For J.5 (a), why is HF in the reactants not broken down into its separate ions H+ and F-?
and for (b) can someone explain how the product becomes (CH3)3NHO3?
HW J.5
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Re: HW J.5
For a) HF is a weak acid, not a strong one, therefore it will not break down completely into its corresponding ions. This means you can't include them as ions in the overall nor net ionic equation.
For b) You are reacting (CH3)3N with HNO3.When the HNO3 acts an acid it will lose its proton becoming H+ and NO3-. The H+ binds to the (CH3)3N at the N causing the formal charge on the N to go from 0 to +1. This results in the new molecule (CH3)3NH+, then in the overall reaction you would be left with (CH3)3NH+ and NO3- which are oppositely charged ions. This means that they will be written in the overall reaction as one overall neutral compound (CH3)3NHNO3. But for the ionic equation you see that the (CH3)3NH+ and NO3- are written separately on the products side to indicate that they are in fact ions. Then in the net ionic equation the NO3- cancels on both sides leaving you with (CH3)3N + H+
(CH3)3NH+.
For b) You are reacting (CH3)3N with HNO3.When the HNO3 acts an acid it will lose its proton becoming H+ and NO3-. The H+ binds to the (CH3)3N at the N causing the formal charge on the N to go from 0 to +1. This results in the new molecule (CH3)3NH+, then in the overall reaction you would be left with (CH3)3NH+ and NO3- which are oppositely charged ions. This means that they will be written in the overall reaction as one overall neutral compound (CH3)3NHNO3. But for the ionic equation you see that the (CH3)3NH+ and NO3- are written separately on the products side to indicate that they are in fact ions. Then in the net ionic equation the NO3- cancels on both sides leaving you with (CH3)3N + H+
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