Question

The Ka for hypochlorous acid, HClO, is 3.0 x 10-8.Calculate the pH after 13.0 mL of 0.100 M NaOH have been added to 30.0 mL of 0.100 M HClO

Answer 1

we have:

Molarity of HClO = 0.1 M

Volume of HClO = 30 mL

Molarity of NaOH = 0.1 M

Volume of NaOH = 13 mL

mol of HClO = Molarity of HClO * Volume of HClO

mol of HClO = 0.1 M * 30 mL = 3 mmol

mol of NaOH = Molarity of NaOH * Volume of NaOH

mol of NaOH = 0.1 M * 13 mL = 1.3 mmol

We have:

mol of HClO = 3 mmol

mol of NaOH = 1.3 mmol

1.3 mmol of both will react

excess HClO remaining = 1.7 mmol

Volume of Solution = 30 + 13 = 43 mL

[HClO] = 1.7 mmol/43 mL = 0.0395M

[ClO-] = 1.3/43 = 0.0302M

They form acidic buffer

acid is HClO

conjugate base is ClO-

Ka = 3*10^-8

pKa = - log (Ka)

= - log(3*10^-8)

= 7.523

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

= 7.523+ log {3.023*10^-2/3.953*10^-2}

= 7.41

Answer: 7.41