Question

A buffer solution contains 0.355 M hypochlorous acid and 0.347 M sodium hypochlorite.

If 0.0243 moles of hydrochloric acid are added to 125 mL of this buffer, what is the pH of the resulting solution ? (Assume that the volume does not change upon adding hydrochloric acid)

pH=?

Answer 1

Ka of hypochlorous acid = 3.5*10^-8

mol of HCl added = 24.3 mmol

ClO- will react with H+ to form HClO

Before Reaction:

mol of ClO- = 0.347 M *125.0 mL

mol of ClO- = 43.375 mmol

mol of HClO = 0.355 M *125.0 mL

mol of HClO = 44.375 mmol

after reaction,

mol of ClO- = mol present initially - mol added

mmol of ClO- = (43.375 - 24.3) mmol

mol of ClO- = 19.075 mmol

mol of HClO = mol present initially + mol added

mol of HClO = (44.375 + 24.3) mmol

mol of HClO = 68.675 mmol

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 {19.07/68.67}

= 6.9

Answer: 6.90