Question

Calculate the pH at the equivalence point in titrating 0.080 M solutions of each of the following with 0.023 M NaOH.

(a) hydroiodic acid (HI)

pH =

(b) hypoiodous acid (HIO), Ka = 2.3e-11

pH =

(c) propionic acid (HC3H5O2), Ka = 1.3e-05

pH =

Answer 1

Solution.

The equivalence point condition is

therefore, the ratio of volumes is

Assuming the total volume of 1 L, V₁ = 0.2233 L, V₂ = 0.7767 L.

Resulting concentration of a salt is

pc = -log(c) = 1.748.

a) HI is a strong acid, NaOH is a strong base, therefore, the medium is neutral, pH = 7.

b) HIO is a weak acid, NaOH is a strong base, therefore, the medium is basic:

pK_b = 14-pK_a = 14+log(2.3e-11) = 3.362;

pOH = (pK_b+pc)/2 = (3.362+1.748)/2 = 2.555;

pH = 14-pOH = 14-2.555 = 11.45.

c) Propionic acid is a weak acid, NaOH is a strong base, therefore, the medium is basic:

pK_b = 14-pK_a = 14+log(1.3e-05) = 9.114;

pOH = (pK_b+pc)/2 = (9.114+1.748)/2 = 5.431;

pH = 14-pOH = 14-5.431 = 8.57.