Write the balanced NET ionic equation for the reaction when CaBr₂ and Na₃PO₄ are mixed in aqueous solution. If no reaction occurs, simply write only NR.
Well, let's see what happens when CaBr₂ and Na₃PO₄ get together in an aqueous solution.
First, let's break them down into their ions. CaBr₂ dissociates into Ca²⁺ and 2 Br⁻ ions. Na₃PO₄ dissociates into 3 Na⁺ ions and PO₄³⁻.
When we mix these together, we need to find out if any insoluble precipitate forms. In this case, since both calcium ions (Ca²⁺) and phosphate ions (PO₄³⁻) can form insoluble precipitates, we should be on the lookout for that.
If a precipitate forms, then a precipitation reaction occurs. So, let's check if there's any chance that a precipitate could form.
When we combine Ca²⁺ and PO₄³⁻, calcium phosphate (Ca₃(PO₄)₂) can be formed, which is insoluble. So, bingo! We have a precipitation reaction happening here!
Now, let's write the balanced net ionic equation:
3 Ca²⁺(aq) + 2 PO₄³⁻(aq) → Ca₃(PO₄)₂(s)
There you have it! Now you know the balanced net ionic equation for the reaction when CaBr₂ and Na₃PO₄ are mixed. Keep in mind that if no reaction occurs, we would simply write NR (Not Roger, I mean No Reaction!).
To write the balanced net ionic equation for the reaction between CaBr₂ (calcium bromide) and Na₃PO₄ (sodium phosphate) in aqueous solution, we need to determine the products of the reaction.
The balanced molecular equation for the reaction is:
3 Na₃PO₄(aq) + CaBr₂(aq) → 6 NaBr(aq) + Ca₃(PO₄)₂(aq)
In order to determine the net ionic equation, we need to break down the compounds into their respective ions:
Na₃PO₄(aq) → 3 Na⁺(aq) + PO₄³⁻(aq)
CaBr₂(aq) → Ca²⁺(aq) + 2 Br⁻(aq)
NaBr(aq) → Na⁺(aq) + Br⁻(aq)
Ca₃(PO₄)₂(aq) → 3 Ca²⁺(aq) + 2 PO₄³⁻(aq)
Now, we can write the net ionic equation by eliminating the spectator ions (the ions that appear on both sides of the equation):
3 Na⁺(aq) + PO₄³⁻(aq) + 3 Ca²⁺(aq) + 2 Br⁻(aq) → 6 Na⁺(aq) + 6 Br⁻(aq) + 3 Ca²⁺(aq) + 2 PO₄³⁻(aq)
The net ionic equation for this reaction is:
PO₄³⁻(aq) + 2 Br⁻(aq) → 6 Br⁻(aq) + 2 PO₄³⁻(aq)
However, since the phosphate ion (PO₄³⁻) and bromide ion (Br⁻) are both spectator ions in this reaction, there is no net change in ions. Therefore, the net ionic equation is:
NR (No reaction)
To write the balanced net ionic equation for the given reaction, you first need to determine the chemical formula of the products formed when CaBr₂ and Na₃PO₄ are mixed.
Let's start by finding the formulas of the reactants:
- CaBr₂: Calcium bromide
- Na₃PO₄: Sodium phosphate
Now, let's determine the products by combining the cation from one reactant with the anion from the other reactant:
- Ca²⁺ (cation from CaBr₂) will combine with PO₄³⁻ (anion from Na₃PO₄) to form Ca₃(PO₄)₂.
- Na⁺ (cation from Na₃PO₄) will combine with Br⁻ (anion from CaBr₂) to form NaBr.
The balanced molecular equation for the reaction is:
3CaBr₂ + 2Na₃PO₄ → Ca₃(PO₄)₂ + 6NaBr
To find the net ionic equation, you need to identify the spectator ions. These are ions that appear on both sides of the equation and do not participate in the reaction. In this case, Na⁺ and Br⁻ ions are spectator ions.
The net ionic equation is obtained by removing the spectator ions from the balanced molecular equation:
Ca²⁺ + PO₄³⁻ → Ca₃(PO₄)₂
Therefore, the balanced net ionic equation for the reaction when CaBr₂ and Na₃PO₄ are mixed is:
Ca²⁺ + PO₄³⁻ → Ca₃(PO₄)₂
I assume you meant when AQUEOUS solutions of those salts are mixed.
balanced molecular equation:
3CaBr2(aq) + 2Na3PO4(aq) ==> Ca3(PO4)2(s) + 6NaBr(aq)
total ionic:
3Ca^2+(aq) + 6Br^-(aq) + 6Na^+(aq) + 2[PO4]^3-(aq) ==> Ca3(PO4)2(s) + 6Na^+(aq) + 6Br^-(aq)
net ionic: Cancel ions that appear on both sides. The remainder is the net ionic equation:
3Ca^2+(aq) + 2[PO4]^3-(aq) ==> Ca3(PO4)2(s)