


Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
Solutions to recitation section problems related to calculating ph of acid-base systems using quadratic equations and henderson-hasselbalch equation. The problems involve chlorous acid, hydronium ions, and methylamine. Students will learn how to set up and solve the equations to determine the ph of the solutions.
Typology: Study notes
1 / 4
This page cannot be seen from the preview
Don't miss anything!
Ka =
[ClO− 2 ][H 3 O+] [HClO 2 ]
[HClO 2 ] [ClO− 2 ] [H 3 O+] initial 1.20 M 0.50 M 0 M change - y y y equilibrium (1.20 -y) M (0.50 + y) M y M
y(y + 0.50)
y^2 + 0. 51 y − 0 .013 = 0
y =
= 2. 5 × 10 −^2 negative solution ignored
[H 3 O+] = 2. 5 × 10 −^2 M pH = − log 10 (2. 5 × 10 −^2 ) = 1. 61
nHClO 2 nClO− 2 nH 3 O+ initial 1.20 0.50 0 M change 0.10 -0.10 y equilibrium 1.30 M 0.40 M y M
y(0.40)
y = 1. 1 × 10 −^2
(
) = 3. 6 × 10 −^2
pH = − log 10 (3. 6 × 10 −^2 ) = 1. 45
nOH−^ added = (0.100 mol L−^1 )(0.0100 L) = 0.0010 mol
pH = pKa + log 10
= 3.36 + log 10