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

2388 Views

See similar Chemistry GCSE tutors

Related Chemistry GCSE answers

All answers ▸

An industrial process converts the alkene ethene into ethanol, according to the following reaction: C2H4 + H2O --> CH3CH20H. What mass of ethanol can be made from 53g of ethane, given that the water is in excess. (2 marks) (6-7 grade)


Why are metals good conductors of electricity?


Calculate the relative formula mass of water.


What is an ionic bond?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences