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

Ammonia (NH3) + acid (HA)  ammonium ion (NH4+) + A- 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) → H+(aq) + Cl-(aq) NaOH(aq) → Na+(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

grams = 58.4 x 5 moles x 0.1 liter = 29.29 g 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

(20.1 g/1000 mEq) x (1000 mg/g) x (5 mEq /L) = 100 mg/L 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) Physiological gram mEq/L 1 Eq Major electrolytes ? 136-145 = 23.1 g Na+ 98-106 - 35.5 g Cl- 3 (24.3)/2 = 12.15 g Mg+2 4.5-6.0 (40.1/2) = ~20.05 g Ca2+ 3.4-5.0 39.1 g K+ 25-29 61 g HCO3- 2 SO3-2 and HPO43-

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

MacidVacid = MbaseVbase 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 Conjugate base CH3COOH CH3COONa (NaCH3COO) H3PO4 NaH2PO4 H2PO4- (or NaH2PO4) Na2HPO4 H2CO3 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 CO2 + H20 H2CO3 H+ + HCO3- Blood (instantaneously) Lungs (within minutes) Excretion 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

Respiratory Acidosis H+ + HCO3-  H2CO3  CO2 + H2O

Respiratory Alkalosis H+ + HCO3-  H2CO3  CO2 + H2O

Metabolic Acidosis H+ + HCO3-  H2CO3  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.

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 but>7.40 Met. Acidosis but<7.40 Met. alkalosis

Partially compensated pH pCO2 HCO3- Res.Acidosis Res.Alkalosis Met. Acidosis Met.Alkalosis

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 pH 7.48 PaCO2 32 HCO3- 24 pH 7.32 PaCO2 48 HCO3- 25

Answers to Practice ABG’s Respiratory alkalosis Respiratory acidosis Metabolic acidosis Compensated Respiratory acidosis Metabolic alkalosis Compensated 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