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I first conducted the following reactions, each in separate test tubes:

  1. $\ce{NaBr}$ + $\ce{H2SO4}$ + $\ce{NaBrO3}$
  2. $\ce{KI}$ + $\ce{H2SO4}$ + $\ce{KIO3}$

...with water and dichloroethane added to each as well.

Then, I removed the aqueous layer from each test tube and added $\ce{NaOH}$ to the dichloroethane layer in each.

  1. (#1) $\ce{NaOH}$ added to dichloroethane
  2. (#2) $\ce{NaOH}$ added to dichloroethane

So far, I have the following equations written:

  1. $\ce{NaBrO3 + 5NaBr + 3H2SO4 -> 3Na2SO4 + 3H2O + 3Br2}$
  2. $\ce{KIO3 + 5KI + 3H2SO4 -> 3K2SO4 + 3H2O + 3I2}$
  3. $\ce{3Br2 + 6OH^{-} -> 5Br^{-} + BrO3^{-} + 3H2O}$
  4. $\ce{3I2 + 6OH -> 5I^{-} + IO3^{-} + 3H2O}$

Please check my work. If without error, I just need to convert these to equations to net ionic.

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  • $\begingroup$ A list of 3 compounds is not a reaction. Is sodium bromate a reactant or a product? You are not being remotely clear about what experiment you did. $\endgroup$ – DavePhD Dec 2 '14 at 20:08
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    $\begingroup$ @DavePhD I corrected the question for clarity. In the experiment, I put a small quantity of sodium bromide in one test tube and a similar quantity of potassium iodide in the other test tube. I then added 1 mL of water, 3 drops of 9M sulfuric acid, and 2 mL of dichloroethane to each. I then added a small quantity of sodium bromate to the first test tube and a similar quantity of potassium iodate to the other. $\endgroup$ – JaxBragg Dec 2 '14 at 20:27
  • $\begingroup$ So, you're just checking if these equations are balanced? $\endgroup$ – John Snow Dec 2 '14 at 22:21
  • $\begingroup$ @JohnSnow The original question was much different $\endgroup$ – DavePhD Dec 3 '14 at 1:21
  • $\begingroup$ @JaxBragg looks almost right now, just missing a minus sign in equation 4, and 6 H2SO4 going to 6 NaHSO4 or 6 KHSO4 is more realistic in equations 1 and 2. $\endgroup$ – DavePhD Dec 3 '14 at 1:27
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(answer to original question)

No, you are not on the right track.

If bromate is a starting material, it should not appear as a product in reaction 1. Instead, consider how bromide and bromate can react with each other.

If iodate is a starting material, it should not appear as a product in reaction 2. Instead, consider how iodide and iodate can react with each other.

For reactions 3 and 4, a hint would be "disproportionation".

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  • $\begingroup$ Okay, so my initial reaction (#1) would be NaBrO3 + 5 NaBr + 3 H2SO4 --> 3 Na2SO4 + 3 H2O + 3 Br2. Then, adding NaOH to the dichloroethane layer, I would have 3 Br2 + 6 OH(-) --> 5 Br(-) + BrO3(-) + H2O. Correct? $\endgroup$ – JaxBragg Dec 2 '14 at 20:58
  • $\begingroup$ @JaxBragg looks much better, the 3 Br2 + 6 OH- reaction could use balancing with respect to O though $\endgroup$ – DavePhD Dec 2 '14 at 21:10

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