first we dissociate the salt
NaOCl ⇒ Na⁺ + OCl⁻
note that [NaOCl] = [OCl⁻]
note that Na⁺ does not undergo hydrolysis so it is a spectator.
OCl⁻ can reform HOCl in an equilibrium, with OCl⁻ acting as the base.
OCl⁻ + H₂O ⇄ OH⁻ + HOCL
Ka for HOCl = 3.0x10^-8, therefore
Kb for OCL⁻ = Kw / (Ka for HOCl)
= 1.0 × 10⁻¹⁴ / 3.0 × 10⁻⁸
= 3.3 × 10⁻⁷
since Kb for OCl⁻ is given, and the pH is given, then [OCl⁻] must be found, and [OCl⁻] = [NaOCl]
convert the pH into [OH⁻]:
pOH = 14 - pH = 14 - 10.3 = 3.7
[OH⁻] = 10^(-3.7) = 1.995 × 10⁻⁴
set up equib table
OCl⁻ + H₂O ⇄ OH⁻ + HOCl
ST x 0 0
+Δ -1.995 × 10⁻⁴ +1.995 × 10⁻⁴ +1.995 × 10⁻⁴
---------------------------------------------------------------------------------
EQ: x -1.995 × 10⁻⁴ 1.995 × 10⁻⁴ 1.995 × 10⁻⁴
Kb = [OH⁻][HOCl] / [OCl⁻]
3.3 × 10⁻⁷ = (1.995 × 10⁻⁴)² / (x -1.995 × 10⁻⁴ )
x = 1.1963 × 10⁻¹ M = [OCl⁻] = [NaOCl]
The concentration of the salt formed, NaOCl, is 1.2 × 10⁻¹ M