What is the acid dissociation constant, Ka of the 0.150 mol dm–3 solution of weak acid HA with pH of 2.34?

First, It would be useful to write the equation for the dissociation of the weak acid HA, which is HA <--> H++ A-. Then, write the Ka expression of the weak acid HA, which is [H+][A-]/[HA]. We know that [HA] = 0.150 mol dm-3 as the concentration is given in the question. We also know the pH of HA is 2.34, we can find [H+] as pH = -log10[H+] = 2.34. To arrange this equation, [H+] = 10-pH= 10-2.34 = 4.57 x 10-3mol dm-3. As this is a weak acid, it means that HA is only weakly dissociated, so only a very small amount of HA is dissociated into H+ and A- ions. Therefore, we can make the assumption that [H+]=[A-] = 4.57 x 10-3mol dm-3. Now, we have found all the concentrations we need, [H+], [A-] and [HA], we can substitute these values into the Ka expression, so Ka = [H+][A-]/[HA] = (4.57 x 10-3)2 / 0.150 = 1.39 x 10-4 mol dm-3 (3 s.f.).

Answered by Venus S. Chemistry tutor

6649 Views

See similar Chemistry A Level tutors

Related Chemistry A Level answers

All answers ▸

Explain how you can prove that C6H6 does not form 1,3,5-Cyclohexatriene but forms Benzene


Predict the relative boiling points of propanal, butane and prop-2-en-1-ol from the highest to the lowest boiling point


Why does the first ionisation energy generally increase across a period? Explain why there are dips in energy between groups 2 and 3 and groups 5 and 6?


What factors affect ionisation energy and how does each of them affect it?


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 2013–2024

Terms & Conditions|Privacy Policy
Cookie Preferences