Asolution naocl was prepared at ph 10.3. calculate the concentration of the salt formed. given ka for hocl=3.0x10^-8

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

Yes i think so not 100 percent sure though

The total number of electrons is 14