Acid-Base Equilibria: Calculation of pH and Buffer Solutions - Prof. Abdo, Lecture notes of Analytical Chemistry

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Acid – Base Equilibria

Prof. Dr. Elham Y. Hashem

Calculation of pH

ً Strong acid {completely ionized }: ًاًًًًًًًًًًًًًًأول

e.g.

ً ًً Strong base completely ionized }: اًًًًًًًًًًًًثانيا

e.g.

 

HCl  H  Cl

 

NaOH  Na  OH

OH  NaOH 



 

 OH

Kb

ً Weak bases :رابعااً

يتم حساب

NH 4 OH

  NH 4  OH

e.g.   

 NH OH

NH OH K

4

4 b

 



 

 NH OH 

OH

4

^2 

 OH^ ^  Kb C



{ partially ionized }

ً Saltsاألمالح :ًًًًخامساًا

انـ يتم حساب pH نهمهحانمائ ًمه معادنة انتحهم

Four Kinds of Salts

1) Salts of strong acid and strong base

HCL + NaOH = NaCl + H2O

متعادل NaCl (ًانمحهىل انمائ (

2) Salts of weak acid and strong base

CH3COOH + NaOH = CH3COONa + H2O

NaAc ًاوظر معادنة انتحهم انمائ قلوى ) ًانمحهىل انمائ )

NaAc  H 2 O  NaOH  HAc

Ac  H 2 O OH  HAc

 

   

 





 Ac

OH HAc Kh

 

 





 Ac

OH

2

K h

  OH  Kh. C



  C K

K

OH

a

w 



Hydrolysis of NaAc

Hydrolysis of NH 4 Cl

NH (^) 4 Cl  H 2 O  NH 4 OH  HCl

  NH 4  H 2 O NH 4 OH H

   

 







4

4 h NH

NH OH H K

 

 



  4

2

NH

H Kh

 H  Kh C



  C K

K H

b

w 



 Calculate the pH of the following solutions:

a- 1  10 -3^ M acetic acid ( CH 3 COOH )

b- 1  10 -3^ M ammonium hydroxide ( NH 4 OH )

c- 0.1 M sodium acetate soln. ( CH 3 COONa )

d- 0.1 M ammonium chloride slon. ( NH 4 CL )

Ka = 1.78 10 -

Kb = 1.78 10 -

Kw = 1 10

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Thank You

HAc

 

H  Ac

Henderson Hassebalch equation :

For ( acid buffer )

( HAC/NaAc ) buffer

  

 HAc 

H Ac

Ka

 

 

 



  Ac

HAc H Ka

 

 HAc 

Ac pH pKa log



 

 Acid 

Salt  p Ka  log

 

proton donnor

protonacceptor  pKa  log

بالضرب فى - log

 

 

 B 

BH . K

K H w

a

  

 

( )

log

pH p OH

BH

B pH pKa pKw

 

   

 



  BH

B pOH pKa log

 

 B 

BH pOH pKb log



 

or

Kw = Ka. Kb Kb = Kw/ Ka 1/ Kb = Ka/ Kw

( NH (^) 4 OH / NH 4 Cl ) Buffer   NH 4 OH NH 4  OH

  

 NH OH

NH OH K

4

4 b

 



 

 

 



 

4

4 b NH

NH OH OH K

 

 NH OH 

NH pOH pKb 4

log^4

  

 base 

Salt pOH  pKb  log

For

Kw = Ka. Kb Kb = Kw/ Ka 1/ Kb = Ka/ Kw

Ex. (2):

Calculate the pH of a soln. prepared by adding 25 ml of 0.1 M

sodium hydroxide to 30ml of 0.2 M acetic acid. (pKa= 4.76 )

m. mol HAc = 30  0.2 = 6 :^ الحــل

m. mol NaOH = 25  0.1 = 2.

Now

HAC + NaOH = NaAC + H 2 O

Initial conc. 6 2.5 --

Final conc. 3.5 -- 2.

 Acid 

Salt pH  pKa  log

4. 61

5. 5

6. 5  4. 76  log 

Tris Buffer

Tris is tris(hydroxymethyl) amino methan

NH 2

C

CH 2 OH

CH CH^2 OH 2 OH

NH 3

C

CH 2 OH

CH CH^2 OH 2 OH

• H

• (^) =

(B)

Tris

(BH+)

Tris hydrochloride

pKa = 8.