Question

HClO is a weak acid (Ka = 4.0 × 10–8) and so the salt NaClO acts as a weak base. What is the pH of a solution that is 0.088 M in NaClO at 25 °C?

Answer 1

NaClO is a soluble ionic compound which completely dissolve to give Na+ and ClO-

ClO- then dissociates in water with kb

ClO- + H2O = HClO + OH- kb

where kb = [HClO][OH-] / [ClO-] = Kw / Ka

before dissociation

[ClO-] = 0.088 M

[HlO] = [OH-] = 0M

at eqlibrium

[ClO-] = (0.088-Y) = 0.088 M assuming that 0.088 >> yM

[HClO] = [OH-] = yM

kb = Kw/Ka = (1*10^-14)(4.0*10&-8) = y^2 / 0.088

y = 1.4*10^-4

[OH-] = 1.4*10^-4

pOH = -log[OH-] = -log[1.4*10^-4) = 3.85

pH = 14 - 3.85 = 10.15