Calculate the percent ionization of a 0.18 m benzoic acid solution in a solution containing 0.10 m sodium benzoate.

% I = 0.083 %

Explanation:

∴ M C6H5COOH = 0.18 mol/L

∴ M NaC6H5CO2 = 0.10 mo/L

⇒ C6H5COOH ↔ H3O+  +  C6H5COO-

⇒ NaC6H5CO2 ↔ Na+  +  C6H5COO-

∴ Ka = 6.4 E-5 = ( [ H3O+ ] * [ C6H5COO- ] ) / [ C6H5COOH ]  

Ka value taken from the literature

mass balance

⇒ 0.18 + 0.1 = [ C6H5COOH ] + [ C6H5COO- ]

⇒ [ C6H5COOH ] = 0.28 - [ C6H5COO- ]  (1)

charge balance:

⇒ [ H3O+ ] + [ Na+ ] = [ C6H5COO- ]

∴ [ Na+ ] ≅ M NaC6H5CO2 = 0.1 M

⇒ [ C6H5COO- ] = [ H3O+ ] + 0.1(2)

(1) and (2) in Ka:

⇒ 6.4 E-5 = ( [ H3O+ ] * ( [ H3O+ ] + 0.1 ) / ( 0.28 - ( [ H3O+ ] + 0.1 ) )

⇒ 6.4 E-5 = [ H3O+ ]² + 0.1 [H3O+ ] / ( 0.18 - [ H3O+ ] )

⇒ 1.17 E-5 - 6.5 E-5 [ H3O+ ] = [ H3O+ ]² + 0.1 [ H3O+ ]

⇒ [ H3O+ ]² + 0.1 [ H3O+ ] - 1.17 E-5 = 0

⇒ [ H3O+ ] = 1.493 E-4 M

⇒ % I = ( 1.493 E-4 / 0.18 ) * 100 = 0.083 %

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