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

Answered by Octavia M. Chemistry tutor

1872 Views

See similar Chemistry GCSE tutors

Related Chemistry GCSE answers

All answers ▸

What mass of CO2 will be produced when 50 g of CaCO3 decomposes?


What is an acid?


What is an isotope?


What does an energy level diagram tell me?


We're here to help

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