Question

When a 26.7 mL sample of a 0.477 M aqueous hypochlorous acid solution is titrated with a 0.300 M aqueous sodium hydroxide solution, what is the pH after 63.7 mL of sodium hydroxide have been added? pH =

Answer 1

Given:

M(HClO) = 0.477 M

V(HClO) = 26.7 mL

M(NaOH) = 0.3 M

V(NaOH) = 63.7 mL

mol(HClO) = M(HClO) * V(HClO)

mol(HClO) = 0.477 M * 26.7 mL = 12.7359 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.3 M * 63.7 mL = 19.11 mmol

We have:

mol(HClO) = 12.7359 mmol

mol(NaOH) = 19.11 mmol

12.7359 mmol of both will react

excess NaOH remaining = 6.3741 mmol

Volume of Solution = 26.7 + 63.7 = 90.4 mL

[OH-] = 6.3741 mmol/90.4 mL = 0.0705 M

use:

pOH = -log [OH-]

= -log (7.051*10^-2)

= 1.1517

use:

PH = 14 - pOH

= 14 - 1.1517

= 12.8483

Answer: 12.85