Biochemistry PPT Part-2

8/11/2019 Biochemistry PPT Part-2

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Acids and bases, pH and buffers

Dr. Mamoun Ahram

Lecture 2


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ACIDS AND BASES


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Acids versus bases

Acid: a substance that produces H+ when dissolved

in water (e.g., HCl, H2SO4)

Base: a substance that produces OH- when dissolvedin water (NaOH, KOH)

What about ammonia (NH3)?


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Brnsted-Lowry acids and bases

The Brnsted-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

Brnsted-Lowry base: any substance that accepts aproton (H+) from an acid

NaOH, NH3, KOH


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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


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Water: acid or base?

Both

Products: hydronium ion (H3O+) and hydroxide


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Amphoteric substances

Example: water

NH3 (g)+ H2O(l) NH4+

(aq)+ OH

(aq)

HCl(g)+ H2O(l) H3O+

(aq)+ Cl-(aq)


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Acid-base reactions

Acid + basesalt + H2O

Exceptions:

Carbonic acid (H2CO3)-Bicarbobate ion (HCO3-)

Ammonia (NH3)-


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Acid/base

strength


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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)


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Equilibrium constant

HA H+ + A-

Ka

: >1 vs.


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Expression

Molarity (M)

Normality (N)

Equivalence (N)


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Molarity of solutions

moles = grams / MW

M = moles / volume (L)

grams = M x vol (L) x MW


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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


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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 themolarity


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Equivalents

The amount of molar mass (g) of hydrogen ions that

an acid will donate

or a base will accept

1M HCl = 1M [H+] = 1 equivalent

1M H2SO4 = 2M [H+] = 2 equivalents


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Exercise

What is the normality of H2SO3solution made by

dissolving 6.5 g into 200 mL? (MW = 98)?


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Example

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

Howework:

Calculate milligrams of Ca+2 in blood if total

concentration of Ca+2 is 5 mEq/L.


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Titration

The concentration of acids and bases can be

determined by titration


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Excercise

A 25 ml solution of 0.5 M NaOH is titrated until

neutralized into a 50 ml sample of HCl. What was 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


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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


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Another method

MacidVacid= MbaseVbase


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Note

What if one mole of acid produces two moles of H+

MacidVacid= 2MbaseVbase


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Homework

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?


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Equivalence point


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Ionization of water

H3O+ = H+


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Equilibrium constant

Keq = 1.8 x 10-16M


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Kw

Kw is called the ion product for water


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PH


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What is pH?


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Acid dissociation constant

Strong acid

Strong bases

Weak acid

Weak bases


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pKa


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What is pKa?


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HENDERSON-HASSELBALCHEQUATION


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The equation

pKa is the pH where 50% of acid is dissociated into

conjugate base


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BUFFERS


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Maintenance of equilibrium


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What is buffer?


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Titration


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Midpoint


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Buffering capacity


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Conjugate bases

Acid Conjugate base

CH3COOH CH3COONa (NaCH3COO)

H3PO4 NaH2PO4

H2PO4- (or NaH2PO4) Na2HPO4

H2CO3 NaHCO3


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How do we choose a buffer?


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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


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Exercise

What is the pH of a buffer containing 0.1M HF and

0.1M NaF? (Ka = 3.5 x 10-4)


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Homework

What is the pH of a solution containing 0.1M HF and

0.1M NaF, when 0.02M NaOH is added to the

solution?


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At the end point of the buffering capacity of a buffer,

it is the moles of H+and OH-that are equal


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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.


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Polyprotic weak acids

Example:


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Hence


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Excercise

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 bufferwhen pH of 7.2 is obtained when 100 ml of 0.1 M

NaH2PO3 is mixed with 100 ml of 0.1 M Na2HPO3?


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Buffers in human body

Carbonic acid-bicarbonate system (blood)

Dihydrogen phosphate-monohydrogen phosphate

system (intracellular)

Proteins


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Blood buffering

CO2 + H20 H2CO3 H+ + HCO3-

Blood (instantaneously)

Lungs(within

minutes)

Excretion viakidneys (hours

to days)


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Roles of lungs and kidneys

Maintaining blood is balanced by the kidneys and the

lungs

Kidneys control blood HCO3 concentration ([HCO3])

Lungs control the blood CO2 concentration (PCO2)


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Calculations


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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)


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Acid-Base Imbalances

pH< 7.35 acidosis

pH > 7.45 alkalosis


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Respiratory Acidosis

H+

+ HCO3-

H2CO3 CO2+ H2O


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Respiratory Alkalosis

H++ HCO3- H2CO3 CO2+ H2O


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Metabolic Acidosis

H+

+ HCO3-

H2CO3 CO2+ H2O


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Metabolic Alkalosis

H++ HCO3- H2CO3 CO2+ H2O

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Biochemistry PPT Part-2

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