JK-TK/Newdata huggingface dataset:
(f)Iron(II)sulphate(VI) Elements making compound Iron sulphate(VI) Symbol of elements/radicals in compound Fe SO4 Assign valencies as superscript Fe2 SO4 2 Exchange/Interchange the valencies as subscript Fe2 SO4 2 Divide by two to get smallest whole number ratio Fe1 SO4 1
21 Chemical formula of Iron(II) sulphate(VI) is thus: FeSO4 This means:1 atom of Iron combine with 1 sulphate(VI) radical. (g)Copper(II)sulphate(VI) Elements making compound Copper sulphate(VI) Symbol of elements/radicals in compound Cu SO4 Assign valencies as superscript Cu2 SO4 2 Exchange/Interchange the valencies as subscript Cu2 SO4 2 Divide by two to get smallest whole number ratio Cu1 SO4 1 Chemical formula of Cu(II)sulphate(VI) is thus: CuSO4 This means:1 atom of Copper combine with 1 sulphate(VI) radical. (h)Aluminium sulphate(VI) Elements making compound Aluminium sulphate(VI) Symbol of elements/radicals in compound Al SO4 Assign valencies as superscript Al3 SO4 2 Exchange/Interchange the valencies as subscript Al2 SO4 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al2(SO4)3 This means:2 atom of Aluminium combine with 3 sulphate(VI) radical. (i)Aluminium nitrate(V) Elements making compound Aluminium nitrate(V) Symbol of elements/radicals in compound Al NO3 Assign valencies as superscript Al3 NO3 1 Exchange/Interchange the valencies as subscript Al1 NO3 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al (NO3)3 This means:1 atom of Aluminium combine with 3 nitrate(V) radical.
(g)Copper(II)sulphate(VI) Elements making compound Copper sulphate(VI) Symbol of elements/radicals in compound Cu SO4 Assign valencies as superscript Cu2 SO4 2 Exchange/Interchange the valencies as subscript Cu2 SO4 2 Divide by two to get smallest whole number ratio Cu1 SO4 1 Chemical formula of Cu(II)sulphate(VI) is thus: CuSO4 This means:1 atom of Copper combine with 1 sulphate(VI) radical. (h)Aluminium sulphate(VI) Elements making compound Aluminium sulphate(VI) Symbol of elements/radicals in compound Al SO4 Assign valencies as superscript Al3 SO4 2 Exchange/Interchange the valencies as subscript Al2 SO4 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al2(SO4)3 This means:2 atom of Aluminium combine with 3 sulphate(VI) radical. (i)Aluminium nitrate(V) Elements making compound Aluminium nitrate(V) Symbol of elements/radicals in compound Al NO3 Assign valencies as superscript Al3 NO3 1 Exchange/Interchange the valencies as subscript Al1 NO3 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al (NO3)3 This means:1 atom of Aluminium combine with 3 nitrate(V) radical. (j)Potassium manganate(VII) Elements making compound Potassium manganate(VII) Symbol of elements/radicals in compound K MnO4 Assign valencies as superscript K 1 MnO4 1 Exchange/Interchange the valencies as subscript K1 MnO4 1 Divide by two to get smallest whole number ratio - -
22 Chemical formula of Potassium manganate(VII) is thus: KMnO4 This means:1 atom of Potassium combine with 4 manganate(VII) radical.
(h)Aluminium sulphate(VI) Elements making compound Aluminium sulphate(VI) Symbol of elements/radicals in compound Al SO4 Assign valencies as superscript Al3 SO4 2 Exchange/Interchange the valencies as subscript Al2 SO4 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al2(SO4)3 This means:2 atom of Aluminium combine with 3 sulphate(VI) radical. (i)Aluminium nitrate(V) Elements making compound Aluminium nitrate(V) Symbol of elements/radicals in compound Al NO3 Assign valencies as superscript Al3 NO3 1 Exchange/Interchange the valencies as subscript Al1 NO3 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al (NO3)3 This means:1 atom of Aluminium combine with 3 nitrate(V) radical. (j)Potassium manganate(VII) Elements making compound Potassium manganate(VII) Symbol of elements/radicals in compound K MnO4 Assign valencies as superscript K 1 MnO4 1 Exchange/Interchange the valencies as subscript K1 MnO4 1 Divide by two to get smallest whole number ratio - -
22 Chemical formula of Potassium manganate(VII) is thus: KMnO4 This means:1 atom of Potassium combine with 4 manganate(VII) radical. (k)Sodium dichromate(VI) Elements making compound Sodium dichromate(VI) Symbol of elements/radicals in compound Na Cr2O7 Assign valencies as superscript Na 1 Cr2O7 2 Exchange/Interchange the valencies as subscript Na2 Cr2O7 1 Divide by two to get smallest whole number ratio - - Chemical formula of Sodium dichromate(VI) is thus: Na2 Cr2O7 This means:2 atom of Sodium combine with 1 dichromate(VI) radical.
(i)Aluminium nitrate(V) Elements making compound Aluminium nitrate(V) Symbol of elements/radicals in compound Al NO3 Assign valencies as superscript Al3 NO3 1 Exchange/Interchange the valencies as subscript Al1 NO3 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al (NO3)3 This means:1 atom of Aluminium combine with 3 nitrate(V) radical. (j)Potassium manganate(VII) Elements making compound Potassium manganate(VII) Symbol of elements/radicals in compound K MnO4 Assign valencies as superscript K 1 MnO4 1 Exchange/Interchange the valencies as subscript K1 MnO4 1 Divide by two to get smallest whole number ratio - -
22 Chemical formula of Potassium manganate(VII) is thus: KMnO4 This means:1 atom of Potassium combine with 4 manganate(VII) radical. (k)Sodium dichromate(VI) Elements making compound Sodium dichromate(VI) Symbol of elements/radicals in compound Na Cr2O7 Assign valencies as superscript Na 1 Cr2O7 2 Exchange/Interchange the valencies as subscript Na2 Cr2O7 1 Divide by two to get smallest whole number ratio - - Chemical formula of Sodium dichromate(VI) is thus: Na2 Cr2O7 This means:2 atom of Sodium combine with 1 dichromate(VI) radical. (l)Calcium hydrogen carbonate Elements making compound Calcium Hydrogen carbonate Symbol of elements/radicals in compound Ca CO3 Assign valencies as superscript Ca 2 HCO3 1 Exchange/Interchange the valencies as subscript Ca1 HCO3 2 Divide by two to get smallest whole number ratio - - Chemical formula of Calcium hydrogen carbonate is thus: Ca(HCO3)2 This means:1 atom of Calcium combine with 2 hydrogen carbonate radical.
(j)Potassium manganate(VII) Elements making compound Potassium manganate(VII) Symbol of elements/radicals in compound K MnO4 Assign valencies as superscript K 1 MnO4 1 Exchange/Interchange the valencies as subscript K1 MnO4 1 Divide by two to get smallest whole number ratio - -
22 Chemical formula of Potassium manganate(VII) is thus: KMnO4 This means:1 atom of Potassium combine with 4 manganate(VII) radical. (k)Sodium dichromate(VI) Elements making compound Sodium dichromate(VI) Symbol of elements/radicals in compound Na Cr2O7 Assign valencies as superscript Na 1 Cr2O7 2 Exchange/Interchange the valencies as subscript Na2 Cr2O7 1 Divide by two to get smallest whole number ratio - - Chemical formula of Sodium dichromate(VI) is thus: Na2 Cr2O7 This means:2 atom of Sodium combine with 1 dichromate(VI) radical. (l)Calcium hydrogen carbonate Elements making compound Calcium Hydrogen carbonate Symbol of elements/radicals in compound Ca CO3 Assign valencies as superscript Ca 2 HCO3 1 Exchange/Interchange the valencies as subscript Ca1 HCO3 2 Divide by two to get smallest whole number ratio - - Chemical formula of Calcium hydrogen carbonate is thus: Ca(HCO3)2 This means:1 atom of Calcium combine with 2 hydrogen carbonate radical. (l)Magnesium hydrogen sulphate(VI) Elements making compound Magnesium Hydrogen sulphate(VI) Symbol of elements/radicals in compound Mg HSO4 Assign valencies as superscript Mg 2 HSO4 1 Exchange/Interchange the valencies as subscript Mg1 HSO4 2 Divide by two to get smallest whole number ratio - - Chemical formula of Magnesium hydrogen sulphate(VI) is thus: Mg(HSO4)2 This means:1 atom of Magnesium combine with 2 hydrogen sulphate(VI) radical. Compounds are formed from chemical reactions. A chemical reaction is formed when atoms of the reactants break free to bond again and form products. A chemical reaction is a statement showing the movement of reactants to form products. The following procedure is used in writing a chemical equations:
23 1. Write the word equation 2.
A chemical reaction is a statement showing the movement of reactants to form products. The following procedure is used in writing a chemical equations:
23 1. Write the word equation 2. Write the correct chemical formula for each of the reactants and products 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. 4. Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal. This is called balancing. Do not change the chemical formula of the products/reactants. 5. Assign in brackets, the physical state/state symbols of the reactants and products after each chemical formula as: (i) (s) for solids (ii) (l) for liquids (iii) (g) for gas (iv) (aq) for aqueous/dissolved in water to make a solution. Practice examples Write a balanced chemical equation for the following (a) Hydrogen gas is prepared from reacting Zinc granules with dilute hydrochloric acid. Procedure 1. Write the word equation Zinc + Hydrochloric acid -> Zinc chloride + hydrogen gas 2. Write the correct chemical formula for each of the reactants and products Zn + HCl -> ZnCl2 + H2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. Number of atoms of Zn on the reactant side is equal to product side One atom of H in HCl on the reactant side is not equal to two atoms in H2 on product side. One atom of Cl in HCl on the reactant side is not equal to two atoms in ZnCl2 on product side. 4. Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal. Multiply HCl by “2” to get “2” Hydrogen and “2” Chlorine on product and reactant side. Zn + 2 HCl -> ZnCl2 + H2 5. Assign in brackets, the physical state/state symbols . Zn(s) + 2 HCl(aq) -> ZnCl2 (aq) + H2(g)
24 (b) Oxygen gas is prepared from decomposition of Hydrogen peroxide solution to water Procedure 1.
Zn + 2 HCl -> ZnCl2 + H2 5. Assign in brackets, the physical state/state symbols . Zn(s) + 2 HCl(aq) -> ZnCl2 (aq) + H2(g)
24 (b) Oxygen gas is prepared from decomposition of Hydrogen peroxide solution to water Procedure 1. Write the word equation Hydrogen peroxide -> Water + oxygen gas 2. Write the correct chemical formula for each of the reactants and products H2O2 -> H2O + O2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. Number of atoms of H on the reactant side is equal to product side Two atom of O in H2O2 on the reactant side is not equal to three atoms (one in H2O and two in O2) on product side. 4. Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal. Multiply H2O2 by “2” to get “4” Hydrogen and “4” Oxygen on reactants Multiply H2O by “2” to get “4” Hydrogen and “2” Oxygen on product side When the “2” Oxygen in O2 and the“2” in H2O are added on product side they are equal to the“4” Oxygen on reactants side. 2H2O2 -> 2H2O + O2 5. Assign in brackets, the physical state/state symbols . 2H2O2(aq) -> 2H2O(l) + O2(g) (c) Chlorine gas is prepared from Potassium manganate(VII) reacting with hydrochloric acid to form potassium chloride solution, manganese(II) chloride solution,water and chlorine gas. Procedure 1. Write the word equation Potassium manganate(VII) + Hydrochloric acid -> potassium chloride + manganese(II) chloride + chlorine +water 2. Write the correct chemical formula for each of the reactants and products KMnO4 + HCl -> KCl + MnCl2 +H2O + Cl2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side.
Write the word equation Potassium manganate(VII) + Hydrochloric acid -> potassium chloride + manganese(II) chloride + chlorine +water 2. Write the correct chemical formula for each of the reactants and products KMnO4 + HCl -> KCl + MnCl2 +H2O + Cl2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. 25 Number of atoms of K and Mn on the reactant side is equal to product side Two atom of H in H2O on the product side is not equal to one atom on reactant side. Four atom of O in KMnO4 is not equal to one in H2O One atom of Cl in HCl on reactant side is not equal to three (one in H2O and two in Cl2) 4. Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal. Multiply HCl by “16” to get “16” Hydrogen and “16” Chlorine on reactants Multiply KMnO4 by “2” to get “2” Potassium and “2” manganese, “2 x4 =8” Oxygen on reactant side. Balance the product side to get: 2 KMnO4 +16 HCl -> 2 KCl + 2 MnCl2 +8 H2O + 5 Cl2 5. Assign in brackets, the physical state/state symbols . 2KMnO4(s) +16 HCl(aq)-> 2 KCl (aq) + 2MnCl2(aq)+8 H2O(l)+5 Cl2(g) (d)Carbon(IV)oxide gas is prepared from Calcium carbonate reacting with hydrochloric acid to form calcium chloride solution, water and carbon(IV)oxide gas. Procedure 1. Write the word equation Calcium carbonate + Hydrochloric acid -> calcium chloride solution+ water +carbon(IV)oxide 2. Write the correct chemical formula for each of the reactants and products CaCO3 + HCl -> CaCl2 +H2O + CO2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. 4.
Write the correct chemical formula for each of the reactants and products CaCO3 + HCl -> CaCl2 +H2O + CO2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. 4. Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal. 5. Assign in brackets, the physical state/state symbols . CaCO3(s) + 2 HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g) (d)Sodium hydroxide solution neutralizes hydrochloric acid to form salt and water. 26 NaOH(aq) + HCl(aq) -> NaCl (aq) + H2O(l) (e)Sodium reacts with water to form sodium hydroxide and hydrogen gas. 2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g) (f)Calcium reacts withwater to form calcium hydroxide and hydrogen gas Ca(s) + 2H2O(l) -> Ca(OH)2(aq) + H2(g) (g)Copper(II)Oxide solid reacts with dilute hydrochloric acid to form copper(II)chloride and water. CuO(s) + 2HCl(aq) -> CuCl2(aq) + H2O(l) (h)Hydrogen sulphide reacts with Oxygen to form sulphur(IV)Oxide and water. 2H2S(g) + 3O2(g) -> 2SO2(g) + 2H2O(l) (i)Magnesium reacts with steam to form Magnesium Oxide and Hydrogen gas. Mg(s) + 2H2O(g) -> MgO(s) + H2(g) (j)Ethane(C2H6) gas burns in air to form Carbon(IV)Oxide and water. 2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(l) (k)Ethene(C2H4) gas burns in air to form Carbon(IV)Oxide and water.
2H2S(g) + 3O2(g) -> 2SO2(g) + 2H2O(l) (i)Magnesium reacts with steam to form Magnesium Oxide and Hydrogen gas. Mg(s) + 2H2O(g) -> MgO(s) + H2(g) (j)Ethane(C2H6) gas burns in air to form Carbon(IV)Oxide and water. 2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(l) (k)Ethene(C2H4) gas burns in air to form Carbon(IV)Oxide and water. C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l) (l)Ethyne(C2H2) gas burns in air to form Carbon(IV)Oxide and water. 2C2H2(g) + 5O2(g) -> 4CO2(g) + 2H2O(l)
27 C.PERIODICITY OF CHEMICAL FAMILES/DOWN THE GROUP. The number of valence electrons and the number of occupied energy levels in an atom of an element determine the position of an element in the periodic table.i.e The number of occupied energy levels determine the Period and the valence electrons determine the Group. Elements in the same group have similar physical and chemical properties. The trends in physical and chemical properties of elements in the same group vary down the group. Elements in the same group thus constitute a chemical family. (a) Group I elements: Alkali metals Group I elements are called Alkali metals except Hydrogen which is a non metal. The alkali metals include: Element Symbol Atomic number Electron structure Oxidation state Valency Lithium Li 3 2:1 Li+ 1 Sodium Na 11 2:8:1 Na+ 1 Potassium K 19 2:8:8:1 K+ 1 Rubidium Rb 37 2:8:18:8:1 Rb+ 1 Caesium Cs 55 2:8:18:18:8:1 Cs+ 1 Francium Fr 87 2:8:18:32:18:8:1 Fr+ 1 All alkali metals atom has one electron in the outer energy level.
Elements in the same group thus constitute a chemical family. (a) Group I elements: Alkali metals Group I elements are called Alkali metals except Hydrogen which is a non metal. The alkali metals include: Element Symbol Atomic number Electron structure Oxidation state Valency Lithium Li 3 2:1 Li+ 1 Sodium Na 11 2:8:1 Na+ 1 Potassium K 19 2:8:8:1 K+ 1 Rubidium Rb 37 2:8:18:8:1 Rb+ 1 Caesium Cs 55 2:8:18:18:8:1 Cs+ 1 Francium Fr 87 2:8:18:32:18:8:1 Fr+ 1 All alkali metals atom has one electron in the outer energy level. They therefore are monovalent. They donate /lose the outer electron to have oxidation state M+ The number of energy levels increases down the group from Lithium to Francium. The more the number of energy levels the bigger/larger the atomic size. e.g. The atomic size of Potassium is bigger/larger than that of sodium because Potassium has more/4 energy levels than sodium (3 energy levels). Atomic and ionic radius The distance between the centre of the nucleus of an atom and the outermost energy level occupied by electron/s is called atomic radius. Atomic radius is measured in nanometers(n).The higher /bigger the atomic radius the bigger /larger the atomic size. The distance between the centre of the nucleus of an ion and the outermost energy level occupied by electron/s is called ionic radius. Ionic radius is also
28 measured in nanometers(n).The higher /bigger the ionic radius the bigger /larger the size of the ion. Atomic radius and ionic radius depend on the number of energy levels occupied by electrons. The more the number of energy levels the bigger/larger the atomic /ionic radius. e.g. The atomic radius of Francium is bigger/larger than that of sodium because Francium has more/7 energy levels than sodium (3 energy levels). Atomic radius and ionic radius of alkali metals increase down the group as the number of energy levels increases. The atomic radius of alkali metals is bigger than the ionic radius. This is because alkali metals react by losing/donating the outer electron and hence lose the outer energy level.
Atomic radius and ionic radius of alkali metals increase down the group as the number of energy levels increases. The atomic radius of alkali metals is bigger than the ionic radius. This is because alkali metals react by losing/donating the outer electron and hence lose the outer energy level. Table showing the atomic and ionic radius of some alkali metals Element Symbol Atomic number Atomic radius(nM) Ionic radius(nM) Lithium Li 3 0.133 0.060 Sodium Na 11 0.157 0.095 Potassium K 19 0.203 0.133 The atomic radius of sodium is 0.157nM .The ionic radius of Na+ is 0.095nM. This is because sodium reacts by donating/losing the outer electrons and hence the outer energy level. The remaining electrons/energy levels experience more effective / greater nuclear attraction/pull towards the nucleus reducing the atomic radius. Electropositivity The ease of donating/losing electrons is called electropositivity. All alkali metals are electropositive. Electropositivity increase as atomic radius increase. This is because the effective nuclear attraction on outer electrons decreases with increase in atomic radius. The outer electrons experience less nuclear attraction and can be lost/ donated easily/with ease. Francium is the most electropositive element in the periodic table because it has the highest/biggest atomic radius. Ionization energy The minimum amount of energy required to remove an electron from an atom of element in its gaseous state is called 1st ionization energy. The SI unit of ionization energy is kilojoules per mole/kJmole-1 .Ionization energy depend on atomic radius. The higher the atomic radius, the less effective the nuclear attraction on outer electrons/energy level and thus the lower the ionization energy. For alkali metals the 1st ionization energy decrease down the group as
29 the atomic radius increase and the effective nuclear attraction on outer energy level electrons decrease. e.g. The 1st ionization energy of sodium is 496 kJmole-1 while that of potassium is 419 kJmole-1 .This is because atomic radius increase and thus effective nuclear attraction on outer energy level electrons decrease down the group from sodium to Potassium. It requires therefore less energy to donate/lose outer electrons in Potassium than in sodium.
e.g. The 1st ionization energy of sodium is 496 kJmole-1 while that of potassium is 419 kJmole-1 .This is because atomic radius increase and thus effective nuclear attraction on outer energy level electrons decrease down the group from sodium to Potassium. It requires therefore less energy to donate/lose outer electrons in Potassium than in sodium. Physical properties Soft/Easy to cut: Alkali metals are soft and easy to cut with a knife. The softness and ease of cutting increase down the group from Lithium to Francium. This is because an increase in atomic radius, decreases the strength of metallic bond and the packing of the metallic structure Appearance: Alkali metals have a shiny grey metallic luster when freshly cut. The surface rapidly/quickly tarnishes on exposure to air. This is because the metal surface rapidly/quickly reacts with elements of air/oxygen. Melting and boiling points: Alkali metals have a relatively low melting/boiling point than common metals like Iron. This is because alkali metals use only one delocalized electron to form a weak metallic bond/structure. Electrical/thermal conductivity: Alkali metals are good thermal and electrical conductors. Metals conduct using the outer mobile delocalized electrons. The delocalized electrons move randomly within the metallic structure. Summary of some physical properties of the 1st three alkali metals Alkali metal Appearance Ease of cutting Melting point (oC) Boiling point (oC) Conductivity 1st ionization energy Lithium Silvery white Not easy 180 1330 Good 520 Sodium Shiny grey Easy 98 890 Good 496 Potassium Shiny grey Very easy 64 774 Good 419 Chemical properties (i)Reaction with air/oxygen On exposure to air, alkali metals reacts with the elements in the air. Example On exposure to air, Sodium first reacts with Oxygen to form sodium oxide. 4Na(s) + O2(g) -> 2Na2O(s) The sodium oxide formed further reacts with water/moisture in the air to form sodium hydroxide solution. 30 Na2O(s) + H2O(l) -> 2NaOH(aq) Sodium hydroxide solution reacts with carbon(IV)oxide in the air to form sodium carbonate.
Example On exposure to air, Sodium first reacts with Oxygen to form sodium oxide. 4Na(s) + O2(g) -> 2Na2O(s) The sodium oxide formed further reacts with water/moisture in the air to form sodium hydroxide solution. 30 Na2O(s) + H2O(l) -> 2NaOH(aq) Sodium hydroxide solution reacts with carbon(IV)oxide in the air to form sodium carbonate. 2NaOH(aq) + CO2(g) -> Na2CO3(g) + H2O(l) (ii)Burning in air/oxygen Lithium burns in air with a crimson/deep red flame to form Lithium oxide 4Li (s) + O2(g) -> 2Li2O(s) Sodium burns in air with a yellow flame to form sodium oxide 4Na (s) + O2(g) -> 2Na2O(s) Sodium burns in oxygen with a yellow flame to form sodium peroxide 2Na (s) + O2(g) -> Na2O2 (s) Potassium burns in air with a lilac/purple flame to form potassium oxide 4K (s) + O2(g) -> 2K2O (s) (iii) Reaction with water: Experiment Measure 500 cm3 of water into a beaker. Put three drops of phenolphthalein indicator. Put about 0.5g of Lithium metal into the beaker. Determine the pH of final product Repeat the experiment using about 0.1 g of Sodium and Potassium.