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PH calculations A guide for A level students KNOCKHARDY PUBLISHING 2008 SPECIFICATIONS.

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1 pH calculations A guide for A level students KNOCKHARDY PUBLISHING 2008 SPECIFICATIONS

2 pH calculations INTRODUCTION This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards. Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available. Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at... Navigation is achieved by... either clicking on the grey arrows at the foot of each page orusing the left and right arrow keys on the keyboard

3 CONTENTS What is pH? - a reminder Calculating the pH of strong acids and bases Calculating the pH of weak acids Calculating the pH of mixtures - strong acid and strong alkali pH calculations

4 Before you start it would be helpful to… know the differences between strong and weak acid and bases be able to calculate pH from hydrogen ion concentration be able to calculate hydrogen ion concentration from pH know the formula for the ionic product of water and its value at 25°C pH calculations

5 What is pH? pH = - log 10 [H + (aq) ] where [H + ] is the concentration of hydrogen ions in mol dm -3 to convert pH into hydrogen ion concentration [H + (aq) ] = antilog (-pH) IONIC PRODUCT OF WATER K w = [H + (aq)] [OH¯(aq)] mol 2 dm -6 = 1 x mol 2 dm -6 (at 25°C)

6 Calculating pH - strong acids and alkalis Strong acids and alkalis completely dissociate in aqueous solution only need to know the concentration It is easy to calculate the pH; you only need to know the concentration. Calculate the pH of 0.02M HCl HCl completely dissociates in aqueous solution HC l H + + C l ¯ One H + is produced for each HCl dissociating so [H + ] = 0.02M = 2 x mol dm -3 pH = - log [H + ] = 1.7 WORKED EXAMPLE WORKED EXAMPLE

7 Calculating pH - strong acids and alkalis Strong acids and alkalis completely dissociate in aqueous solution only need to know the concentration It is easy to calculate the pH; you only need to know the concentration. Calculate the pH of 0.02M HCl HCl completely dissociates in aqueous solution HC l H + + C l ¯ One H + is produced for each HCl dissociating so [H + ] = 0.02M = 2 x mol dm -3 pH = - log [H + ] = 1.7 Calculate the pH of 0.1M NaOH NaOH completely dissociates in aqueous solution NaOH Na + + OH¯ One OH¯ is produced for each NaOH dissociating [OH¯] = 0.1M = 1 x mol dm -3 The ionic product of water (at 25°C) K w = [H + ][OH¯] = 1 x mol 2 dm -6 therefore [H + ] = K w / [OH¯] = 1 x mol dm -3 pH = - log [H + ] = 13 WORKED EXAMPLE WORKED EXAMPLE

8 Calculating pH - weak acids A weak acid, HA, dissociates as followsHA (aq) H + (aq) + A¯ (aq) (1) A weak acid is one which only partially dissociates in aqueous solution

9 Calculating pH - weak acids A weak acid, HA, dissociates as followsHA (aq) H + (aq) + A¯ (aq) (1) Applying the Equilibrium Law K a = [H + (aq) ] [A¯ (aq) ] mol dm -3 (2) [HA (aq) ] A weak acid is one which only partially dissociates in aqueous solution

10 Calculating pH - weak acids A weak acid, HA, dissociates as followsHA (aq) H + (aq) + A¯ (aq) (1) Applying the Equilibrium Law K a = [H + (aq) ] [A¯ (aq) ] mol dm -3 (2) [HA (aq) ] The ions are formed in equal amounts, so [H + (aq) ] = [A¯ (aq) ] therefore K a = [H + (aq) ] 2 (3) [HA (aq) ] A weak acid is one which only partially dissociates in aqueous solution

11 Calculating pH - weak acids A weak acid, HA, dissociates as followsHA (aq) H + (aq) + A¯ (aq) (1) Applying the Equilibrium Law K a = [H + (aq) ] [A¯ (aq) ] mol dm -3 (2) [HA (aq) ] The ions are formed in equal amounts, so [H + (aq) ] = [A¯ (aq) ] therefore K a = [H + (aq) ] 2 (3) [HA (aq) ] Rearranging (3) gives[H + (aq) ] 2 = [HA (aq) ] K a therefore[H + (aq) ] = [HA (aq) ] K a A weak acid is one which only partially dissociates in aqueous solution

12 Calculating pH - weak acids A weak acid, HA, dissociates as followsHA (aq) H + (aq) + A¯ (aq) (1) Applying the Equilibrium Law K a = [H + (aq) ] [A¯ (aq) ] mol dm -3 (2) [HA (aq) ] The ions are formed in equal amounts, so [H + (aq) ] = [A¯ (aq) ] therefore K a = [H + (aq) ] 2 (3) [HA (aq) ] Rearranging (3) gives[H + (aq) ] 2 = [HA (aq) ] K a therefore[H + (aq) ] = [HA (aq) ] K a pH = [H + (aq) ] A weak acid is one which only partially dissociates in aqueous solution

13 Calculating pH - weak acids A weak acid, HA, dissociates as followsHA (aq) H + (aq) + A¯ (aq) (1) Applying the Equilibrium Law K a = [H + (aq) ] [A¯ (aq) ] mol dm -3 (2) [HA (aq) ] The ions are formed in equal amounts, so [H + (aq) ] = [A¯ (aq) ] therefore K a = [H + (aq) ] 2 (3) [HA (aq) ] Rearranging (3) gives[H + (aq) ] 2 = [HA (aq) ] K a therefore[H + (aq) ] = [HA (aq) ] K a pH = [H + (aq) ] ASSUMPTION HA is a weak acid so it will not have dissociated very much. You can assume that its equilibrium concentration is approximately that of the original concentration. A weak acid is one which only partially dissociates in aqueous solution

14 Calculating pH - weak acids Calculate the pH of a weak acid HX of concentration 0.1M ( K a = 4x10 -5 mol dm -3 ) HX dissociates as followsHX (aq) H + (aq) + X¯ (aq) WORKED EXAMPLE WORKED EXAMPLE

15 Calculating pH - weak acids Calculate the pH of a weak acid HX of concentration 0.1M ( K a = 4x10 -5 mol dm -3 ) HX dissociates as followsHX (aq) H + (aq) + X¯ (aq) Dissociation constant for a weak acid K a = [H + (aq) ] [X¯ (aq) ] mol dm -3 [HX (aq) ] WORKED EXAMPLE WORKED EXAMPLE

16 Calculating pH - weak acids Calculate the pH of a weak acid HX of concentration 0.1M ( K a = 4x10 -5 mol dm -3 ) HX dissociates as followsHX (aq) H + (aq) + X¯ (aq) Dissociation constant for a weak acid K a = [H + (aq) ] [X¯ (aq) ] mol dm -3 [HX (aq) ] Substitute for X¯ as ions are formed in [H + (aq) ] = [HX (aq) ] K a mol dm -3 equal amounts and then rearrange equation WORKED EXAMPLE WORKED EXAMPLE

17 Calculating pH - weak acids Calculate the pH of a weak acid HX of concentration 0.1M ( K a = 4x10 -5 mol dm -3 ) HX dissociates as followsHX (aq) H + (aq) + X¯ (aq) Dissociation constant for a weak acid K a = [H + (aq) ] [X¯ (aq) ] mol dm -3 [HX (aq) ] Substitute for X¯ as ions are formed in [H + (aq) ] = [HX (aq) ] K a mol dm -3 equal amounts and the rearrange equation ASSUMPTION HA is a weak acid so it will not have dissociated very much. You can assume that its equilibrium concentration is approximately that of the original concentration WORKED EXAMPLE WORKED EXAMPLE

18 Calculating pH - weak acids Calculate the pH of a weak acid HX of concentration 0.1M ( K a = 4x10 -5 mol dm -3 ) HX dissociates as followsHX (aq) H + (aq) + X¯ (aq) Dissociation constant for a weak acid K a = [H + (aq) ] [X¯ (aq) ] mol dm -3 [HX (aq) ] Substitute for X¯ as ions are formed in [H + (aq) ] = [HX (aq) ] K a mol dm -3 equal amounts and the rearrange equation ASSUMPTION HA is a weak acid so it will not have dissociated very much. You can assume that its equilibrium concentration is approximately that of the original concentration [H + (aq) ] = 0.1 x 4 x mol dm -3 = 4.00 x mol dm -3 = 2.00 x mol dm -3 ANSWER pH = - log [H + (aq) ] = WORKED EXAMPLE WORKED EXAMPLE

19 CALCULATING THE pH OF MIXTURES The method used to calculate the pH of a mixture of an acid and an alkali depends on... whether the acids and alkalis are STRONG or WEAK which substance is present in excess STRONG ACID and STRONG BASE - EITHER IN EXCESS

20 pH of mixtures Strong acids and strong alkalis (either in excess) 1. Calculate the initial number of moles of H + and OH¯ ions in the solutions 2. As H + and OH¯ ions react in a 1:1 ratio; calculate unreacted moles species in excess 3. Calculate the volume of solution by adding the two original volumes 4. Convert volume to dm 3 (divide cm 3 by 1000) 5. Divide moles by volume to find concentration of excess the ion in mol dm Convert concentration to pH If the excess isH + pH = - log[H + ] If the excess is OH¯ pOH = - log[OH¯] then pH + pOH = 14 or use K w = [H + ] [OH¯] = 1 x at 25°C therefore [H + ] = K w / [OH¯] then pH = - log[H + ]

21 Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l pH of mixtures Strong acids and alkalis (either in excess) WORKED EXAMPLE WORKED EXAMPLE

22 Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l 1. Calculate the number of moles of H + and OH¯ ions present pH of mixtures Strong acids and alkalis (either in excess) 25cm 3 of 0.1M NaOH 20cm 3 of 0.1M HC l 2.5 x moles 2.0 x moles moles of OH ¯ = 0.1 x 25/1000 = 2.5 x moles of H + = 20 x 20/1000 = 2.0 x WORKED EXAMPLE WORKED EXAMPLE

23 Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l 1. Calculate the number of moles of H + and OH¯ ions present 2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species pH of mixtures Strong acids and alkalis (either in excess) The reaction taking place is… HC l + NaOH NaC l + H 2 O or in its ionic form H + + OH¯ H 2 O (1:1 molar ratio) 25cm 3 of 0.1M NaOH 20cm 3 of 0.1M HC l 2.5 x moles 2.0 x moles WORKED EXAMPLE WORKED EXAMPLE

24 Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l 1. Calculate the number of moles of H + and OH¯ ions present 2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species pH of mixtures Strong acids and alkalis (either in excess) The reaction taking place is… HC l + NaOH NaC l + H 2 O or in its ionic form H + + OH¯ H 2 O (1:1 molar ratio) 2.0 x moles of H + will react with the same number of moles of OH¯ this leaves 2.5 x x = 5.0 x moles of OH¯ in excess 5.0 x moles of OH¯ UNREACTED 25cm 3 of 0.1M NaOH 20cm 3 of 0.1M HC l 2.5 x moles 2.0 x moles WORKED EXAMPLE WORKED EXAMPLE

25 Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l 1. Calculate the number of moles of H + and OH¯ ions present 2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species 3. Calculate the volume of the solution by adding the two individual volumes pH of mixtures Strong acids and alkalis (either in excess) the volume of the solution is = 45cm 3 WORKED EXAMPLE WORKED EXAMPLE

26 Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l 1. Calculate the number of moles of H + and OH¯ ions present 2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species 3. Calculate the volume of the solution by adding the two individual volumes 4. Convert volume to dm 3 (divide cm 3 by 1000) pH of mixtures Strong acids and alkalis (either in excess) the volume of the solution is = 45cm 3 there are 1000 cm 3 in 1 dm 3 volume = 45/1000 = 0.045dm 3 WORKED EXAMPLE WORKED EXAMPLE

27 Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l 1. Calculate the number of moles of H + and OH¯ ions present 2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species 3. Calculate the volume of the solution by adding the two individual volumes 4. Convert volume to dm 3 (divide cm 3 by 1000) 5. Divide moles by volume to find concentration of excess ion in mol dm -3 pH of mixtures Strong acids and alkalis (either in excess) WORKED EXAMPLE WORKED EXAMPLE [OH¯]= 5.0 x / = 1.11 x mol dm -3

28 Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH is added to 20cm 3 of 0.1M HC l 1. Calculate the number of moles of H + and OH¯ ions present 2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species 3. Calculate the volume of the solution by adding the two individual volumes 4. Convert volume to dm 3 (divide cm 3 by 1000) 5. Divide moles by volume to find concentration of excess ion in mol dm As the excess is OH¯ usepOH = - log[OH¯] then pH + pOH = 14 or K w = [H + ][OH¯] so [H + ] = K w / [OH¯] pH of mixtures Strong acids and alkalis (either in excess) K w = 1 x mol 2 dm -6 (at 25°C) WORKED EXAMPLE WORKED EXAMPLE [OH¯]= 5.0 x / = 1.11 x mol dm -3 [H + ] = K w / [OH¯] = 9.00 x mol dm -3 pH = - log[H + ] = 12.05

29 REVISION CHECK What should you be able to do? Calculate pH from hydrogen ion concentration Calculate hydrogen ion concentration from pH Write equations to show the ionisation in strong and weak acids Calculate the pH of strong acids and bases knowing their molar concentration Calculate the pH of weak acids knowing their K a and molar concentration Calculate the pH of mixtures of strong acids and strong bases CAN YOU DO ALL OF THESE? YES NO

30 You need to go over the relevant topic(s) again Click on the button to return to the menu

31 WELL DONE! Try some past paper questions

32 pH calculations THE END © 2009 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


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