For the following,

2HBr (aq) + Ba(OH)2 (aq) ---> 2H2O (l) + BaBr2 (aq)
write the net ionic equation including the phases

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1. I showed you how to do the last one. You try this one. Just follow the rule.
weak acids, weak bases, water, insoluble substances(solids; i.e., ppts), gases all written as molecules.
All others written as ions.

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2. 2HBr (aq) +OH^-(aq)----> 2H2O (l)+Br2^-(aq)???

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3. 2HBr (aq) + Ba(OH)2 (aq) ---> 2H2O (l) + BaBr2 (aq)
No, HBr is a strong acid. Ba(OH)2 is a strong base and you've handled that ok. Bromide exists as Br^- and not Br2^- so the way you do that is 2Br^-
Why don't we do this the long way?
You have the balanced molecular equation. Next step is to convert it to the total ionic e3quation like this.
2H^+ + 2Br^- + 2K^+ + 2OH^- ==> H2O + Ba^2+ + 2Br^-
Now cancel the ions common to both sides; i.e., 2H^+ does not cancel, 2Br^- does, 2K^+ does, 2OH^- does not.. Net ionic equation is
2H^+(aq) + 2OH^-(aq) ==> 2H2O(l)

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4. Consider the following precipitation reaction (balanced).
precipitation reaction:
2NH_{4}Br(aq) + Pb(C_{2}H_{3}O_{2})_{2}(aq) -> 2NH_{4}C_{2}H_{3}O_{2}(aq) + PbBr_{2}(s)
2
NH
4
Br
(
aq
)
+
Pb
(
C
2
H
3
O
2
)
2
(
aq
)

2
NH
4
C
2
H
3
O
2
(
aq
)
+
PbBr
2
(
s
)

Enter the balanced net ionic equation, including phases, for this reaction.
net ionic equation:

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