6A.9

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Daniel_Frees_1L
Posts: 40
Joined: Fri Sep 28, 2018 12:18 am

6A.9

Postby Daniel_Frees_1L » Fri Dec 07, 2018 7:32 pm

Will we have any problems like 6A.9 on the final? If so, what is the process for writing net ionic equations (I get what they are, but not how to consistently know where to break up big compounds into two ions)?

For example, for part C I thought I could break CH3COOH to CH3CO+ and OH- and then NH3 was donating a H to OH-. But apparently this is wrong.

Joaquin Andrade
Posts: 31
Joined: Fri Sep 28, 2018 12:27 am

Re: 6A.9

Postby Joaquin Andrade » Fri Dec 07, 2018 10:46 pm

Acetat (CH3COOH) is a weak acid and thus will give off the H+ proton to NH3, thereby the products are NH4+ (ammonium) and CH3COO-.

Daniel_Frees_1L
Posts: 40
Joined: Fri Sep 28, 2018 12:18 am

Re: 6A.9

Postby Daniel_Frees_1L » Fri Dec 07, 2018 11:20 pm

@Joaquin

I should've included that the products are given as well. (-->CH3CONH2 + H2O). The book says no proton is transferred. These problems we have to break up into ionic parts and then get rid of ions that are unchanged to find the net ionic equation. We then look at whether a H+ ion moved. I tried to break it up the way I originally posted, and that made it appear like an H+ moved, but the answer says none did. All in all a kinda confusing and slightly ambiguous (to my lack of memorization) problem that we haven't seen before in class.

I wonder if a mod could tell us if these will be on the test?


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