dissociated into
conjugate base
BUFFERS
Maintenance of equilibrium
What is buffer?
Titration
Midpoint
Buffering capacity
Conjugate bases
Acid
CH3COOH
H3PO4
H2PO4- (or NaH2PO4)
H2CO3
Conjugate base
CH3COONa (NaCH3COO)
NaH2PO4
Na2HPO4
NaHCO3
How do we choose a buffer?
Problems and solutions
• A solution of 0.1 M acetic acid and 0.2 M acetate ion. The pKa of
acetic acid is 4.8. Hence, the pH of the solution is given by
• Similarly, the pKa of an acid can be calculated
Exercise
• What is the pH of a buffer containing 0.1M HF and
0.1M NaF? (Ka = 3.5 x 10-4)
Homework
• What is the pH of a solution containing 0.1M HF and
0.1M NaF, when 0.02M NaOH is added to the
solution?
At the end point of the buffering capacity of a buffer,
it is the moles of H+ and OH- that are equal
Exercise
• What is the concentration of 5 ml of acetic acid
knowing that 44.5 ml of 0.1 N of NaOH are needed to
reach the end of the titration of acetic acid? Also,
calculate the normality of acetic acid.
Polyprotic weak acids
• Example:
Hence
Excercise
• What is the pH of a lactate buffer that contain 75%
lactic acid and 25% lactate? (pKa = 3.86)
• What is the pKa of a dihydrogen phosphae buffer
when pH of 7.2 is obtained when 100 ml of 0.1 M
NaH2PO3 is mixed with 100 ml of 0.1 M Na2HPO3?
Buffers in human body
• Carbonic acid-bicarbonate system (blood)
• Dihydrogen phosphate-monohydrogen phosphate
system (intracellular)
• Proteins
Blood buffering
Blood (instantaneously)
CO2 + H20
H2CO3
Lungs
(within
minutes)
H+ + HCO3Excretion via
kidneys (hours
to days)
Roles of lungs and kidneys
• Maintaining blood is balanced by the kidneys and the
lungs
• Kidneys control blood HCO3 concentration ([HCO3])
• Lungs control the blood CO2 concentration (PCO2)
Calculations…
Acidosis and alkalosis
• Can be either metabolic or respiratory
• Acidosis:
– Metabolic: production of ketone bodies (starvation)
– Respiratory: pulmonary (asthma; emphysema)
• Alkalosis:
– Metabolic: administration of salts or acids
– Respiratory: hyperventilation (anxiety)
Acid-Base Imbalances
• pH< 7.35 acidosis
• pH > 7.45 alkalosis
62
Respiratory Acidosis
H+ + HCO3-  H2CO3 
CO2 + H O
2
Respiratory Alkalosis
H+ + HCO3-  H2CO3  CO2 + H2O
Metabolic Acidosis
+
H + HCO -  H CO
3
2
3
 CO2 + H2O
Metabolic Alkalosis
H+
+ HCO3-  H2CO3  CO2 + H2O