Question

When a 17.5 mL sample of a 0.359 M aqueous hypochlorous acid solution is titrated with a 0.460 M aqueous sodium hydroxide solution, what is the pH after 20.5 mL of sodium hydroxide have been added? pH =

Answer 1

Ka of HClO = 3.0*10^-8

Given:

M(HClO) = 0.359 M

V(HClO) = 17.5 mL

M(NaOH) = 0.46 M

V(NaOH) = 20.5 mL

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

mol(HClO) = 0.359 M * 17.5 mL = 6.2825 mmol

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

mol(NaOH) = 0.46 M * 20.5 mL = 9.43 mmol

We have:

mol(HClO) = 6.2825 mmol

mol(NaOH) = 9.43 mmol

6.2825 mmol of both will react

excess NaOH remaining = 3.1475 mmol

Volume of Solution = 17.5 + 20.5 = 38 mL

[OH-] = 3.1475 mmol/38 mL = 0.0828 M

use:

pOH = -log [OH-]

= -log (8.283*10^-2)

= 1.08

use:

PH = 14 - pOH

= 14 - 1.08

= 12.92

Answer: 12.92