## 7th Edition 6B #9

taywebb
Posts: 60
Joined: Fri Sep 28, 2018 12:15 am

### 7th Edition 6B #9

When I fill in the table to find the [OH-] using the already given [H3O+] I keep getting 6.7x10^-14, but the solutions manual says the answer is 1.5x10^-14. I am just dividing 10^-14 by the given molarity (1.5). This is for the i part of the table. Can anyone explain what I am doing wrong?

Chem_Mod
Posts: 17842
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 406 times

### Re: 7th Edition 6B #9

The answer in the book is actually incorrect. You're answer is correct and the book is almost correct but they made a small mistake. If instead of using [H3O+]* [OH-] =1 x 10-14 to find [OH-], you first convert [H3O+] into pH you would get the pH = -log[1.5]= -0.176. The book says pH=0.176, so the book forgot to take the - of the log. Having the negative pH and then using pH+pOH=14 you can calculate that the pOH is actually 14.176, not 13.824 like in the book. Then if you take the inverse -log of the pOH, so 10-14.176 you get that [OH-] =6.67 x 10-15, just like you would get if you divide Kw by 1.5. I think the book makes the error because this is such a strong acid that it doesn't follow the general scale that pH should be between 0 and 14.

Return to “Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions”

### Who is online

Users browsing this forum: No registered users and 1 guest