J.5a&b

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vivianndo_1L
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Joined: Fri Apr 06, 2018 11:02 am

J.5a&b

Postby vivianndo_1L » Fri Jun 08, 2018 12:47 am

J.5a

HF(aq) + NaOH(aq) -> NaF(aq) + H2O(l)

-Why is the complete ionic equation for this problem HF + Na+ + OH- -> Na+ + F- + H2O and not H+ + F- + Na+ + OH- -> Na+ + F- + H2O?

J.5b

(CH3)3N(aq) + HNO3(aq) -> (CH3)N3HNO3(aq)

-I don't understand why (CH3)3N(aq) + HNO3(aq) just combines to become (CH3)N3HNO3(aq). Why doesn't the proton just transfer to (CH3)3N so that the equation looks like this:

(CH3)3N(aq) + HNO3(aq) -> (CH3)3NH(aq)+ + NO3-(aq)

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Re: J.5a&b

Postby Chem_Mod » Fri Jun 08, 2018 10:03 am

For J.5 a) HF is a weak acid, so it does not completely dissociate into H+ and F-. NaOH is a strong base, so it does completely dissociate into Na+ and OH- so this is shown in the net ionic equation. When writing the complete ionic equation, strong acids and strong bases are written as their ions but weak acids and weak bases are not.

For J.5 b) The first equation that you wrote, where everything is attached is going to be the overall equation, but the second equation that you wrote out is the complete ionic equation. You are correct in thinking that the (CH3)3N is going to act as the base, and it will receive a proton that will attach to the N. But you will notice that you now have (CH3)3HN+ and NO3-. Generally, these two species will bind to each other forming a salt, so they would be written in the overall equation as being bound together, similar to how NaCl would be. But in the complete ionic they are broken down into their ions.


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