C4 - Chemical changes Flashcards

Question

How do you predict the names of the salt produced?

Answer 1

**Salts** * A salt is a compound that is formed when the **hydrogen atom i**n an acid is replaced by a **metal** * For example if we replace the H in HCl with a potassium atom, then the salt potassium chloride is formed, KCl * Salts are an important branch of chemistry due to the varied and important uses of this class of compounds * These uses include fertilisers, batteries, cleaning products, healthcare products and fungicides **Naming salts** * The name of a salt has two parts * The first part comes from the **metal, metal** oxide or **metal** carbonate used in the reaction * The second part comes from the **acid** * The name of the salt can be determined by looking at the reactants * For example hydrochloric acid always produces salts that end in chloride and contain the **chloride** ion, Cl^– * Other examples: * **Sodium** hydroxide reacts with hydrochloric acid to produce sodium chloride. * **Zinc** oxide reacts with **sulfuric** acid to produce **zinc sulfate** * A list of the common ions and their formulae is shown below Salts have no overall charge since the sum of the charges on the ions is equal to **zero** * From the table of the common ions it is clear the charges on each the Group I elements is always +1, Group II is +2, Group VI is -2 and Group VII is -1 * If you know the ions present in a salt you can identify the formula from balancing the charges

Answer 2

**Answer** **Step 1:** Write out the formulae of each ion, including their charges Mg²⁺ and PO₄^3- **Step 2:** Balance the charges by multiplying them out: (Mg²⁺) x **3** = +6 and (PO₄^3-) x **2** = -6; so +6 – 6 = 0. So the formula is **Mg₃(PO₄)₂**

Answer 3

Naming salts The name of a salt has two parts. The first part comes from the base, alkali or metal carbonate. The second part comes from the acid: * **hydrochloric acid produces chloride salts HCL** * **nitric acid produces nitrate salts HNO3** * **sulfuric acid produces sulfate salts H2SO4**

Answer 4

Making soluble salts from insoluble substances A **soluble** **salt** can be prepared by reacting an **acid** with a suitable **insoluble** **reactant** including: * a metal * a metal oxide * a metal hydroxide * a carbonate The insoluble reactant chosen depends upon the particular salt required. For example, copper does not react with dilute acids, so this metal cannot be used. On the other hand, sodium is too reactive to be used safely. As the reaction between metals and acids produces **flammable** hydrogen, chemists usually make salts by reacting a metal oxide or a metal carbonate with an acid. Choosing reactants The table shows some examples of the salts produced by different combinations of insoluble reactants and acids. Making a salt To make a soluble salt from an acid and an insoluble reactant: 1. Add powdered insoluble reactant to acid in a beaker, one spatula at a time, stirring to mix. Continue adding powder until it is in **excess** (some unreacted powder is left over). All the acid has now reacted. 2. Filter the mixture in the beaker to remove the excess solid. The filtrate now contains only the salt and water. 3. Heat the solution in an evaporating dish over a water bath. Stop heating when small crystals start to appear around the edge of the evaporating basin. The solution is now **saturated**. 4. Leave the saturated solution at room temperature for a day or two. This gives time for large **crystals** to form. 5. If necessary, dry the crystals by dabbing gently with filter paper.

Answer 5

* A soluble salt can be made from the reaction of an acid with an insoluble base * During the preparation of soluble salts, the insoluble reactant is added in excess to ensure that **all** of the **acid** has **reacted** * If this step is not completed, any unreacted acid would become **dangerously** concentrated during evaporation and crystallisation * The excess reactant is then removed by **filtration** to ensure that only the salt and water remain * Since all of the acid has reacted and the excess solid base has been removed then the solution left can only be salt and water * If a carbonate was used as the solid base instead of an oxide or hydroxide, then any carbon dioxide gas produced would have been released into the atmosphere * A common example is the preparation of copper(II) sulfate which can be made with copper(II) oxide and dilute sulfuric acid: **CuO (s) + H₂SO₄(aq) ⟶ CuSO₄(s) + H₂O (l)** * The acid could also be reacted with a metal to produce the salt, as long as the metal is above hydrogen in the reactivity series and not too reactive so that a dangerous reaction does not take place

Answer 6

**Strong and weak acids** **Acids** in solution are a source of hydrogen **ions**, H⁺. The hydrogen ions are produced when the acid **dissociates** or breaks down to form ions. **Strong acids** Strong acids completely dissociate into ions in solution. For example, hydrochloric acid is a strong acid. It ionises completely to form hydrogen ions and chloride ions: HCl(aq) → H⁺(aq) + Cl^-(aq) Nitric acid and sulfuric acid are also strong acids. **Weak acids** Weak acids only partially dissociate in solution. For example, ethanoic acid is a weak acid. It is only partially ionised to form hydrogen ions and ethanoate ions: CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO^-(aq) The ⇌ symbol is used in the equation to show that the reaction is a **reversible reaction** and does not go to completion.

Answer 7

pH and hydrogen ion concentration The **pH** of a solution is a measure of its concentration of hydrogen ions: * the higher the concentration of H⁺ ions in an acidic solution, the lower the pH * the lower the concentration of H⁺ ions in an acidic solution, the higher the pH This means that, for a given concentration in aqueous solution, the stronger an acid, the lower the pH. **The more concentrated the solution of an acid, the lower its pH will be.** If the hydrogen ion concentration in a solution increases by a factor of 10, the pH of the solution decreases by 1 unit. pH of alkaline solutions The higher the concentration of OH^- ions in an alkaline solution, the higher the pH.

Answer 8

* We have already seen that pH is a measure of the concentration of H⁺ ions in solution * The pH scale is **logarithmic**, meaning that each change of 1 on the scale represents a change in concentration by a **factor** of **10** * Therefore an acid with a pH of 3 has ten times the concentration of H⁺ ions than an acid of pH 4 * An acid with a pH of 2 has 10 x 10 = 100 times the concentration of H⁺ ions than an acid with a pH of 4 * From this we can summarize that for two acids of **equal concentration**, where one is **strong** and the other is **weak**, then the strong acid will have a lower pH due to its capacity to dissociate more and hence put more H⁺ ions into solution than the weak acid

Answer 9

* Titrations are a method of analysing the **concentration** of solutions * Acid-base titrations are one of the most important kinds of titrations * They can determine exactly how much alkali is needed to neutralise a quantity of acid – and vice versa * You may be asked to calculate the **moles** present in a given amount, the **concentration** or **volume** required to **neutralise** an acid or a base * Titrations can also be used to prepare **salts** or other precipitates and in **redox** **reactions** * **Indicators** are used to show the endpoint in a titration * Wide range indicators such as litmus are not suitable for a titration as they do not give a sharp enough colour change at the end point * Some of the most common indicators with their corresponding colours are shown below

Answer 10

* Once a titration is completed and the average titre has been calculated, you can now proceed to calculate the unknown variable using the formula triangle as shown below

Answer 11

A **soluble** **salt** can be prepared by reacting an **acid** with a dilute **solution** of an **alkali** such as sodium hydroxide or ammonia. The main steps are: 1. Carry out a **titration**. This is to determine the volumes of acid and alkali that must be mixed to obtain a solution containing only salt and water. 2. Mix the acid and alkali in the correct proportions, as determined in step 1, but this time without including an **indicator**. Pure dry **crystals** can be produced by **crystallisation**, followed by drying on a watch glass or in a warm oven. Carrying out a titration to find out volumes of acid and alkali solutions that reactApparatus The apparatus needed includes: * a **pipette** to accurately measure the volume of a reactant before transferring it to a conical flask * a **burette** to add small, measured volumes of one **reactant** to the other reactant Method This is an outline method for carrying out a titration in which an acid is added to an alkali. The method is the same for sulfuric acid, hydrochloric acid and nitric acid. 1. Use the pipette and pipette filler to add a measured volume of sodium hydroxide solution to a clean conical flask. 2. Add a few drops of indicator and put the conical flask on a white tile. 3. Fill the burette with hydrochloric acid and note the starting volume. 4. Slowly add the acid from the burette to the alkali in the conical flask, swirling to mix. 5. Stop adding the acid when the **end-point** is reached (when the indicator first permanently changes colour). Note the final volume reading. 6. Repeat steps 1 to 5 until **concordant titres** are obtained. More accurate results are obtained if acid is added drop by drop near to the end-point. Readings should be recorded to two decimal places, ending in 0 or 5 (where the liquid level is between two graduations on the burette). The **titre** is the volume added (the difference between the end and start readings). **Analysis** At least two concordant titres should be ticked (✔) in the table above. These are titres within 0.20 cm³ (or sometimes 0.10 cm³) of each other. **Worked example** Calculate the mean titre from the table above. Ignore the rough run, and run 2 (because they are not concordant): Mean titre = (24.60 + 24.70)/2 = 24.65 cm³

Answer 12

* Covalent compounds **cannot conduct electricity** hence they do not undergo electrolysis * An electrolytic cell is the name given to the set-up used in electrolysis and which consists of the following: * **Electrode:** a rod of metal or graphite through which an electric current flows into or out of an electrolyte * **Electrolyte:** ionic compound in molten or dissolved solution that conducts the electricity * **Anode:** the positive electrode of an electrolysis cell * **Anion:** negatively charged ion which is attracted to the anode * **Cathode:** the negative electrode of an electrolysis cell * **Cation:** positively charged ion which is attracted to the cathode

Answer 13

* During electrolysis the electrons move from the power supply towards the **cathode** * Electron flow in electrochemistry thus occurs in **alphabetical** order as electrons flow from the **anode** to the **cathode** * Positive ions within the electrolyte migrate towards the **negatively** charged electrode which is the **cathode** * Negative ions within the electrolyte migrate towards the **positively** charged electrode which is the **anode**

Answer 14

Products of electrolysis When ions reach an electrode, they gain or lose **electrons**. As a result, they form **atoms** or **molecules** of **elements**: * positive ions gain electrons from the negatively charged cathode * negative ions lose electrons at the positively charged anode Molten lead bromide, PbBr₂(l), is an electrolyte. During electrolysis: * Pb²⁺ ions gain electrons at the cathode and become Pb atoms * Br^- ions lose electrons at the anode and become Br atoms, which pair up to form Br₂ molecules So lead forms at the negative electrode and bromine forms at the positive electrode. **Example** Predict the products of electrolysis of molten calcium chloride. Positively charged calcium ions move to the negative electrode. Here, they gain electrons to form calcium atoms, so calcium is formed at the negative electrode. Negatively charged chloride ions move to the positive electrode. Here, they lose electrons to form chlorine atoms. The atoms join up in pairs to form Cl₂ molecules, so chlorine gas is formed at the positive electrode. **During the electrolysis of molten salts, a metal forms at the cathode and a non-metal forms at the anode.**

Answer 15

* Lead(II) bromide is a binary ionic compound meaning that it is a compound consisting of just two elements joined together by ionic bonding * When these compounds are heated beyond their melting point, they become molten and can conduct electricity as their ions can move freely and carry the charge * These compounds undergo electrolysis and always produce their corresponding element * To predict the products of any binary molten compound first identify the ions present * The **positive** ion will migrate towards the **cathode** and the **negative** ion will migrate towards the **anode** * Therefore the **cathode** product will always be the metal and the product formed at the **anode** will always be the **non-metal** **Method:** * Add lead(II) bromide into a crucible and heat so it will turn molten, allowing ions to be free to move and conduct an electric charge * Add two graphite rods as the electrodes and connect this to a power pack or battery * Turn on the power pack or battery and allow electrolysis to take place * Negative bromide ions move to the positive electrode (anode) and lose two electrons to form bromine molecules. There is bubbling at the anode as brown bromine gas is given off * Positive lead ions move to the negative electrode (cathode) and gain electrons to form grey lead metal which deposits on the bottom of the electrode **Electrode Products:** **Anode:** Bromine gas **Cathode:** Lead metal

Answer 16

* The position of the metal on the reactivity series determines the method of extraction * Higher placed metals (above carbon) have to be extracted using **electrolysis** as they are too reactive and cannot be reduced by carbon * Lower placed metals can be extracted by heating with carbon which **reduces** * Electrolysis is very expensive as **large amounts of energy** are required to melt the ores and produce the electrical current * The reactivity series of metals is shown below with the corresponding method of extraction

Answer 17

**Raw Materials:** * Aluminium Ore (bauxite) * Cryolite (sodium aluminium fluoride) **Explanation:** * The bauxite is first purified to produce aluminium oxide, Al₂O₃ * Aluminium oxide has a very high melting point so it is first dissolved in molten **cryolite** producing an electrolyte with a lower melting point, as well as a better conductor of electricity than molten aluminium oxide * This reduces the costs considerably making the process more efficient * The electrolyte is a solution of aluminium oxide in molten cryolite at a temperature of about 1000 °C * The molten aluminium is siphoned off from time to time and fresh aluminium oxide is added to the cell * The cell operates at 5-6 volts and with a current of 100,000 amps * The heat generated by the huge current keeps the electrolyte molten * A lot of electricity is required for this process of extraction which is a major expense * The overall equation is: **2Al₂O₃(l) ⟶ 4Al (l) + 3O₂(g)** * Some of the oxygen produced at the positive electrode then reacts with graphite (carbon) electrode to produce carbon dioxide gas: **C (s) + O₂(g) ⟶ CO₂(g)** * This causes the carbon anodes to burn away, so they must be replaced regularly

Answer 18

* Aqueous solutions will always have water (H₂O) * In the electrolysis of aqueous solutions, the water molecules dissociate producing H⁺ and OH^– ions: **H₂O ⇌ H⁺ + OH^–** * These ions are also involved in the process and their chemistry must be considered * We now have an electrolyte that contains ions from the compound plus ions from the water * Which ions get discharged and at which electrode depends on the **relative reactivity** of the elements involved * Concentrated and dilute solutions of the **same** compound give **different** products * For anions, the **more concentrated** ion will tend to get discharged over a more dilute ion

Answer 19

**Positive Electrode (anode)** * Negatively charged OH^– ions and nonmetal ions are attracted to the positive electrode * Either OH^– or nonmetal ions will lose electrons and oxygen gas or the gas of the nonmetal in question is released e.g. chlorine, or bromine * Therefore at the anode, **oxygen gas will be produced unless the ionic compound contains halide ions, in which case the halogen will be produced**

Answer 20

**Negative Electrode (cathode)** * Positively charged H⁺ and metal ions are attracted to the negative electrode but only one will gain electrons * Either hydrogen gas or the metal will be produced * **If the metal is above hydrogen in the reactivity series, then hydrogen will be produced and bubbling will be seen at the cathode** * This is because the **more reactive ions will remain in solution**, causing the least reactive ion to be discharged. * Therefore at the cathode, **hydrogen gas** will be produced unless the positive ions from the ionic compound are less reactive than hydrogen, in which case the **metal** is produced

Answer 21

**Determining what Gas is Produced** * The gas produced can be tested to determine its identity * If the gas produced at the **cathode** burns with a ‘pop’ with a lighted splint then the gas is **hydrogen** * If the gas produced at the **anode** relights a glowing splint dipped into the gas then the gas is **oxygen** * The halogen gases all produce their own colours (bromine is **red-brown**, chlorine is **yellow-green**)

Answer 22

**Hypothesis:** A metal will be produced at the negative electrode because metal ions are positive **Materials:** * Test tubes * Electrolyte solutions * 100 cm³ beaker * Stand and clamp * Two carbon rod electrodes * Two crocodile / 4 mm plug leads * Low voltage power supply * Blue litmus paper **Practical Tip:** Make sure the test tubes do not cover the electrodes completely and fall to the bottom of the cell or the conductivity will fall considerably and the rate of electrolysis will be very slow **Method:** 1. Set up the apparatus as shown in the diagram 2. Add the aqueous solution to the beaker 3. Add two graphite rods as the electrodes and connect this to a power pack or battery 4. Turn on the power pack or battery and allow electrolysis to take place 5. Record the results in a suitable table (see below) and repeat for another solution, checking the electrodes in between runs to see if any metal has been deposited 6. The following aqueous solutions are suitable for this investigation: copper chloride, copper sulfate, sodium chloride, sodium bromide, sodium nitrate 7. The gases produced can be collected in the test tubes to be tested later **Results:** Record your results in a suitable table: **Electrolyte:** Dilute NaCl **Observation at Anode and gas test results:** Effervescence, no colour, split relights therefore gas is O2. **Observation at Cathode and gas test results:** Effervescence, no colour, squeaky pop therefore gas is H2. **Evaluation:** The gases and corresponding tests are: * Hydrogen – lighted splint goes out with a squeaky pop * Oxygen – a glowing splint relights * Chlorine – damp blue litmus paper turns red and is then bleached white **Conclusion:** Describe how the results obtained compare with the expected results based on the hypothesis

Answer 23

**Electrolysis of molten lead(II)bromide** * In the electrolysis of molten lead(II) bromide the half equation at the negative electrode (cathode) is: **Pb²⁺ + 2e^– ⟶ Pb** * At the positive electrode (anode) bromine gas is produced by the discharge of bromide ions: **2Br^- – 2e^–⟶ Br₂** OR **2Br^-⟶ Br₂ + 2e^–**

Answer 24

* Aluminium ions are discahrged at the negative electrode (cathode) and the aluminium is collected at the bottom of the cell: **Al³⁺ + 3e^– ⟶ Al** * At the positive electrode (anode) oxygen gas is produced: **2O^2- – 4e^–⟶ O₂** OR **2O^2- ⟶ O₂ + 4e^–** * Half equations illustrate the **transfer** of **electrons** during a chemical process as they provide a more detailed picture of the redox processes taking place * Half equations combine to give the **ionic equation** for an electrolytic cell. * The example below illustrates how this is done for the sodium chloride:

C4 - Chemical changes Flashcards

(48 cards)