0.8 Reactions in aqueous solutions  (Page 7/10)

Method:

1. Prepare 2 test tubes with approximately 5 ml of dilute Cu(II) chloride solution in each
2. Prepare 1 test tube with 5 ml sodium carbonate solution
3. Prepare 1 test tube with 5 ml sodium sulphate solution
4. Carefully pour the sodium carbonate solution into one of the test tubes containing copper(II) chloride and observe what happens
5. Carefully pour the sodium sulphate solution into the second test tube containing copper(II) chloride and observe what happens

Results:

1. A light blue precipitate forms when sodium carbonate reacts with copper(II) chloride
2. No precipitate forms when sodium sulphate reacts with copper(II) chloride

It is important to understand what happened in the previous demonstration. We will look at what happens in each reaction, step by step.

1. Reaction 1: Sodium carbonate reacts with copper(II) chloride.
   When these compounds react, a number of ions are present in solution: ${\mathrm{Cu}}^{2+}$ , ${\mathrm{Cl}}^{-}$ , ${\mathrm{Na}}^{+}$ and ${\mathrm{CO}}_3^{2-}$ . Because there are lots of ions in solution, they will collide with each other and may recombine in different ways. The product that forms may be insoluble, in which case a precipitate will form, or the product will be soluble, in which case the ions will go back into solution. Let's see how the ions in this example could have combined with each other:
   ${\mathrm{Cu}}^{2+}+{\mathrm{CO}}_3^{2-}\to {\mathrm{CuCO}}_3$
   ${\mathrm{Cu}}^{2+}+2{\mathrm{Cl}}^{-}\to {\mathrm{CuCl}}_2$
   ${\mathrm{Na}}^{+}+{\mathrm{Cl}}^{-}\to \mathrm{NaCl}$
   $2{\mathrm{Na}}^{+}+{\mathrm{CO}}_3^{2-}\to {\mathrm{Na}}_2{\mathrm{CO}}_3$
   You can automatically exclude the reactions where sodium carbonate and copper(II) chloride are the products because these were the initial reactants. You also know that sodium chloride ( $\mathrm{NaCl}$ ) is soluble in water, so the remaining product (copper carbonate) must be the one that is insoluble. It is also possible to look up which salts are soluble and which are insoluble. If you do this, you will find that most carbonates are insoluble, therefore the precipitate that forms in this reaction must be ${\mathrm{CuCO}}_3$ . The reaction that has taken place between the ions in solution is as follows:
   $2{\mathrm{Na}}^{+}+{\mathrm{CO}}_3^{2-}+{\mathrm{Cu}}^{2+}+2{\mathrm{Cl}}^{-}\to {\mathrm{CuCO}}_3+2{\mathrm{Na}}^{+}+2{\mathrm{Cl}}^{-}$
2. Reaction 2: Sodium sulphate reacts with copper(II) chloride.
   The ions that are present in solution are ${\mathrm{Cu}}^{2+}$ , ${\mathrm{Cl}}^{-}$ , ${\mathrm{Na}}^{+}$ and ${\mathrm{SO}}_4^{2-}$ . The ions collide with each other and may recombine in different ways. The possible combinations of the ions are as follows:
   ${\mathrm{Cu}}^{2+}+{\mathrm{SO}}_4^{2-}\to {\mathrm{CuSO}}_4$
   ${\mathrm{Cu}}^{2+}+2{\mathrm{Cl}}^{-}\to {\mathrm{CuCl}}_2$
   ${\mathrm{Na}}^{+}+{\mathrm{Cl}}^{-}\to \mathrm{NaCl}$
   ${\mathrm{Na}}^{+}+{\mathrm{SO}}_4^{2-}\to {\mathrm{Na}}_2{\mathrm{SO}}_4$
   If we look up which of these salts are soluble and which are insoluble, we see that most chlorides and most sulphates are soluble. This is why no precipitate forms in this second reaction. Even when the ions recombine, they immediately separate and go back into solution. The reaction that has taken place between the ions in solution is as follows:
   $2{\mathrm{Na}}^{+}+{\mathrm{SO}}_4^{2-}+{\mathrm{Cu}}^{2+}+2{\mathrm{Cl}}^{-}\to 2{\mathrm{Na}}^{+}+{\mathrm{SO}}_4^{2-}+{\mathrm{Cu}}^{2+}+2{\mathrm{Cl}}^{-}$

[link] shows some of the general rules about the solubility of different salts based on a number of investigations: