Question

calculate the pH of a solution that contains 35.5 g of sodium hypochlorite in 275 mL of solution. Ka= 3.5 x 10^-8 for HClO (aq)

Answer 1

K_a for HClO = 3.5 x 10^-8

To calculate K_b for ClO^-1,conjugate base for HClO

K_a x K_b = 10^-14

Therefore,K_b=2.85 x 10^-7

Molar mass of Sodium Hypochlorite = 74.44 g/mol

Number of moles of Sodium Hypochlorite = 35.5/74.44=0.4768

Molarity = moles/Volume of solution in litres = 0.4768/0.275 = 1.734 M

ClO^-1+ H₂O ------> HClO + OH^-1
[1.734 - X ] X. X

Kb = [HClO] [OH^-]/ [ClO^-]

2.85 X 10^-7= [X] [X] / [1.734]

X²= 4.94 x 10^-7

X = [OH-] = 7.02 x 10^-4

pOH = 3.15

since pH + pOH = 14
pH = 10.85