Question

The Ka of acetic acid is 1.8 x 10^-5 at 25C.

a) What is the pH of a 0.5 M solution of acetic acid at 25C?

b)What is the pH of a solution made up of 20 mL of 0.5 M solution of acetic acid and 10 mL of 0.5 M sodium acetate?

c) If 10 mL of 0.1 M NaOH is added to the solution in part b, what is the final pH?

d) What is the pH of a solution containing equal concentrations of acetic acid and sodium acetate?

Answer 1

(a)

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

    0.5                              0                0

    0.5 - x                          x                x

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

1.8 x 10^-5 = x^2 / ( 0.5 - x)

1.8 x 10^-5 = x^2 / 0.5   ( Since x is very small )

x^2 = 9 x 10^-6

x = 3 x 10^-3 M

[H+] = 3 x 10^-3 M

pH = - log( 3 x 10^-3) = 3 - log3 = 2.523

(b)

pKa of acetic acid = - log ( 1.8 x 10^-5) = 4.74

millimoles of acetic acid = Molarity * Volume = 20 * 0.5 = 10 millimol

millimoles of sodium acetate = 10 * 0.5 = 5 millimol

pH = pKa + log ( base / acid)

pH = 4.74 + log ( 5 / 10)

pH = 4.74 - 0.301 = 4.439

(c)

NaOH added will lead to the addition of OH- ions which increases the millimoles of base and decreases the millimoles of acid because acetic acid reacts with NaOH to form sodium acetate

Millimoles of NaOH = 10 * 0.1 = 1 millimol

millimoles of acetic acid = 10 - 1 = 9 millimol

millimoles of sodium acetate = 5 + 1 = 6 millimol

pH = pKa + log ( base / acid)

pH = 4.74 + log( 6 / 9 )

pH = 4.74 - 0.176 = 4.564

(d)

pH = 4.74 + log( x / x)

pH = 4.74 + log(1)

pH = 4.74    ( Since log1 = 0)

If there are equal concentrations then pH = pKa