Question
When a solution of nickel (II) chloride is mixed with a stoichiometric amount of colorless sodium hydroxide solution, a green, gelatinous precipitate forms.
(a) Write the balanced molecular equation and the net-ionic equation for this reaction.
molecular: Ni(Cl)2 (aq) + NaOH (aq) -> Ni(OH)2 (aq) + NaCl
net-ionic: Cl^- -> OH^-
I am pretty sure that I did this wrong. Could someone please help?
(b) What would you expect to observe if some hydrobromic acid solution was added to the reaction mixture in (a)? Be specific. Include equations for any secondary reactions occurring.
(a) Write the balanced molecular equation and the net-ionic equation for this reaction.
molecular: Ni(Cl)2 (aq) + NaOH (aq) -> Ni(OH)2 (aq) + NaCl
net-ionic: Cl^- -> OH^-
I am pretty sure that I did this wrong. Could someone please help?
(b) What would you expect to observe if some hydrobromic acid solution was added to the reaction mixture in (a)? Be specific. Include equations for any secondary reactions occurring.
Answers
You didn't balance the molecular equation. Balanced it is
molecular: Ni(Cl)2 (aq) + 2NaOH (aq) -> Ni(OH)2 (aq) + 2NaCl
The net ionic equation is
Ni^2+(aq) + 2OH^-(aq) ==> Ni(OH)2(s)
You get that by writing the molecular equation as ions UNLESS the material is a solid, gas, or slightly ionized.
Ni^2+(aq) + 2Cl^-(aq) + 2Na^+(aq) + 2Cl^-(aq) ==>Ni(OH)2(s) + 2Na^+(aq) + 2Cl^-(aq)
Now cancel the spectator ions; those are the ions that appear on both sides of the equation. The 2Cl^-(aq) cancel; the two 2Na^+(aq) ions cancel to leave the net ionic equation I wrote above.
molecular: Ni(Cl)2 (aq) + 2NaOH (aq) -> Ni(OH)2 (aq) + 2NaCl
The net ionic equation is
Ni^2+(aq) + 2OH^-(aq) ==> Ni(OH)2(s)
You get that by writing the molecular equation as ions UNLESS the material is a solid, gas, or slightly ionized.
Ni^2+(aq) + 2Cl^-(aq) + 2Na^+(aq) + 2Cl^-(aq) ==>Ni(OH)2(s) + 2Na^+(aq) + 2Cl^-(aq)
Now cancel the spectator ions; those are the ions that appear on both sides of the equation. The 2Cl^-(aq) cancel; the two 2Na^+(aq) ions cancel to leave the net ionic equation I wrote above.
If HBR was added, would water form?
HBr (aq) + NaOH (aq) -> H2O (l) + NaBr (aq)
HBr (aq) + NaOH (aq) -> H2O (l) + NaBr (aq)
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