In aqueous solution, sulphuric acid dissociates into ions in 2 stages. The pKa for the first dissociation is -3. Calculate the pH of a 0.025 mol dm-3 solution of sulphuric acid using the pKa value of the 1st dissociation.

Ka = 10-pKa = 10--3 = 1000[H+] = √ Ka [0.025] = √1000 x 0.025 = 0.791pH = -log [H] = -log(0.791) = 0.102 = 0.1

OM
Answered by Octavia M. Chemistry tutor

2754 Views

See similar Chemistry GCSE tutors

Related Chemistry GCSE answers

All answers ▸

Explain in terms of structure and bonding why graphite conducts electricity.


Diamond is very hard because:


How can potassium form an ionic compound with sulphur?


what are isotopes and why do they have the same properties?


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning