A salt can be defined as a substance formed when the hydrogen in an acid is replaced by a metal ion. You will use some of the reactions of acids to form salts and then isolate them from the reaction mixture.
The reaction of copper(II) carbonate with hydrochloric acid
- Open Yenka file Model 1.
- Acids react with metal carbonates to form metal salts, water, and one other product. Name this other product.
- Copper(II) carbonate is an insoluble substance, and when it reacts with hydrochloric acid it forms copper(II) chloride which is soluble and dissolves. In this experiment you will use a large quantity of copper(II) carbonate (an excess) and only a small amount of acid. Which of the two reactants will be left over at the end of the reaction, and which will be used up completely?
Reactant left over Reactant used upAnswerCopper(II) carbonateAnswerHydrochloric acid
- What method can you use to remove the excess copper(II) carbonate, remembering that it is insoluble?
- Add the two reactants together in the beaker. Since this reaction takes some time to complete, select 'Simulation time' in the main toolbar and then use the slider to increase the rate ×10. What is the colour of the copper(II) chloride solution formed?
- Click on the reaction details for the beaker and wait until there is no hydrochloric acid left. Close the reaction details. Place a new beaker under the filter funnel and pour the reaction mixture through the filter paper. What is the residue and what is the filtrate?
Residue FiltrateAnswerCopper(II) carbonateAnswerCopper(II) chloride and water
or copper(II) chloride solution
- If you left the filtrate in a warm room for a long time, the water would evaporate leaving crystals of copper(II) chloride in the beaker. How can you speed up this process?
- Try out your suggestion above, and write down the mass of copper(II) chloride that is produced at the end of the experiment.
Hydrochloric acid reacts with metal carbonates to produce metal chlorides. Provided that the initial metal carbonate is insoluble and the metal chloride is soluble, it is possible to filter off the unreacted excess metal carbonate leaving the desired salt solution. Evaporation of the salt solution produces the pure salt.
- Open Yenka file Model 2
- You are provided with some general information in this model. Use it and the chemicals provided in Scene 2 (by clicking on the button in the main toolbar which says 'Scenes' when hovered over) to answer the following questions and perform the tasks required.
Write a word equation for the reaction of copper(II) oxide with hydrochloric acid.AnswerCopper(II) oxide + hydrochloric acid → copper(II) chloride + water
- Which reactant would you have in excess in order to remove it easily from the final mixture? Explain your answer.
AnswerCopper(II) oxide, because it is insoluble and can be filtered off at the end of the experiment
- Click on Scene 2. Use the chemicals provided to produce a dry sample of copper(II) chloride (hint: use heat as in task 1).
Write a balanced formula equation for the reaction between copper(II) oxide and hydrochloric acid.AnswerCuO + 2HCl → CuCl2 + H2O
- When you have completed the preparation you should notice a small amount of another solid produced in addition to the desired product. What is the formula of this impurity?
The method of adding excess insoluble reactant to a small quantity of acid can also be used for forming soluble metal salts from insoluble metal oxides, which are known as bases.
- The first task takes the student through the practical details of a salt preparation from an insoluble metal carbonate.
- The second task asks the student to use the similarities between metal carbonates and oxides to work out an analogous method for producing copper(II) chloride from copper(II) oxide.
- Exercise Salt Preparation Methods 2 deals with titrations to produce soluble salts from soluble reactants.