Question

Calculate the pH of each of the following solutions. (a) 0.17 M KNO2 (b) 0.37 M NaOCl (c) 0.50 M NH4ClO4

Answer 1

a) 0.17 M KNO2

C =0.17 M

Ka of HNO2 = 5.6×10^–4

pKa = 3.25

KNO2 is salt of strong base and weak acid . so

pH = 7 + 1/2 (pKa + log C)

= 7 + 1/2 (3.25 + log 0.17)

   = 8.24

pH = 8.24

(b) 0.37 M NaOCl

Ka of HClO = 4.0×10^–8

pKa = 7.4

NaClO is salt of strong base and weak acid . so

pH = 7 + 1/2 (pKa + log C)

= 7 + 1/2 (7.4 + log 0.37)

   = 10.48

pH = 10.48

(c) 0.50 M NH4ClO4

pKb of NH3 = 4.74

NH4ClO4 is salt of stong acid and weak base . so

pH = 7 - 1/2 (pKb + log C)

= 7 - 1/2 (4.74 + log 0.5)

   = 4.78

pH = 4.78