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Finding the pH of weak acids

Finding the pH of weak acids. What is a “weak acid”? What is a “weak base”? Calculate the pH of a weak acid. What is pH?. pH = - log 10 [H + (aq) ] where [H + ] is the concentration of hydrogen ions in mol dm -3 to convert pH into

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Finding the pH of weak acids

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  1. Finding the pH of weak acids • What is a “weak acid”? • What is a “weak base”? • Calculate the pH of a weak acid.

  2. 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 WATERKw= [H+(aq)] [OH¯(aq)] mol2 dm-6 = 1 x 10-14 mol2 dm-6 (at 25°C)

  3. 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

  4. 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 solutionNaOH 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

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

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

  7. Calculating pH - weak acids A weak acid is one which only partially dissociates in aqueous solution A weak acid, HA, dissociates as followsHA(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) [HA(aq)]

  8. Calculating pH - weak acids A weak acid is one which only partially dissociates in aqueous solution A weak acid, HA, dissociates as followsHA(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) [HA(aq)] Rearranging (3) gives[H+(aq)]2 = [HA(aq)]Ka therefore [H+(aq)] = [HA(aq)]Ka

  9. Calculating pH - weak acids A weak acid is one which only partially dissociates in aqueous solution A weak acid, HA, dissociates as followsHA(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) [HA(aq)] Rearranging (3) gives[H+(aq)]2 = [HA(aq)]Ka therefore [H+(aq)] = [HA(aq)]Ka pH = [H+(aq)]

  10. Calculating pH - weak acids A weak acid is one which only partially dissociates in aqueous solution A weak acid, HA, dissociates as followsHA(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) [HA(aq)] Rearranging (3) gives[H+(aq)]2 = [HA(aq)]Ka therefore [H+(aq)] = [HA(aq)]Ka pH = [H+(aq)] ASSUMPTIONHA 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.

  11. 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)

  12. 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)]

  13. 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

  14. 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

  15. 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 10-5mol dm-3 = 4.00 x 10-6mol dm-3 = 2.00 x 10-3mol dm-3 ANSWERpH = - log [H+(aq)] = 2.699

  16. 6. What is the pH of the following weak acids? a) Ka = 3.7 × 10-8 mol dm-3 [HA] = 0.01 mol dm-3 b) Ka = 5.8 × 10-10 mol dm-3 [HA] = 0.01 mol dm-3 c) Ka = 5.6 × 10-4 mol dm-3 [HA] = 0.01 mol dm-3 d) Ka = 1.74 × 10-5 mol dm-3 [HA] = 0.10 mol dm-3 e) Ka = 5.62 × 10-4 mol dm-3 [HA] = 0.20 mol dm-3 6. a) 4.72 b) 5.62 c) 2.68 d) 2.88 e) 1.97

  17. 0.1 mol dm-3 solution of HA has a pH of 5.10. What is its Ka value?

  18. Exercise

  19. Finding the pH of weak acids • What is a “weak acid”? • What is a “weak base”? • Calculate the pH of a weak acid.

  20. 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

  21. 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 useKw = [H+] [OH¯] = 1 x 10-14 at 25°C therefore [H+] = Kw / [OH¯]then pH = - log[H+]

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

  23. WORKED EXAMPLE pH of mixtures Strong acids and alkalis (either in excess) 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

  24. 25cm3 of 0.1M NaOH 20cm3 of 0.1M HCl 2.5 x 10-3 moles 2.0 x 10-3 moles WORKED EXAMPLE pH of mixtures Strong acids and alkalis (either in excess) 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 The reaction taking place is… HCl + NaOH NaCl + H2O or in its ionic form H+ + OH¯ H2O (1:1 molar ratio)

  25. 25cm3 of 0.1M NaOH 20cm3 of 0.1M HCl 5.0 x 10-4 moles of OH¯ UNREACTED 2.5 x 10-3 moles 2.0 x 10-3 moles WORKED EXAMPLE pH of mixtures Strong acids and alkalis (either in excess) 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 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

  26. WORKED EXAMPLE pH of mixtures Strong acids and alkalis (either in excess) 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

  27. WORKED EXAMPLE pH of mixtures Strong acids and alkalis (either in excess) 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

  28. WORKED EXAMPLE pH of mixtures Strong acids and alkalis (either in excess) 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

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