3 Flashcards

Question

Test for carbon dioxide

Answer 1

If you make a solution of calcium hydroxide in water (called limewater) and bubble gas through it, the solution will turn cloudy if there's carbon dioxide in the gas. The cloudiness is caused by the formation of calcium carbonate.

Answer 2

To make a pure, dry sample of an insoluble salt, you can use a precipitation reaction. Precipitation occurs when an insoluble solid (known as a precipitate) forms in a solution. You just need to pick the right two soluble salts and react them together to get your insoluble salt. You can make lead chloride (an insoluble salt) by mixing together lead nitrate and sodium chloride (both soluble salts). lead nitrate + sodium chloride → lead chloride + sodium nitrate Pb(NO3)2(aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3(aq)

Answer 3

1. Add one spatula of lead nitrate to a test tube. Add some deionised water in order to dissolve it. Shake the test tube thoroughly to ensure that all the lead nitrate has dissolved. 2. Repeat step 1 in a separate test tube, except this time use one spatula of sodium chloride. 3. Tip the two solutions into a small beaker, and stir thoroughly to make sure the solutions are fully mixed. The lead chloride should precipitate out 4. Fold a piece of filter paper and place it into a filter funnel. Place the funnel into the top of a conical flask. 5. Pour the contents of the beaker into the middle of the filter paper. Make sure that the solution doesn't go above the top of the filter paper- otherwise some of the solid lead chloride could end up in the filtrate. 6. Swill out the beaker with more deionised water, and tip this into the filter paper. This will ensure all the precipitate is taken from the beaker. 7. Rinse the contents of the filter paper with deionised water to make sure that all the soluble sodium nitrate has been washed away. 8. Scrape the lead chloride onto fresh filter paper and leave it to dry in an oven or a desiccator.

Answer 4

Soluble salts (salts that dissolve in water) can be made by reacting an acid with a metal or an insoluble base (such as a metal oxide, metal hydroxide metal carbonate). When making a soluble salt, the first thing you need to do is choose appropriate reagents to produce that particular salt. You should be able to work out which are the right reagents to choose from the name of the salt you want to produce.

Answer 5

You can make copper sulfate (a soluble salt) by adding copper oxide (an insoluble reagent) to warm sulfuric acid. CuO (s) + H2SO4(aq) → CuSO4 (aq) + H2O(l) The copper oxide is added in excess and separated out at the end of the reaction using filter paper. Many metal oxides and some metal carbonates are insoluble, so this is also the method you'd use if you're using any of those.

Answer 6

1. Put the sulfuric acid in a beaker. Gently warm the dilute acid by placing the beaker in a water bath (see p.326 for more on water baths) speeds up the reaction between the acid and the insoluble base. Carry out this step in a fume cupboard to avoid releasing acid fumes into the room. -this 2. Add the copper oxide (the base) to the acid and stir - they will react with each other to form copper sulfate (a soluble salt) and water. 3. Keep adding the copper oxide until it is in excess. When it reaches excess, the oxide will start to sink to the bottom and won't react - this shows that all the acid has been neutralised and the reaction has finished. It's important that the base is in excess, as this ensures that you won't have any leftover acid in your product. 4. Then you need to filter out the excess copper oxide to separate it from the copper sulfate solution. This ensures that the copper sulfate salt isn't contaminated with copper oxide once it's dried. This stage is done using filter paper and a filter funnel 5. At the end of the filtration, you'll be left with a solution of copper sulfate salt dissolved in water. You can convert this into pure, solid crystals of copper sulfate using crystallisation. To do this you first need to heat the salt solution using a Bunsen burner and an evaporating dish (see p.105), in order to evaporate some of the water and make the solution more concentrated, 6. Then, stop heating it and leave the solution to cool. Blue crystals of hydrated copper sulfate should form, which can be filtered out of the solution and then dried..

Answer 7

Soluble salts can be made by reacting an acid with an alkali. It is not always obvious when the reaction has finished, as there's no signal that all the acid has been neutralised. You also can't just add an excess of alkali to the acid, because the salt is soluble and would be contaminated with the excess alkali. In order to overcome these problems you need to work out exactly the right amount of alkali to neutralise the acid. For this, you need to do a titration using an indicator.

Answer 8

1. Measure out a set amount of acid into a conical flask using a pipette. Add a few drops of indicator, then place the flask on a white tile. 2. Fill a burette with alkali and record the volume. 3. Slowly add alkali to the acid, gently swirling the flask all the time, until reach the end point- this is when the you acid's been exactly neutralised and the indicator changes colour 4. Read the final volume of alkali left in the burette. Calculate the amount of alkali that was needed to neutralise the acid by subtracting the final volume of alkali in the burette from the initial volume. 5. Then, carry out the reaction using exactly the same volumes of alkali and acid but with no indicator, so the salt won't be contaminated with indicator. 6. The solution that remains when the reaction is complete contains only the salt and water. This is because all the alkali has reacted with all the acid. 7. Slowly evaporate off some of the water and then leave the solution to crystallise. Filter off the solid and dry it - you'll be left with a pure, dry salt.

Answer 9

Electrolysis requires a liquid to conduct the electricity, called the electrolyte. Electrolytes contain free ions - they're usually a molten or dissolved ionic compound. In either case, it's the free ions which conduct the electricity and allow the whole thing to work.

Answer 10

a solid that conducts electricity and is submerged in the electrolyte.

Answer 11

In electrolysis, the electrodes are placed into the electrolyte and ions move from one electrode to the other - this allows the conduction of electricity through the circuit. The positive ions (cations) in the electrolyte will move towards the negative electrode (the cathode) and gain electrons. The negative ions (anions) in the electrolyte will move towards the positive electrode (the anode) and lose electrons. As ions gain or lose electrons they become atoms or molecules and are discharged (released) from the electrolyte. These atoms or molecules are the products of electrolysis.

Answer 12

Electrolysis always involves an oxidation reaction and a reduction reaction. Reduction is occurring at the negative electrode (the cathode) as the positive ions (cations) are gaining electrons. Oxidation occurs at the positive electrode (the anode) as the negative ions (anions) are losing electrons.

Answer 13

Electrolysis of molten lead bromide Lead bromide (PbBr2) is an ionic compound, so when it is molten, it will conduct electricity. Electrolysis of lead bromide breaks it down into lead (Pb) and bromine (Br2). The positive ions are attracted towards the negative electrode. Here, each lead ion gains two electrons and becomes a lead atom. Pb2+ + 2e- -> Pb The negative ions are attracted towards the positive electrode. Here, bromide ions lose one electron each and form bromine molecules (Br₂). 2Br- -> Br2 + 2e-

Answer 14

If a salt is dissolved in water there will be some H+ and OH- ions as well as the ions from the salt in the solution. In this situation, the products of electrolysis depend on how reactive the elements involved are. At the negative electrode (the cathode), if the metal is more reactive than hydrogen the metal ions will stay in solution (e.g. sodium). This is because the more reactive an element, the more likely it is to stay as ions. So, hydrogen will be produced unless the metal is less reactive than it. If the metal is less reactive than hydrogen (e.g. copper), then a solid layer of pure metal will be produced instead. At the positive electrode (the anode), if OH- and halide ions (Cl-, Br-, I-) are present then molecules of chlorine, bromine or iodine will be formed. If no halide is present (e.g. NO3-, SO4 2-), then the OH- ions are discharged and oxygen and water will be formed - 4OH- -> 2H2O + O2 + 4e-

Answer 15

A solution of copper sulphate (CuSO4) contains four different ions: Cu 2+, SO4 2-, H+ and OH-. When you electrolyse a solution of CuSO4 with inert electrodes: Copper metal is less reactive than hydrogen. So, at the cathode, copper metal is produced and coats the electrode: Cu2+ + 2e- → Cu There aren't any halide ions present so at the anode oxygen and water are produced. The oxygen can be seen as bubbles: 4OH- → О₂ + 2H₂O + 4е-

Answer 16

As the reaction continues, the mass of the anode will decrease and the mass of the cathode will increase. This is because copper is transferred from the anode to the cathode. The reaction is quite slow, so the cell should be left for around 30 minutes in order to achieve a measurable change in mass. You can measure how the mass of the electrodes has changed during the experiment by finding the difference between the masses of the electrodes before and after the experiment. Anode – Cu → Cu 2+ + 2e- Cathode – Cu 2+ + 2e- → Cu

Answer 17

Copper can be extracted from its ore by reduction with carbon, but copper made in this way is impure. Electrolysis can be used to purify the copper using an electrochemical cell with copper electrodes. When copper is purified using electrolysis, the anode starts off as a big lump of impure copper and the cathode starts off as a thin piece of pure copper. The electrolyte is copper sulfate solution (which contains Cu²+ ions). Here's what happens during the process: 1. Copper in the impure copper anode forms copper ions which dissolve into the electrolyte – Cu → Cu 2+ + 2e- 2. The copper ions move to the pure copper cathode, and react to form a layer of pure copper – Cu 2+ + 2e- → Cu 3. Any impurities from the impure copper anode sink to the bottom of the cell, forming a sludge.