Asked by Anonymous
I need some help finding the net ionic equation for these two problems:
1. Bromine is added to a dilute solution of sodium hydroxide.
2. Sulfuric acid is added to a solution of lithium hydrogen carbonate.
This is what I have for reactants:
Br2 + NaOH and H2SO4 + LiHCO3.
I'm not sure about 1, since I don't think the rxn makes bromine hydroxide (activity series?)
And for 2, I thought it was a double replacement, but...it isn't? Help plz?
Thanks!
1. Bromine is added to a dilute solution of sodium hydroxide.
2. Sulfuric acid is added to a solution of lithium hydrogen carbonate.
This is what I have for reactants:
Br2 + NaOH and H2SO4 + LiHCO3.
I'm not sure about 1, since I don't think the rxn makes bromine hydroxide (activity series?)
And for 2, I thought it was a double replacement, but...it isn't? Help plz?
Thanks!
Answers
Answered by
DrBob222
I would write #1 as
Br2 + 2NaOH ==> NaOBr + NaBr + H2O
This is the balanced molecular equation.
Now convert to ions using the following rules.
solids written as molecules
liquids written as molecules
solutions written as ions unless they are weak electrolytes which are written as molecules.
Br2 + 2Na^+ + 2OH^- ==> NaOBr + Na^+ + Br^-^- + H2O
Now cancel ions common to both sides.
1Na^+ can be canceled on each side.
What is left is the net ionic equation.
Br2 + Na^+ + 2OH^- ==>NaOBr + Br^- + H2O
Check my thinking. Check my work.
Br2 + 2NaOH ==> NaOBr + NaBr + H2O
This is the balanced molecular equation.
Now convert to ions using the following rules.
solids written as molecules
liquids written as molecules
solutions written as ions unless they are weak electrolytes which are written as molecules.
Br2 + 2Na^+ + 2OH^- ==> NaOBr + Na^+ + Br^-^- + H2O
Now cancel ions common to both sides.
1Na^+ can be canceled on each side.
What is left is the net ionic equation.
Br2 + Na^+ + 2OH^- ==>NaOBr + Br^- + H2O
Check my thinking. Check my work.
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