Q. 133.8( 5 Votes )

What will be the

Answer :

Ionization of CH3COOH occurs as :


CH3COOH + H2O H3O+ + CH3COO-


Initial concentration: 0.01 0 0


Equilibrium concentration: 0.01 – x x x


ionization constant =


since , x>>0.01 hence, 0.01 – x ~ 0.01 .


x2/0.01 = 1.74 × 10-5 (given)


X2 = 1.74 × 10-5 x 0.01


X = 4.2 x 10-4


Or, pH = -log (4.2 x 10-4 ) = 3.4

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