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

pH calculations www.knockhardy.org.uk/sci.htm 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... www.knockhardy.org.uk/sci.htm Navigation is achieved by... either clicking on the grey arrows at the foot of each page or using the left and right arrow keys on the keyboard

pH calculations 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

Before you start it would be helpful to… pH calculations 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

where [H+] is the concentration of hydrogen ions in mol dm-3 What is pH? pH = - log10 [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 Kw = [H+(aq)] [OH¯(aq)] mol2 dm-6 = 1 x 10-14 mol2 dm-6 (at 25°C)

Calculating pH - strong acids and alkalis WORKED EXAMPLE Calculating pH - strong acids and alkalis Strong acids and alkalis completely dissociate in aqueous solution 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 HCl H+ + Cl¯ One H+ is produced for each HCl dissociating so [H+] = 0.02M = 2 x 10-2 mol dm-3 pH = - log [H+] = 1.7

Calculating pH - strong acids and alkalis WORKED EXAMPLE Calculating pH - strong acids and alkalis Strong acids and alkalis completely dissociate in aqueous solution 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 HCl H+ + Cl¯ One H+ is produced for each HCl dissociating so [H+] = 0.02M = 2 x 10-2 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 10-1 mol dm-3 The ionic product of water (at 25°C) Kw = [H+][OH¯] = 1 x 10-14 mol2 dm-6 therefore [H+] = Kw / [OH¯] = 1 x 10-13 mol dm-3 pH = - log [H+] = 13

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

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

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

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

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

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

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

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

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

Calculating pH - weak acids WORKED EXAMPLE Calculating pH - weak acids Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 ) HX dissociates as follows HX(aq) H+(aq) + X¯(aq) Dissociation constant for a weak acid Ka = [H+(aq)] [X¯(aq)] mol dm-3 [HX(aq)] Substitute for X¯ as ions are formed in [H+(aq)] = [HX(aq)] Ka 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

Calculating pH - weak acids WORKED EXAMPLE Calculating pH - weak acids Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 ) HX dissociates as follows HX(aq) H+(aq) + X¯(aq) Dissociation constant for a weak acid Ka = [H+(aq)] [X¯(aq)] mol dm-3 [HX(aq)] Substitute for X¯ as ions are formed in [H+(aq)] = [HX(aq)] Ka 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 10-5 mol dm-3 = 4.00 x 10-6 mol dm-3 = 2.00 x 10-3 mol dm-3 ANSWER pH = - log [H+(aq)] = 2.699

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

Strong acids and strong alkalis (either in excess) 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 dm3 (divide cm3 by 1000) 5. Divide moles by volume to find concentration of excess the ion in mol dm-3 6. Convert concentration to pH If the excess is H+ pH = - log[H+] If the excess is OH¯ pOH = - log[OH¯] then pH + pOH = 14 or use Kw = [H+] [OH¯] = 1 x 10-14 at 25°C therefore [H+] = Kw / [OH¯] then pH = - log[H+]

Strong acids and alkalis (either in excess) pH of mixtures Strong acids and alkalis (either in excess) WORKED EXAMPLE Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl

Strong acids and alkalis (either in excess) pH of mixtures Strong acids and alkalis (either in excess) WORKED EXAMPLE Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl 1. Calculate the number of moles of H+ and OH¯ ions present 25cm3 of 0.1M NaOH 20cm3 of 0.1M HCl 2.5 x 10-3 moles 2.0 x 10-3 moles moles of OH ¯ = 0.1 x 25/1000 = 2.5 x 10-3 moles of H+ = 20 x 20/1000 = 2.0 x 10-3

Strong acids and alkalis (either in excess) pH of mixtures Strong acids and alkalis (either in excess) WORKED EXAMPLE Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl 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 25cm3 of 0.1M NaOH 20cm3 of 0.1M HCl 2.5 x 10-3 moles 2.0 x 10-3 moles The reaction taking place is… HCl + NaOH NaCl + H2O or in its ionic form H+ + OH¯ H2O (1:1 molar ratio)

Strong acids and alkalis (either in excess) pH of mixtures Strong acids and alkalis (either in excess) WORKED EXAMPLE Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl 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 5.0 x 10-4 moles of OH¯ UNREACTED 25cm3 of 0.1M NaOH 20cm3 of 0.1M HCl 2.5 x 10-3 moles 2.0 x 10-3 moles The reaction taking place is… HCl + NaOH NaCl + H2O or in its ionic form H+ + OH¯ H2O (1:1 molar ratio) 2.0 x 10-3 moles of H+ will react with the same number of moles of OH¯ this leaves 2.5 x 10-3 - 2.0 x 10-3 = 5.0 x 10-4 moles of OH¯ in excess

Strong acids and alkalis (either in excess) pH of mixtures Strong acids and alkalis (either in excess) WORKED EXAMPLE Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl 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 the volume of the solution is 25 + 20 = 45cm3

Strong acids and alkalis (either in excess) pH of mixtures Strong acids and alkalis (either in excess) WORKED EXAMPLE Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl 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 dm3 (divide cm3 by 1000) the volume of the solution is 25 + 20 = 45cm3 there are 1000 cm3 in 1 dm3 volume = 45/1000 = 0.045dm3

Strong acids and alkalis (either in excess) pH of mixtures Strong acids and alkalis (either in excess) WORKED EXAMPLE Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl 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 dm3 (divide cm3 by 1000) 5. Divide moles by volume to find concentration of excess ion in mol dm-3 [OH¯] = 5.0 x 10-4 / 0.045 = 1.11 x 10-2 mol dm-3

Strong acids and alkalis (either in excess) pH of mixtures Strong acids and alkalis (either in excess) WORKED EXAMPLE Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl 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 dm3 (divide cm3 by 1000) 5. Divide moles by volume to find concentration of excess ion in mol dm-3 6. As the excess is OH¯ use pOH = - log[OH¯] then pH + pOH = 14 or Kw = [H+][OH¯] so [H+] = Kw / [OH¯] [OH¯] = 5.0 x 10-4 / 0.045 = 1.11 x 10-2 mol dm-3 [H+] = Kw / [OH¯] = 9.00 x 10-13 mol dm-3 pH = - log[H+] = 12.05 Kw = 1 x 10-14 mol2 dm-6 (at 25°C)

What should you be able to do? 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 Ka and molar concentration Calculate the pH of mixtures of strong acids and strong bases CAN YOU DO ALL OF THESE? YES NO

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

Try some past paper questions WELL DONE! Try some past paper questions

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