Unformatted text preview: 0.1104 mol/0.405 L = 0.273 M
HC2H3O2 + H2O
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
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.
- Fall '10