Transcript Acids and bases, pH and buffers Dr. Mamoun Ahram Lecture 2

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
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
Exercise
• What is the normality of H2SO3 solution made by
dissolving 6.5 g into 200 mL? (MW = 98)?
But…
• Molarity (and normality) is not useful for
understanding neutralization reactions.
1M HCL neutralizes 1M NaOH
But…
1M HCl does not neutralize 1M H2SO3
• Why?
Equivalents
• The amount of molar mass (g) of hydrogen ions that
an acid will donate
– or a base will accept
• 1 mole HCl = 1 mole [H+] = 1 equivalent
• 1 mole H2SO4 = 2 mole [H+] = 2 equivalents
Examples
• 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
Exercise
• calculate the number of equivalents of:
40g of Mg+2
16g of Al+3
Mg++ : 40g x (1mol/24g) x (2eq/1mol) = 3.3 eq
Al3+: 16g x (27g/1mol) x (3eq/1mol) = 1.8 eq
Exercises
• Calculate milligrams of Ca+2 in blood if total
concentration of Ca+2 is 5 mEq/L. (MW = 40.1)
(20.1 g/1000 mEq) x (1000 mg/g) x (5 mEq /L) = 100 mg/L
• What is the normality of H2SO3 made by dissolving
6.5g in 200 ml? equivalents? (MW = 98)
Examples (calculate grams)
Major electrolytes
Na+
1 Eq
mEq/L
= 23.1 g
136-145
?
Cl-
- 35.5 g
98-106
?
Mg+2
Ca2+
(24.3)/2 = 12.15 g
3
?
(40.1/2) = ~20.05 g
4.5-6.0
?
K+
39.1 g
3.4-5.0
?
HCO3-
61 g
25-29
?
2
?
SO3-2 and HPO43-
Physiological gram
Titration and equivalence point
• 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 is the
concentration of the HCl?
Step 1 - Determine [OH-]
Step 2 - Determine the number of moles of OHStep 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+
• Consider the charges (or normality)
Modified equation
Na x Va= Nb x Vb
Na = normality of acid
Va = volume of acid
Nb = normality of base
Vb = volume of base
NaOH=1
H2SO3=2
H3PO4=3
Exercises
• 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?
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?
Exercise
• What is the pH of 0.01 M HCl?
• What is the pH of 0.01 N H2SO3?
• What is the pH of a solution of 1 x 10-11 HCl?
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
Le Châtelier’s principle
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)
• 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
Equivalence
point
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
Excercises
• 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
Bicarbonate buffer
CO2 + H20
H2CO3
H+ + HCO3-
Blood buffering
Blood (instantaneously)
CO2 + H20
Lungs
(within
minutes)
H2CO3
H+ + HCO3Excretion via
kidneys (hours
to days)
Arterial blood gases (ABG)
Calculations…
• The ratio of bicarbonate to carbonic acid determines
the pH of the blood
– Normally the ratio is about 20:1 bicarbonate to
carbonic acid
• Blood pH can be calculated from this equation:
– pH = pK + log (HCO3-/H2CO2)
– pK is the dissociation constant of the buffer, 6.10
– H2CO3 =0.03 x pCO2
Titration curve of bicarbonate buffer
Note pKa
Why is this buffer effective?
• Even though the normal blood pH of 7.4 is outside
the optimal buffering range of the bicarbonate
buffer, which is 6.1, this buffer pair is important due
to two properties:
– bicarbonate is present in a relatively high
concentration in the ECF (24mmol/L)
– the components of the buffer system are effectively
under physiological control: the CO2 by the lungs, and
the bicarbonate by the kidneys
– It is an open system (not a closed system like in
laboratory)
Open system
• An open system is a system that continuously
interacts with its environment.
Exercise
H+ + HCO3-  H2CO3  CO2 + H2O
• Blood plasma contains a total carbonate (HCO3- and
CO2) of 2.52 x 10-2 M. What is the HCO3-/CO2 ratio
and the concentration of each buffer component at
pH 7.4?
Exercise (continued)
H+ + HCO3-  H2CO3  CO2 + H2O
• What would the pH be if 10-2 M H+ is added and CO2
is eliminated (closed system)?
Exercise (continued)
H+ + HCO3-  H2CO3  CO2 + H2O
• What would the pH be if 10-2 M H+ is added under
physiological conditions (open system)?
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
72
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
Causes of respiratory acid-base
disorders
Causes of metabolic acid-base
disorders
Compensation
• Compensation: The change in HCO3- or pCO3 that
results from the primary event
• If underlying problem is metabolic, hyperventilation
or hypoventilation can help : respiratory
compensation.
• If problem is respiratory, renal mechanisms can bring
about metabolic compensation.
78
Complete vs. partial compensation
• May be complete if brought back within normal
limits
• Partial compensation if range is still outside norms.
79
Acid-Base Disorder
Change
Respiratory acidosis
Respiratory alkalosis
Metabolic acidosis
Metabolic alkalosis
Primary Change
pCO2 up
pCO2 down
HCO3- down
HCO3- up
Compensatory
HCO3- up
HCO3- down
PCO2 down
PCO2 up
FULLY COMPENSATED
pH
Resp. acidosis Normal
But<7.40
Resp. alkalosis Normal
but>7.40
Met. Acidosis Normal
but<7.40
Met. alkalosis Normal
but>7.40
pCO2
HCO3-
Partially compensated
pH
Res.Acidosis
Res.Alkalosis
Met. Acidosis
Met.Alkalosis
pCO2
HCO3-
EXAMPLES
Example 1
• Mr. X is admitted with severe attack of asthma. Her
arterial blood gas result is as follows:
– pH : 7.22
– PaCO2 : 55
– HCO3- : 25
– pH is low – acidosis
– paCO2 is high – in the opposite direction of the pH.
– HCO3- is Normal
• Respiratory Acidosis
Example 2
• Mr. D is admitted with recurring bowel obstruction has
been experiencing intractable vomiting for the last
several hours. His ABG is:
– pH : 7.5
– PaCO2 :42
– HCO3- : 33
• Metabolic alkalosis
Example 3
• Mrs. H is kidney dialysis patient who has missed his last 2
appointments at the dialysis centre. His ABG results:
– pH: 7.32
– PaCO2 : 32
– HCO3-: 18
• Partially compensated metabolic Acidosis
Example 4
• Mr. K with COPD.His ABG is:
– pH: 7.35
– PaCO2 : 48
– HCO3- :28
• Fully compensated Respiratory Acidosis
Example 5
• Mr. S is a 53 year old man presented to ED with the
following ABG.
– pH: 7.51
– PaCO2 : 50
– HCO3- : 40
• Metabolic alkalosis
Practice ABG’s
1. pH 7.48
2. pH 7.32
3. pH 7.30
4. pH 7.38
5. pH 7.49
6. pH 7.35
7. pH 7.45
8. pH 7.31
9. pH 7.30
10. pH 7.48
PaCO2 32
PaCO2 48
PaCO2 40
PaCO2 48
PaCO2 40
PaCO2 48
PaCO2 47
PaCO2 38
PaCO2 50
PaCO2 40
HCO3HCO3HCO3HCO3HCO3HCO3HCO3HCO3HCO3HCO3-
24
25
18
28
30
27
29
15
24
30
Answers to Practice ABG’s
1. Respiratory alkalosis
2. Respiratory acidosis
3. Metabolic acidosis
4. Compensated Respiratory acidosis
5. Metabolic alkalosis
6. Compensated Respiratory acidosis
7. Compensated Metabolic alkalosis
8. Metabolic acidosis
9. Respiratory acidosis
10. Metabolic alkalosis
Salivary buffers
• Salivary pH  6.3
• Main buffers:
– Bicarbonate
– Phosphate
– Proteins
• Below pH 5.5, demineralization usually follows
Flow rate
• [H2CO3] = 1.3 mM/L –almost constant, but [HCO3-] is not
• The greater the salivary flow, the more bicarbonate ions
available for combining with free hydrogen ions
– Normal salivary flow rates = 0.1 and 0.6 mL/minute
Buffering saliva
Salivary
carbonic
anhydrase