3 Acids versus basesAcid: 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)?
4 Brønsted-Lowry acids and bases The Brønsted-Lowry acid: any substance able to give a hydrogen ion (H+-a proton) to another moleculeMonoprotic acid: HCl, HNO3, CH3COOHDiprotic acid: H2SO4Triprotic acid: H3PO3Brønsted-Lowry base: any substance that accepts a proton (H+) from an acidNaOH, NH3, KOH
5 Ammonia (NH3) + acid (HA) ammonium ion (NH4+) + A- Acid-base reactionsA proton is transferred from one substance (acid) to another moleculeAmmonia (NH3) + acid (HA) ammonium ion (NH4+) + A-Ammonia is baseHA is acidAmmonium ion (NH4+) is conjuagte acidA- is conjugate base
6 Water: acid or base?BothProducts: hydronium ion (H3O+) and hydroxide
13 Molarity of solutions moles = grams / MW M = moles / volume (L) grams = M x vol (L) x MW
14 grams = 58.4 x 5 moles x 0.1 liter = 29.29 g ExerciseHow 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 = g
15 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
16 ExerciseWhat is the normality of H2SO3 solution made by dissolving 6.5 g into 200 mL? (MW = 98)?
17 But…Molarity (and normality) is not useful for understanding neutralization reactions.1M HCL neutralizes 1M NaOHBut…1M HCl does not neutralize 1M H2SO3 Why?
18 EquivalentsThe amount of molar mass (g) of hydrogen ions that an acid will donateor a base will accept1 mole HCl = 1 mole [H+] = 1 equivalent1 mole H2SO4 = 2 mole [H+] = 2 equivalents
19 Examples One equivalent of Na+ = 23.1 g One equivalent of Cl- - 35.5 g One equivalent of Mg+2 = (24.3)/2 = g
20 Exercisecalculate the number of equivalents of: 40g of Mg+2 16g of Al+3Mg++ : 40g x (1mol/24g) x (2eq/1mol) = 3.3 eqAl3+: 16g x (27g/1mol) x (3eq/1mol) = 1.8 eq
21 (20.1 g/1000 mEq) x (1000 mg/g) x (5 mEq /L) = 100 mg/L ExercisesCalculate 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/LWhat is the normality of H2SO3 made by dissolving 6.5g in 200 ml? equivalents? (MW = 98)
23 Titration and equivalence point The concentration of acids and bases can be determined by titration
24 ExcerciseA 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
25 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 VolumeMoles base = moles of acidMolarity of acid= moles/volume
26 MacidVacid = MbaseVbase Another methodMacidVacid = MbaseVbase
27 Note What if one mole of acid produces two moles of H+ Consider the charges (or normality)
28 Modified equation Na x Va= Nb x Vb Na = normality of acid Va = volume of acidNb = normality of baseVb = volume of baseNaOH=1H2SO3=2H3PO4=3
29 ExercisesIf 19.1 mL of M HCl is required to neutralize 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 mL of a sulfuric acid, H2SO4, solution, what is the molarity of the sulfuric acid?
53 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 bySimilarly, the pKa of an acid can be calculated
54 ExerciseWhat 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?
55 At the end point of the buffering capacity of a buffer, it is the moles of H+ and OH- that are equal Equivalencepoint
56 ExerciseWhat 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.
59 ExcercisesWhat 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?
60 Buffers in human body Carbonic acid-bicarbonate system (blood) Dihydrogen phosphate-monohydrogen phosphate system (intracellular)Proteins
64 Calculations…The ratio of bicarbonate to carbonic acid determines the pH of the bloodNormally the ratio is about 20:1 bicarbonate to carbonic acidBlood pH can be calculated from this equation: pH = pK + log (HCO3-/H2CO2)pK is the dissociation constant of the buffer, 6.10H2CO3 =0.03 x pCO2
66 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 kidneysIt is an open system (not a closed system like in laboratory)
67 Open systemAn open system is a system that continuously interacts with its environment.
68 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?
69 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)?
70 Exercise (continued) H+ + HCO3- H2CO3 CO2 + H2O What would the pH be if 10-2 M H+ is added under physiological conditions (open system)?
71 Acidosis and alkalosis Can be either metabolic or respiratoryAcidosis:Metabolic: production of ketone bodies (starvation)Respiratory: pulmonary (asthma; emphysema)Alkalosis:Metabolic: administration of salts or acidsRespiratory: hyperventilation (anxiety)
78 CompensationCompensation: The change in HCO3- or pCO3 that results from the primary eventIf 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 limitsPartial compensation if range is still outside norms.
80 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
81 FULLY COMPENSATED pH pCO2 HCO3- Resp. acidosis Normal But<7.40 Resp. alkalosisbut>7.40Met. Acidosisbut<7.40Met. alkalosis
84 Example 1Mr. X is admitted with severe attack of asthma. Her arterial blood gas result is as follows:pH : 7.22PaCO2 : 55HCO3- : 25pH is low – acidosispaCO2 is high – in the opposite direction of the pH.HCO3- is NormalRespiratory Acidosis
85 Example 2Mr. D is admitted with recurring bowel obstruction has been experiencing intractable vomiting for the last several hours. His ABG is:pH : 7.5PaCO2 :42HCO3- : 33Metabolic alkalosis
86 Example 3Mrs. H is kidney dialysis patient who has missed his last 2 appointments at the dialysis centre. His ABG results:pH: 7.32PaCO2 : 32HCO3-: 18Partially compensated metabolic Acidosis
87 Example 4 Mr. K with COPD.His ABG is: pH: 7.35PaCO2 : 48HCO3- :28Fully compensated Respiratory Acidosis
88 Example 5Mr. S is a 53 year old man presented to ED with the following ABG.pH: 7.51PaCO2 : 50HCO3- : 40Metabolic alkalosis
90 Answers to Practice ABG’s Respiratory alkalosisRespiratory acidosisMetabolic acidosisCompensated Respiratory acidosisMetabolic alkalosisCompensated Metabolic alkalosis
91 Salivary buffers Salivary pH 6.3 Main buffers: BicarbonatePhosphateProteinsBelow pH 5.5, demineralization usually follows
92 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 ionsNormal salivary flow rates = 0.1 and 0.6 mL/minute
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