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

8216 Views

See similar Chemistry A Level tutors

Related Chemistry A Level answers

All answers ▸

Explain the trend in reactivity of group 2 elements with water as you go down the group.


Explain what happens to the boiling and solubility of alcohols as their chain length increases


what is the shape and bond angle of NH3 and use VSEPR theory to explain the bond angle.


A sample of hydrochloric acid has a pH of 2.34. 
Write an expression for pH and calculate the concentration of this acid.


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