Answer to Question #171961 in General Chemistry for Irish

Question

The pH of an acetic and sodium acetate buffer is 3.5 and its ka = 1.7 4x10-5. Compute for the molar ratio and percent ionization of the acid and the salt. 

Answer 1

CH3COOH -> CH3COO- + H+

Ka = (CH3COO-) (H+)/CH3COOH

Therefore the mole ratio of acetate ion to acetic acid is

= Ka/ H+ = CH3COO-/CH3COOH

The ratio of concentration provided similarly corresponds to the molar concentration since the volume is constant

PH provided is 3.5

-log(H+) = 3.5

H+ = 3.16 × 10^-4

CH3COO-/CH3COOH = nCH3COO-/nCH3COOH =

1.74 × 10^-5/3.16 × 10^-4

= 0.0551

% ionization = 0.0551 × 100 = 5.51%