Question: What is the pH of a monoprotic acid with a concentration of 0.022 M?
Answer:
1. Assume this is a strong monoprotic acid so that it fully dissociates: HA → H+ + A-
2. [HA] = [H+] = 0.022 M
3. pH = -log10[H+] = -log10[0.022] = 1.7
Find out more, and see questions with worked solutions, at https://www.ausetute.com.au/phstronga.html
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment