Question

If you start with 200 mL of a 0.250 M hypochlorous acid / 0.300 M potassium hypochlorite buffer, (HClO/ KClO), and add 20.0 mL of a 0.400 M sodium hydroxide, NaOH, what will be the final pH of the solution? (For HClO, Ka = 3.0 x 10−8)

Answer 1

mol of NaOH added = 0.4M *20.0 mL = 8.0 mmol

HClO will react with OH- to form ClO-

Before Reaction:

mol of ClO- = 0.3 M *200.0 mL

mol of ClO- = 60 mmol

mol of HClO = 0.25 M *200.0 mL

mol of HClO = 50 mmol

after reaction,

mol of ClO- = mol present initially + mol added

mol of ClO- = (60 + 8.0) mmol

mol of ClO- = 68 mmol

mol of HClO = mol present initially - mol added

mol of HClO = (50 - 8.0) mmol

mol of HClO = 42 mmol

Ka = 3*10^-8

pKa = - log (Ka)

= - log(3*10^-8)

= 7.523

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

use:

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

= 7.523+ log {68/42}

= 7.732

Answer: 7.73