Acids and bases, pH and buffers

Dr. Mamoun AhramLecture 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: waterNH3 (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

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 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

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=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?

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 equilibriumLe Châtelier’s principle

What is buffer?

Titration

Midpoint

Buffering capacity

Conjugate bases

Acid Conjugate baseCH3COOH 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

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.

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 bufferNote 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.

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)

Acid-Base Imbalances

• pH< 7.35 acidosis

• pH > 7.45 alkalosis

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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

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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.

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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

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

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

Buffering saliva

Salivary carbonic

anhydrase