Question

What is the pH of a solution that is 0.30M HOCl (hypochlorous acid) and 0.15M NaOCl after 0.070 mol HCl/L has been bubbled into the solution? K_a​ (HOCl) = 3.5x10^-8

HOCl_(aq) + H₂O <---> H₃O⁺_(aq) + OCl^-_(aq)

Answer 1

mol of HCl added = 0.07 mol

OCl- will react with H+ to form HOCl

Before Reaction:

mol of OCl- = 0.15 M *1.0 L

mol of OCl- = 0.15 mol

mol of HOCl = 0.3 M *1.0 L

mol of HOCl = 0.3 mol

after reaction,

mol of OCl- = mol present initially - mol added

mol of OCl- = (0.15 - 0.07) mol

mol of OCl- = 0.08 mol

mol of HOCl = mol present initially + mol added

mol of HOCl = (0.3 + 0.07) mol

mol of HOCl = 0.37 mol

Ka = 3.5*10^-8

pKa = - log (Ka)

= - log(3.5*10^-8)

= 7.456

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 7.456+ log {8*10^-2/0.37}

= 6.791

Answer: 6.79