The pH of a soln obtained by mixing 100 ml of 0.2 M CH3COOH with 100ml of 0.2 M NaOHwill be
( pKafor CH3COOH= 4.74 AND LOG 2 = 0.301)

Moles of NaOH added to the mixture =  0.1 X 0.2 =  0.02 moles 
CH3COOH is a weak acid and does not completely dissociate in aqueous solution. 
Ionization reaction  CH3COOH + H2O  ---->    CH3COO-  +  H3 +
Acid dissociation constant, Ka = [CH3COO-] [H3 +O] / [CH3COOH]  (1) 
The initial and equilibrium constant for the reactants and products is 
  CH3COOH   CH3COO-  H3 +
Initial  0.2M   0  0 
Assuming x moles are dissociated at equilibrium 
Equilibrium  0.2-x or 0.2  x  x 
As acetic acid is a weak acid with a degree of dissociation of less than 5%,  
value of x <<< 0.2 and therefore 0.2-x = 0.2 
 pKa  = 4.74 
-log Ka = 4.74 
Ka = antilog (-4.74) = 1.82 X 10-5 
Substituting values in equation 1, 
1.82 X 10-5  =  x  X  x/0.2 
x2 = 3.64 X 10-6 
x = 0.001908M 
Therefore, [H3O+] = 0.001908 M 
Moles of hydronium ions in solution = 0.001908 X 0.1 = 0.0001908mol 
These moles of hydronium ion neutralize the same number of hydroxyl ions. 
Number of hydroxyl ions that contribute to pH = 
(total moles of OH-) – moles of OH-that reacted with hydronium ions 
=0.02 – 0.0001908 = 0.019809mol 
[OH-] in final mixture = 0.019809/0.2 = 0.099M 
(Volume of mixture = 100ml acid solution + 100ml base = 200ml or 0.2L) 
[H3O+] =  1 X 10-14/0.099 =  1.01 X 10-13
pH = -log(1.01 X 10-13) = 12.99 or  13 
 

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