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The following instructions were used to prepare magnesium sulfate crystals, $\ce{MgSO4 . 7H2O}$.

  1. Measure $50~ \mathrm{cm^3}$ of dilute sulfuric acid into a beaker and warm the solution.

  2. Using a spatula, add some magnesium oxide and stir the mixture. Continue adding the magnesium oxide until excess is present.

  3. Separate the excess magnesium oxide from the solution of magnesium sulfate.
  4. Heat the solution until crystals form. Obtain the crystals and dry them.

The question I have an issue with is this: How would you know when excess magnesium oxide is present in step 2?

The answer key says that I would know when no more solid (that is, no more magnesium oxide dissolves). However, since we are talking about dilute sulfuric acid, there should be water, and I read on the internet that $\ce{MgO}$ is soluble in water.

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    $\begingroup$ Check the solubility of magnesium oxide. $\endgroup$ – Brinn Belyea Oct 8 '14 at 1:57
  • $\begingroup$ In what sense? As in how much can dissolve in water or if it can dissolve in water at all? Or are you talking about acids $\endgroup$ – dadadok Oct 8 '14 at 2:00
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$\ce{MgO}$ is insoluble in water (solubility: 0.0086 g/100 mL). On the other hand, $\ce{MgSO4}$ is soluble in water (solubility: 25.5 g/100 mL).

The reaction that is taking place is an acid-base reaction:

$$\ce{MgO + H2SO4 -> MgSO4 + H2O}$$

Now for your question, how would you know when excess $\ce{MgO}$ is present? Here you are trying to make $\ce{H2SO4}$ the limiting reagent. When all the $\ce{H2SO4}$ has reacted, no more $\ce{MgO}$ can further react. Therefore, you will see solid in your reaction mixture that cannot be further dissolved.

At the end of your reaction,

$$\ce{MgO_{excess} + H2SO4 -> MgSO4 + H2O + MgO_{remaining}}$$

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  • $\begingroup$ Thank you! But I am a little confused as to why other sites ( BBC Bitesize, for one) say it is soluble in water to form MgOH $\endgroup$ – dadadok Oct 8 '14 at 2:05
  • $\begingroup$ @dadadok Both MgO and Mg(OH)2 are insoluble - you can easily filter them. Note that for MgSO4, if the solution is too basic it will react with OH- to form Mg(OH)2, leading to low yield. $\endgroup$ – t.c Oct 8 '14 at 2:08
  • $\begingroup$ Thanks!This is pretty much the kind of response I was looking for... and goes some way towards relieving exam stress. Thanks again. $\endgroup$ – dadadok Oct 8 '14 at 2:10
  • $\begingroup$ @dadadok Mg(OH)2 is just barely soluble in water, that's why you're finding some sources saying it's insoluble and others saying it is soluble. The same is true for MgO, which will dissolve to a very small extent in water. "Very small" and "barely" here mean only a few thousandths of a gram per 100mL of water. $\endgroup$ – Jason Patterson Oct 8 '14 at 12:30

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