2. Acids and Alkalis  Forming Acids  Acids are formed from soluble non metal oxides.  They have a ph range between 1 and 7  They turn pH paper of universal indicator red/orange/yellow

3. Forming Alkalis  Alkalis are formed from soluble metal oxides.  They have a pH range between 8 and 14  They turn universal indicator and pH paper Dark green/blue/purple

4. Examples of Acids and Alkalis AcidsAlkalis Hydrochloric AcidSodium Hydroxide Nitric AcidToothpaste VinegarAmmonia Solution Orange JuiceCalcium hydroxide Sulphuric AcidPotassium hydroxide Carbonic AcidBicarbonate of soda

5. Neutral Substances  A neutral substance is neither acid nor alkali.  They have a pH =7  They turn universal indicator and pH paper green. Example – Water, Tea, Alcohol!

6. Acid Rain AA cid rain is formed when acidic gases dissolve in rain water. TT he main gas responsible is Sulphur dioxide! OO xides of Nitrogen can also contribute to the problem. AA cid rain damages the environment: EE ats away buildings, kills plants and fish, corrodes iron structures.

7. Preventing Acid Rain!  Remove Sulphur from fuels before you burn them.  Reduce the amount of fossil Fuels you burn.

8. Dissociation of Water  Water exists mainly as molecules – however it sometimes splits into 2 ions – Hydrogen and Hydroxide.  H 2 O(l) H + (aq) + OH - (aq)  Water will conduct electricity – slightly because it contains these ions.  Water is neutral because it contains equal numbers of each ion.

9. Conductivity  Acids and Alkalis can conduct electricity because they contain ions.  Acids = H+ H+ ion  Alkalis = OH - ion  The stronger the acid or alkali is – the more ions it contains – the greater the conductivity!

10. Diluting!  When we add water to an acid or alkali– we dilute it – reduce the concentration of the ions!  Acids – the concentration oh H+ ions gets less – the pH will increase – moving towards pH7  Alkalis – the concentration of OH- will get less – the pH will move down -towards pH 7

11. Electrolysis of Acids  Acids will split up when you pass an electric current through them.  This is because they contain ions – Hydrogen ions.  When we electrolyse Sulphuric Acid the following happens:  Bubbles of Hydrogen gas appear at the negative electrode.  The H + ions move to the – electrode – they gain electrons to form H 2 gas  2H + (aq) + 2 e —> H 2 (g)

12. Metal Hydroxides WW e can find out if a compound is soluble using p5 in the data book. TT rends MM etal hydroxides in Group 1 and 2 are soluble. E.g NaOH, KOH, Ca( OH ) 2 AA ll other metal hydroxides are insoluble. e.g. Cu(OH) 2, Zn(OH) 2

13. Formula Mass  This is the total mass of a compound.  It takes into consideration the number and mass of each element in a compound.  Steps  Correct Formula – Use valency rules!  Write down numbers of each elements in compound.  Add in mass of element – use data book.  Add up total mass.

14. Formula Mass – example!  Work out the formula mass of Calcium hydroxide. 1. Formula – Ca 2+ OH - Cross over – Ca (OH) 2 2. Numbers of each element (1 x Ca) + (2 x O) + ( 2 x H) 3. Add in mass ( 1 x 40) + ( 2 x 16) + ( 2 x 1) 4. Add total mass 40 + 32 + 2 = 74 amu.

15. The Mole! – Gram Formula Mass TT he mole is basically the formula mass in grams. TT he steps are required. EE xample – Calculate the mass of 1 Mole of Water SS ymbol – H 2 O NN umbers of elements – ( 2 x H) + ( 1 x O) AA dd in Mass - (2x1) + (1x 16) TT otal Mass – (2x1) + 16 = 18 g 22 moles of water would be- 2 x18g = 32g

16. More Calculations! n C V n = Number of Moles C = Concentration (mol/l) V = Volume (l) n = C x V C = n/V V = n/C

17. Examples  1) Calculate – The number of moles if you have a solution of with 2 mol/l HCl in 100cm 3 of water.  n = CxV 2 x 100/1000 2 x 0.1 = 0.2 moles

18. More!  2) Calculate the volume of water required to make a 2 mol/l solution of NaOH using 3 moles of the alkali.  V = n/C =3/2 = 1.5 litres

19. Even More!  Calculate the concentration of solution made when 2 mol/l of acid dissolves in 50 cm 3 of water.  C = n/V = 2/(50/1000) = 2/ 0.05 = 40 mol/l