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I am currently taking a general chemistry course and during a lab they had us do the following procedure:

  1. Place 2mL of 5% $ \ce{NaClO} $ solution in a test tube.
  2. Add 10 drops of cyclohexane.
  3. Add ~10 drops $ 6~\mathrm M\ \ce{HCl} $
  4. In another test tube, add a pinch of solid $ \ce{KBr} $
  5. Use a dropping pipette to withdraw the chlorine/cyclohexane layer from above and add it to the test tube containing $ \ce{KBr} $

When I did this, the mixture turned dark red and bubbled with a hint of purple to it. In my lab report, my teacher wants me to write the net ionic equation(s) of this reaction, but I have no idea how to do that. Any help would be great.

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  • $\begingroup$ This guide might help you. $\endgroup$ – Alex Mar 28 '13 at 18:39
  • $\begingroup$ Do you have any idea of what could be the reaction(s) that took place? $\endgroup$ – Jerry Mar 28 '13 at 19:07
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I'd like to add some things to what Tanith Rosenbaun said.

First, the point of net ionic equations is not just to write out the equations in terms of ions, but to focus on only those ions that experience a change in charge or valence. Ions which don't change in charge don't really participate in the reaction and called spectator ions.

Here's the last equation from Rosenbaum's answer:

$\ce{2H3O+ + 2Cl- + 2Na+ + CO3^2- -> 2H2O + 2Cl- + 2Na+ + CO2}$

Notice that the sodium and chloride ions are unchanged in this reaction. So we drop them, and just write out the ions that change:

$\ce{2H3O+ + CO3^2- -> 2H2O + CO2}$

This is the net ionic equation: it focuses precisely on what is interesting, which in this case is the neutralization reaction.

Now let's go back to your problem: People have been asking what you know about it. You seem to know that you are transferring a solution of chlorine dissolved in cyclohexane. You got that chlorine from the following reaction:

$\ce{2HCl + NaClO -> Cl2 + NaCl + H2O}$

and the chlorine preferentially dissolves in the cyclohexane. Now it's important to understand that the cyclohexane doesn't play any role in the second part of your experiment, it's just how you transport the chlorine. (And it's why you need to be careful with acids around bleach: chlorine gas is poisonous!)

The key question is: what happens when you put elemental chlorine in the presence of dissolved potassium bromide?

$\ce{Cl2 + KBr -> ?}$ (Note: this is not balanced yet!)

In particular, do you know what you produced when that solution turned dark red? Hint: this is a very important property that halogens like chlorine and bromine have towards each other, which is indeed seen best via a net ionic equation. I'll happily provide the answer if you've tried some stuff and are still stuck.

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  • $\begingroup$ Good point on the net equation. I forgot that bit. $\endgroup$ – Tanith Rosenbaum Mar 30 '13 at 15:43
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A ionic equation of a reaction isn't much different from a regular reaction equation, except that you write the ions separately. As perusual, reactants go left, products go right.

As an example, lets let hydrochloric acid react with sodium carbonate in aqueous solution. First you figure out what ions your reactants dissociate into. For $HCl$ that would be $H^{+}$ and $Cl^{-}$, and for $Na_{2}CO_{3}$ that's $2 Na^{+}$ and $CO_{3}^{2-}$, and lastly you got water, $H_{2}O$. That's the left side of this example equation. Which in sum would look like this:

$H^{+}+Cl^{-}+2 Na^{+}+CO_{3}^{2-}+H_{2}O \longrightarrow$

Next you need to figure out what ions react and in what way. In this example, first, $H^{+}$ and $H_{2}O$ react to form $H_{3}O^{+}$, and after that, $H_{3}O^{+}$ and $CO_{3}^{2-}$ react to form $H_{2}O$ and $CO_{2}$. You would write the two parts of the reaction like this:

$H^{+}+Cl^{-}+2 Na^{+}+CO_{3}^{2-}+H_{2}O \longrightarrow Cl^{-}+2 Na^{+}+CO_{3}^{2-}+H_{3}O^{+}$ $Cl^{-}+2 Na^{+}+CO_{3}^{2-}+H_{3}O^{+}\longrightarrow Cl^{-}+2 Na^{+}+CO_{2}+H_{2}O$

Now for the last bit, stoichiometry. Stoichiometry simply means making sure the numbers of atoms left and right are the same, and the ratio of atoms needed to make the reaction work is right. The first line works as is, but in the second line you'll notice that we're using a single positive charge in $H_{3}O^{+}$ to react with two negative charges in $CO_{3}^{2-}$. Something's fishy there, as charges need to match too. And indeed, we need two $H_{3}O^{+}$ for one $CO_{3}^{2-}$. After we changed that in the second line, we need to alter the first line too to match, like so:

$2H^{+}+2Cl^{-}+2 Na^{+}+CO_{3}^{2-}+2H_{2}O \longrightarrow 2Cl^{-}+2 Na^{+}+CO_{3}^{2-}+2H_{3}O^{+}$ $2Cl^{-}+2 Na^{+}+CO_{3}^{2-}+2H_{3}O^{+}\longrightarrow 2Cl^{-}+2 Na^{+}+CO_{2}+3H_{2}O$

That's what your teacher wants you to do for the reaction you cited. Good luck!

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