Time remaining:
AP Chemistry Acid Bases Problem

Science
Tutor: None Selected Time limit: 1 Day

What is the pH of a solution prepared by adding .490g of ammonium iodide (NH4I) to 155 ml of water? Kb of NH3 is 1.8 *10^-5.  (Hint: NH4I immediately dissociates, so you are starting with .490g NH4+)

Apr 10th, 2015

Kb = [NH3]^2 / [NH4+] = x/0.022=1.8*10^-5, x=3.96*10^-7

[NH4I]=0.49g/(145g/mol)/0.144L=0.022

so [H+]=10^-14/(3.96*10^-7)=2.53*10^-8

so pH=-log[H+]=8

Let me know if you have questoins

Apr 10th, 2015

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
Click to visit
The Notebank
...
Apr 10th, 2015
...
Apr 10th, 2015
Dec 7th, 2016
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer