lecture notes2

# 048 mol0405 l 012 m c2h3o2 01104 mol0405 l 0273

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 0.1104 mol/0.405 L = 0.273 M HC2H3O2 + H2O init 0.12 M equil ~0.12 X H3O+ + C2H3O2~10-7 M x x = 7.52 x 10-6 M = [H3O+] 5. 0.273 M ~0.273 Y pH = 5.12 Calculate the pH of 10-5 M HCl. This is a solution of a strong acid and [H3O+] ~ [acid]. [H3O+] = 10-5 M Y pH = 5.0 Chem 102 D. Miller Solutions to Review Problems for Acid/Base Chemistry 6. If 0.050 mL of 6.0 M HCl is added to 400 mL of 10-5 M HCl, what is the resulting pH? The added HCl is combines with the initial HCl to form a new strong acid solution. mol H3O+ added = (6.0 mol/L)(0.000050 L) = 0.00030 mol init mol H3O+ = (10-5 mol/L)(0.400 L) = 0.000004 mol total mol H3O+ = 0.000304 mol [H3O+] = 0.000304 mol/0.40005 L = 7.6 x 10-4 M pH = 3.12 (Notice that the pH of this unbuffered solution decreased by almost 2 pH units with this small addition of acid . Adding 100 times this volume of acid to a buffer (Prob. 4) changed the buffer pH by only 0.34 unit.) C...
View Full Document

## This document was uploaded on 03/06/2014 for the course CHEM 102 at CSU Northridge.

Ask a homework question - tutors are online