Respuesta :
Answer : The net ionic equation will be,
[tex]3Sr^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Sr_3(PO_4)_2(s)[/tex]
Explanation :
In the net ionic equations, we are not include the spectator ions in the equations.
Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.
The given balanced ionic equation will be,
[tex]3Sr(OH)_2(aq)+2Li_3PO_4(aq)\rightarrow 6LiOH(aq)+Sr_3(PO_4)_2(s)[/tex]
The ionic equation in separated aqueous solution will be,
[tex]3Sr^{2+}(aq)+6OH^{-}(aq)+6Li^{+}(aq)+2PO_4^{3-}(aq)\rightarrow Sr_3(PO_4)_2(s)+6Li^+(aq)+6OH^{-}(aq)[/tex]
In this equation, [tex]Li^+\text{ and }OH^-[/tex] are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
[tex]3Sr^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Sr_3(PO_4)_2(s)[/tex]
The complete ionic equation for the reaction between aqueous solutions of strontium hydroxide and lithium phosphate is:
3Sr²⁺(aq) + 6OH¯(aq) + 6Li⁺(aq) + 2PO₄³¯(aq) —> Sr₃(PO₄)₂(s) + 6Li⁺(aq) + 6OH¯(aq)
The complete ionic equation is simply an equation showing all available entities including the spectator ions present in the solution.
With the above information in mind, we shall write the complete ionic equation for reaction between aqueous solutions of strontium hydroxide and lithium phosphate. This can be obtained as follow:
In solution, strontium hydroxide, Sr(OH)₂ and lithium phosphate, Li₃PO₄ will dissociate as follow:
Sr(OH)₂(aq) —> Sr²⁺(aq) + 2OH¯(aq)
Li₃PO₄(aq) —> 3Li⁺(aq) + PO₄³¯(aq)
The reaction will proceed as follow:
Sr(OH)₂(aq) + Li₃PO₄(aq) —>
Sr²⁺(aq) + 2OH¯(aq) + 3Li⁺(aq) + PO₄³¯(aq) —> Sr₃(PO₄)₂(s) + 3Li⁺(aq) + 2OH¯(aq)
Balance the equation by writing 3 before Sr²⁺, 6 before OH¯, 6 before Li⁺ and 2 before PO₄³¯, to obtain the complete ionic equation as shown:
3Sr²⁺(aq) + 6OH¯(aq) + 6Li⁺(aq) + 2PO₄³¯(aq) —> Sr₃(PO₄)₂(s) + 6Li⁺(aq) + 6OH¯(aq)
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