For the following,
2HBr (aq) + Ba(OH)2 (aq) ---> 2H2O (l) + BaBr2 (aq)
write the net ionic equation including the phases
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)
Alright, hold on to your lab goggles, because it's time for some chemical circus tricks!
So, we start off with our brave molecules: 2HBr (aq) + Ba(OH)2 (aq). They jump into the reaction ring, ready to perform some chemistry magic. And, voila! Out comes 2H2O (l) and BaBr2 (aq), leaving the audience in awe!
Now, let's get serious and break it down into the net ionic equation, including the phases, because no circus act is complete without a bit of fancy notation.
First, we'll want to focus on the "spectacular" (a.k.a the species that actually undergoes a change) ions. In this case, those are H+ from HBr and OH- from Ba(OH)2.
2H+(aq) + 2OH-(aq) ---> 2H2O (l)
Ta-da! That's the net ionic equation, in all its glory. The H+ and OH- ions gather together to form good old water (H2O).
Hope that clears things up, and remember, always keep your chemistry sense of humor finely balanced!
To write the net ionic equation for the given chemical reaction, you need to first write the balanced equation and then identify the spectator ions to eliminate them. Here's how you can do it step by step:
Step 1: Write the balanced equation.
2HBr(aq) + Ba(OH)2(aq) ---> 2H2O(l) + BaBr2(aq)
The equation above shows the reactants and products with their respective coefficients. It states that two moles of hydrobromic acid (HBr) react with one mole of barium hydroxide (Ba(OH)2) to form two moles of water (H2O) and one mole of barium bromide (BaBr2).
Step 2: Identify the spectator ions.
To determine the spectator ions, we need to identify the ions that do not participate in the actual chemical reaction. Spectator ions remain unchanged throughout the reaction and are present on both the reactant and product sides.
In this equation, the ions that do not participate in the actual reaction are Ba^2+ and OH^-.
Step 3: Write the net ionic equation.
The net ionic equation only includes the ions or compounds that are directly involved in the reaction. Spectator ions are not included.
HBr(aq) + OH^-(aq) ---> H2O(l) + Br^-(aq)
In the net ionic equation above, we remove the spectator ions (Ba^2+ and OH^-) from the balanced equation. Thus, the net ionic equation emphasizes the reaction between hydrogen bromide (HBr) and hydroxide ions (OH^-), which produce water (H2O) and bromide ions (Br^-).
Note that the phases (aqueous and liquid) are included in the net ionic equation.
2HBr (aq) +OH^-(aq)----> 2H2O (l)+Br2^-(aq)???
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:
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.