The pH of a solution obtained by mixing 100 ml of 0.2 M CH3COOH with 100 ml of 0.2 M NaOH will be??(given pKa for 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+O
Acid dissociation constant, Ka = [CH3COO-] [H3+O] / [CH3COOH]  (1)
The initial and equilibrium constant for the reactants and products is
CH3COOH   CH3COO-  H3+O
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 10-13M
pH = -log(1.01 X 10-13) = 12.99 or  13

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