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12.9 d)

Posted: Fri Dec 08, 2017 1:04 pm
by Paula Dowdell 1F
Can someone walk me throught the steps (including the net ionic equation) of figuring out whether NH4I(am) + KNH2(am) -> KI(am) + 2NH3(l)
can be classified as a rxn between bronsted acids and bases?

Re: 12.9 d)

Posted: Fri Dec 08, 2017 4:17 pm
by Harrison Wang 1H
First, you have to figure out the oxidation states of each species in each compound for the reactants and products. For the reactants, in NH4I: NH4 is +1 and I is -1 and in KNH2: K is +1 and NH2 is -1. For the products, in KI: K is +1 and I is -1, and NH3 is netural. Because K and I did not change oxidation states throughout the reaction we can disregard them in the net ionic equation. So the net ionic equation would be NH4+ + NH2- -> 2NH3. Here, you can clearly see that NH4 is donating a H to NH2 and is therefore the bronsted acid, while NH2 is accepting an H from NH4 and is therefore the bronsted base.