1 ILO of the current topic: By the end of this topic, the student will be able to: Identify...

1

ILO of the current topic:By the end of this topic, the student will be able to: Identify different pH solutions.

pH and Buffers

Our Goals for today

1-Kw2-pH

Dissociation Of Water

Faculty of Medicine Ain Shams University

Medical Biochemistry & Molecular Biology Department

Dissociation Of WaterDissociation Of Water

H2O H+ + OH-

Water has limited capacity to dissociate (ionize) into H+ &

OH-



Ka Is The Acid Dissociation Constant Of

Water

Since water is too little too little dissociated[H2O]

[H+] [OH-]= Ka

[H+] [OH-]So, Kw

[H+] [OH-]= Kw

Kw = The Ion Product Of Water

10-7 X 10-7 =Kw 10-14 mol/L

Since those values involving such –ve powers of 10 which is difficult to use in

calculations

So

Sorensen (1909) suggested that the

[H+]concentration of DILUTEDILUTE aqueous solutions

would be better expressed as PH valuePH value



What is the pH?

• The pH of a solution is the negative of the logarithm to the base 10 of its hydrogen ion concentration in (mol/L)

pH = -log[H+]



pH scale 2 3 4 5 6 7 8 9 10 11 12

neutral @ 25oC(H+) = (OH-)

distilled water

acidic(H+) > (OH-)

basic or alkaline(H+) < (OH-)

ranges from 0 – 14pH < 7 acidpH = 7 neutralpH > 7 base



P= -Log

H= H+ Concentration

pH= -log of H+ Concentration



P= -Log

OH= [OH- Concentration]

pOH= -log of [OH- Concentration]

[OH-] may be represented as pOH value



Since [H+] x [OH-] =10 -14

–log [H+] + -log [OH-] =14

i.e. pH + pOH =14

By –log of both sides



Q- What is the pH of 0.0001 N HCL sol.?A- [H+] = 0.0001 = 10-4 mol/L pH = - log 10-4 = 4Q - What is the pH of 0.001 N NaOH sol.? A- [OH-] = 0.001 = 10-3 mol/L pOH = - log 10-3 = 3 Since pH+ pOH=14 So, pH = 14 - pOH= 14 -3= 11

Problem 2Problem 2



What is the pOH of a 0.010 M solution of Ba(OH)2:

?A. 1.0

B. 2.0

C. 4.0

D. 5.0

Problem 3Problem 3



The pH of a sample of human blood was measured to be 7.41 at 25°C. Calculate pOH, [H+] and [OH-] for the sample.

Problem 4 Problem 4

Answer

pH + pOH = 14.00pOH = 14.00-7.4 = 6.6[H+] = antilog(-pH)= antilog (-7.41) = 3.9x10-8M[OH-] = antilog(-pOH) = antilog (-6.6) = 2.51 x 10–7 M

1-Definition2-Properties3-Classification4-Neuteralization

Acids and bases

Acids and bases

• Taste sour

• React w/ metals & carbonates-produce H gas

• Turn blue litmus paper red

• form hydrogen ions (H+) in water

• Taste bitter• Feel slippery• Don’t react w/

carbonates• Bases turn red

litmus paper blue.• form hydroxoxide ions

in water

Acids Bases



2-17

Acids cause H+ to increaseBases cause OH- to increase



Amphoteric SubstanceAmphoteric Substance

They can act as Proton DONORS DONORS with bases

& Proton ACCEPTORSACCEPTORS with acids

Eg. Amino acid –H2O



In dilute aqueous solution, as [H+] increases: A. pH decreases B. pOH increases C. [OH–] decreases D. all of the above

Problem 5 Problem 5

Neutralization

HCl + NaOH → H2O + NaCl acid base water salt

• reaction between an acid & base

• produces water & a salt

an ionic compound



Strength Of Strength Of Acids Acids & & BasesBases

It depends on the degree of IonizationIonization in an aqueous solution

i.e.i.e.

The efficacy with which The efficacy with which

an an acid acid acts as a proton donor acts as a proton donor

& & a basea base acts as a proton acceptor acts as a proton acceptor



HA Let’s examine the behavior of an acid, HA, in aqueous solution.

What happens to the HA molecules in solution?



HA

H+

A-

Strong Acid

100% dissociation of HA



HA

H+

A-

Weak Acid

Partial dissociation of HA

Would the solution be conductive?

HA

H+

A-

Weak Acid

HA H+ + A-

At any one time, only a fraction of the molecules are

dissociated.

1-Calculation2-Uses

HENDERSON-HUSSELBALCH Equation


Medical Biochemistry & Molecular Biology DepartmentHow can you calculate the How can you calculate the pH pH of of

Strong Acids & Bases ??Strong Acids & Bases ??

From the Molarity or Normality of them

i.e.i.e.

pH of Acid pH of Acid = - log of acid conc = - log of acid conc [H[H++]]

pH of Base pH of Base = 14 - pOH= 14 - pOH



pH pKa

= [HA] [A-]

+Log

HENDERSON-HUSSELBALCHHENDERSON-HUSSELBALCH Equation



USES OF USES OF HENDERSON-HUSSELBALCHHENDERSON-HUSSELBALCH

EquationEquation

1. For Calculation of pH of a weak Acid.

2. Preparation Of Buffer Of Known pH



When [A[A--]= [HA]]= [HA]

So PKa Of an acid is the pH at which

[The Acid conc.The Acid conc.] = [Its Conjugate base conc.Its Conjugate base conc.]

pH pKa

= + [HA] [A-]

Log



Summary



33

Strong AcidWeak Acid

Transfers all of its protons to waterTransfers only a small fraction of its

protons to water

Completely IonizedPartly Ionized

Ka largeKa is small

smallerpKAlarger pKA

Eg. HCl and H2SO4 Eg. Organic acids e.g. H2CO3

1-Indicators2-pH Meter3-Titeration

Determining pH

1-Indicator• Compound that changes color when in

contact w/ an acid or base

• They are weak acids

• ex. litmus paper

• Red litmus paper

• Blue litmus paper

• universal indicators

•Red Cabbage Juice

Mechanism of action of the litmus paper indicator

Nameof indicator

Colour acid

Colour base

pH range

Methyl OrangeRedYellow 3 - 4

Bromothymol BlueYellowBlue6 - 8

PhenolphthaleinClearRed8 - 10

Universal indicators

Universal indicators made by mixing several different indicators

Paper soaked in universal indicator pH paper

pH meter determines the pH of a solution by measuring voltage between two electrodes that are placed in solution

Voltage changes as hydronium ion concentration in solution changes

2-pH meter

1-Definition2-Principle of buffering3-Application

BUFFER



Buffers

It is a solution of A Weak Acid (H(H22COCO33))

& its Conjugate Base (NaHCO(NaHCO33))

Or Weak Base (NH(NH44OHOH) &

its Conjugate Acid (NH(NH44CL)CL)



BuffersIt resists changes in pH in a solution

when moderate amount of strong acids

or bases are added

Acid““HH22COCO33””

Conjugate base“NaHCO3”

OH-

OH-

OH-

OH-H+

H+

H+

H+



Principles of Buffering:AH ↔ H+ + A-



The Extent Of Buffer Capacity Depends on 3 factors

Any buffer exerts maximal buffering capacity when

the pH is equal to its pKa.

If [A-]= [HA ] i.e the acid is half dissociated (half-neutralization state)

Log [A-]/ [HA ] = log 1= zeropH = pKa + 0



pH= pKa 1

Is Considered as pH range

For

SATISFACTORY Buffering SATISFACTORY Buffering

CapacityCapacity



Physiological buffers1. H2CO3/ HCO3 (pKa= 6.1) This is the most important

buffer system in the body

2. H2PO4/ HPO4 (pKa= 6.8).

3. Plasm Proteins (20% of buffering capacity “NH“NH22+ + , ,

COOCOO--””))4. Hb. (60% of buffering capacity “Histidine”“Histidine”)5. Free amino acids

But Theoretically

Bicarbonate buffer Bicarbonate buffer is < efficient than is < efficient than Phosphate buffer Phosphate buffer

in intracellular fluid.in intracellular fluid.



• The pKa of acetic acid is 4.76, its buffer mixture is most effective at pH:

• a) 2.3

• b) 4.5

• c) 3.7

Problem 6Problem 6

pH = -log[H+]pOH= -log of

[OH]

pH + pOH =14pH= pKa +log [A-]

[HA-]


Medical Biochemistry & Molecular Biology DepartmentLab activities include :



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1 ILO of the current topic: By the end of this topic, the student will be able to: Identify...

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