Acids and bases, pH and buffers Dr. Mamoun Ahram Lecture 2.
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Transcript of Acids and bases, pH and buffers Dr. Mamoun Ahram Lecture 2.
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
Amphoteric substances
• Example: waterNH3 (g) + H2O(l) ↔ NH4
+(aq) + OH–
(aq)
HCl(g) + H2O(l) → H3O+(aq) + Cl-
(aq)
Rule
• The stronger the acid, the weaker the conjugate base
HCl(aq) → H+(aq) + Cl-
(aq)
NaOH(aq) → Na+(aq) + OH-
(aq)
HC2H3O2 (aq) ↔ H+(aq) + C2H3O2
-(aq)
NH3 (aq) + H2O(l) ↔ NH4+
(aq) + OH-(aq)
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
But…
• Molarity (and normality) is not useful for understanding neutralization reactions.
1M HCL neutralizes 1M NaOHBut…
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 1 Eq mEq/L Physiological gram
Na+ = 23.1 g 136-145 ?
Cl- - 35.5 g 98-106 ?
Mg+2 (24.3)/2 = 12.15 g 3 ?
Ca2+ (40.1/2) = ~20.05 g 4.5-6.0 ?
K+ 39.1 g 3.4-5.0 ?
HCO3- 61 g 25-29 ?
SO3-2 and HPO43- 2 ?
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 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
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=1H2SO3=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?
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?
Conjugate bases
Acid Conjugate baseCH3COOH CH3COONa (NaCH3COO)
H3PO4 NaH2PO4
H2PO4- (or NaH2PO4) Na2HPO4
H2CO3 NaHCO3
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
Equivalencepoint
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.
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
Blood buffering
CO2 + H20 H2CO3 H+ + HCO3-
Blood (instantaneously)
Lungs
(within minutes)
Excretion via kidneys (hours to days)
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
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)
ExerciseH+ + 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)
78
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.
79
Complete vs. partial compensation
• May be complete if brought back within normal limits
• Partial compensation if range is still outside norms.
Acid-Base Disorder Primary Change Compensatory Change
Respiratory acidosis pCO2 up HCO3- up
Respiratory alkalosis pCO2 down HCO3- down
Metabolic acidosis HCO3- down PCO2 down
Metabolic alkalosis HCO3- up PCO2 up
FULLY COMPENSATED
pH pCO2 HCO3-
Resp. acidosis Normal
But<7.40
Resp. alkalosis Normal
but>7.40
Met. Acidosis Normal
but<7.40
Met. alkalosis Normal
but>7.40
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 PaCO2 32 HCO3- 24
2. pH 7.32 PaCO2 48 HCO3- 25
3. pH 7.30 PaCO2 40 HCO3- 18
4. pH 7.38 PaCO2 48 HCO3- 28
5. pH 7.49 PaCO2 40 HCO3- 30
6. pH 7.35 PaCO2 48 HCO3- 27
7. pH 7.45 PaCO2 47 HCO3- 29
8. pH 7.31 PaCO2 38 HCO3- 15
9. pH 7.30 PaCO2 50 HCO3- 24
10. pH 7.48 PaCO2 40 HCO3- 30
Answers to Practice ABG’s
1. Respiratory alkalosis2. Respiratory acidosis3. Metabolic acidosis4. Compensated Respiratory acidosis5. Metabolic alkalosis6. Compensated Respiratory acidosis7. Compensated Metabolic alkalosis8. Metabolic acidosis9. Respiratory acidosis10. 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