Question

5.)A.)A solution contains 5.21×10-2 M potassium acetate and 0.300 M acetic acid. The pH of this solution is____

B.)A solution contains 0.215 M potassium hypochlorite and 0.480 M hypochlorous acid.
The pH of this solution is  .

C.)A solution contains 0.466 M sodium nitrite and 0.302 M nitrous acid.

The pH of this solution is

Answer 1

Q5.

Potassium Acetate = KCH3COO --> K+ + CH3COO- in solution

Acetic acid = CH3COOH manly, so, this is abuffer since conjugate base ( acetate) and the acid ( acetic acid) are present

so, apply Hendesron haselbach equation

pH = pKA + log(A-/HA)

pKa for acetic acid = 4.75

pH = 4.75 + log(((5.21*10^-2) / (0.300))

pH = 3.9897

B)

note that this is again a buffer since there is ClO- ad HClO:

pKa = 7.53

so

pH = pKa + log(ClO-/HClO)

pH = 7.53 + log(0.215 / 0.48 )

pH = 7.18119

c)

once again a buffer:

Nitrite ion = NO2-

nitrous acid = HNO2

so weak acid + conjugate base are present

pH = pKa + log(NO2-/HNO2)

pKa = 3.39

so

pH = 3.39 + log(0.466 / 0.302 ) = 3.5783