Question

Calculate the pH of a solution made by adding 54 g of sodium acetate, NaCH3COO, to 12 g of acetic acid, CH3COOH, and dissolving in water to make 400. mL of solution.

Answer 1

The pH calculation is as given below.

Let us calculate Molarity (M) of CH3COOH and CH3COONa

Molarity = moles/L = g/molar mass x L

M of CH3COONa = 54/82.0343 x 0.4

                             = 1.65 M

M of CH3COOH = 12/60.05 x 0.4 L

                           = 0.50 M

Now we have the equation,

              CH3COOH <======> H+     +      CH3COO-

Initial           0.5 M                       0M                 1.65 M

change         -x                            +x                      +x

Equilibrium (0.5-x)M                   xM                 (1.65+x)M

--------------------------------------------------------------------------

Thus, we can write,

Ka for CH3COOH = [H+][CH3COO-] / [CH3COOH]

Ka for CH3COOH = 1.8 x 10^-5 [from standard literature]

Thus,

1.8 x 10^-5 = (x)(1.65+x)/(0.5-x)

Considering x be a small amount dissociated, we get,

1.8 x 10^-5 = 1.65x/0.5

x = [H+] = 5.46 x 10^-6 M

pH = -log[H+]

      = -log(5.46 x 10^-6)

      = 5.26

Hence pH of this solution will be 5.26