If a 100 mL sample of potassium hypochlorite (KOCl(aq)) has a pH of 9.75 at 25.0 ° C?

Refer to: https://depts.washington.edu/eooptic/links/acidstr...

Ka for HOCl = 2.9 × 10⁻⁸

(Ka values from different sources may be slightly different, and this may lead to slight different answers.)

Then, Kb for HOCl = Kw / Ka(HOCl) = (1.0 × 10⁻¹⁴) / (2.9 × 10⁻⁸) = 3.45 × 10⁻⁷

________________ OCl⁻(aq) + H₂O(l) ⇌ HOCl⁻(aq) + OH⁻ ____ Kb = 3.45 × 10⁻⁷

Initial (mol/L): _____ x ________________ 0 _______ 0

Change (mol/L): ___ -y _______________ +y ______ +y

Eqm (mol/L): _____ x - y ______________ y _______ y

pOH at equilibrium = -log(y) = 14.00 - 9.75 = 4.25

Hence, y = 10⁻⁴˙²⁵ mol/L

At equilibrium: [HOCl] [OH⁻] / [OCl⁻] = Kb

y² / (x - y) = 3.45 × 10⁻⁷

(10⁻⁴˙²⁵)² / (x - 10⁻⁴˙²⁵) = 3.45 × 10⁻⁷

x - 10⁻⁴˙²⁵ = (10⁻⁴˙²⁵)² / (3.45 × 10⁻⁷)

x = {(10⁻⁴˙²⁵)² / (3.45 × 10⁻⁷)} + 10⁻⁴˙²⁵

x = 9.2 × 10⁻³

Hence, concentration of KOCl = x mol/L = 9.2 × 10⁻³ mol/L

Hence,

a = 9

b = 2

c = 3