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pH calculations

pH calculations. 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

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pH calculations

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  1. pH calculations

  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. Calculate the pH of • 0.01 M HCl • 0.1 M HNO3 • 0.02 M HCl 0.2mol dm-3 nitric acid [H+] = 1.2 x 10-2mol dm-3 [H+] = 6.7 x 10-2mol dm-3 [H+] = 9.1 x 10-2mol dm-3 [H+] = 6.8 x 10-3mol dm-3 [H+] = 1.0 x 10-7mol dm-3 [H+] = 5.4 x 10-9mol dm-3 [H+] = 9.9 x 10-6mol dm-3

  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)

  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)]

  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)]

  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

  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)]

  11. 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 = -log [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.

  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)

  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)]

  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 then rearrange equation

  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

  16. 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

  17. 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 WEAK ACID and EXCESS STRONG BASE STRONG BASE and EXCESS WEAK ACID

  18. 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+]

  19. 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

  20. 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

  21. 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)

  22. 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

  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 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

  24. 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

  25. 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

  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 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)

  27. pH of mixtures Weak acid and EXCESS strong alkali 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 of the excess OH¯ 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 OH¯ in mol dm-3 6. Convert concentration to pH either using Kw = [H+] [OH¯] = 1 x 10-14 at 25°C therefore [H+] = Kw / [OH¯]then pH = - log[H+] or pOH = - log[OH¯] and pH + pOH = 14

  28. WORKED EXAMPLE pH of mixtures Weak acid and EXCESS strong alkali Calculate the pH of a mixture of 25cm3 of 0.1M NaOH and 22cm3 of 0.1M CH3COOH 1. Calculate the number of moles of H+ and OH¯ ions present 25cm3 of 0.1M NaOH 22cm3 of 0.1M CH3COOH 2.2 x 10-3 moles 2.5 x 10-3 moles moles of OH¯ = 0.1 x 25/1000 = 2.5 x 10-3 moles of H+ = 22 x 20/1000 = 2.2 x 10-3

  29. WORKED EXAMPLE pH of mixtures Weak acid and EXCESS strong alkali Calculate the pH of a mixture of 25cm3 of 0.1M NaOH and 22cm3 of 0.1M CH3COOH 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 excess OH¯ 25cm3 of 0.1M NaOH 22cm3 of 0.1M CH3COOH 3.0 x 10-4 moles of OH¯ UNREACTED 2.2 x 10-3 moles 2.5 x 10-3 moles The reaction taking place is CH3COOH + NaOH CH3COONa + H2O or in its ionic form H+ + OH¯ H2O (1:1 molar ratio) 2.2 x 10-3 moles of H+ will react with the same number of moles of OH¯ this leaves 2.5 x 10-3 - 2.2 x 10-3 = 3.0 x 10-4 moles of OH¯ in excess

  30. WORKED EXAMPLE pH of mixtures Weak acid and EXCESS strong alkali Calculate the pH of a mixture of 25cm3 of 0.1M NaOH and 22cm3 of 0.1M CH3COOH 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 excess OH¯ 3. Calculate the volume of solution by adding the two individual volumes the volume of the solution is 25 + 22 = 47cm3

  31. WORKED EXAMPLE pH of mixtures Weak acid and EXCESS strong alkali Calculate the pH of a mixture of 25cm3 of 0.1M NaOH and 22cm3 of 0.1M CH3COOH 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 excess OH¯ 3. Calculate the volume of solution by adding the two individual volumes 4. Convert volume to dm3 (divide cm3 by 1000) the volume of the solution is 25 + 22 = 47cm3 there are 1000 cm3 in 1 dm3 volume = 47/1000 = 0.047dm3

  32. WORKED EXAMPLE pH of mixtures Weak acid and EXCESS strong alkali Calculate the pH of a mixture of 25cm3 of 0.1M NaOH and 22cm3 of 0.1M CH3COOH 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 excess OH¯ 3. Calculate the volume of 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 the volume of the solution is 25 + 22 = 47cm3 there are 1000 cm3 in 1 dm3 volume = 47/1000 = 0.047dm3 [OH¯] = 3.0 x 10-4 /0.047 = 6.38 x 10-3 mol dm-3

  33. WORKED EXAMPLE pH of mixtures Weak acid and EXCESS strong alkali Calculate the pH of a mixture of 25cm3 of 0.1M NaOH and 22cm3 of 0.1M CH3COOH 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 excess OH¯ 3. Calculate the volume of 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¯] = 3x 10-4 /0.045 = 6.38 x 10-3 mol dm-3 [H+] = Kw / [OH¯] =1.57 x 10-12 mol dm-3 pH = - log[H+] =11.8

  34. pH of mixtures EXCESS Weak monoprotic acid and strong alkali This method differs from the others because the excess substance is weak and as such is only PARTIALLY DISSOCIATED into ions. It is probably the hardest calculation to understand. 1. Calculate the initial number of moles of acid and OH¯ ions in the solutions 2. As H+ and OH¯ ions react in a 1:1 ratio, calculate unreacted moles of the excess acid 3. Calculate moles of salt anion formed; 1 mol of anion is formed for every H+ removed 4. Obtain the value of Ka for the weak acid and substitute the other values 5. Re-arrange the expression and calculate the value of [H+] 6. Convert concentration to pH usingpH = - log[H+] The following example shows you how to calculate the pH of the solution produced by adding 20cm3 of 0.1M NaOH to 25cm3 of 0.1M CH3COOH

  35. WORKED EXAMPLE pH of mixtures EXCESS Weak monoprotic acid and strong alkali 1. Calculate the initial number of moles of acid and OH¯ ions in the solutions 20cm3 of 0.1M NaOH 25cm3 of 0.1M CH3COOH 2.0 x 10-3 moles 2.5 x 10-3 moles moles of OH¯ = 0.1 x 20/1000 = 2.0 x 10-3 moles of H+ = 25 x 20/1000 = 2.5 x 10-3

  36. WORKED EXAMPLE pH of mixtures EXCESS Weak monoprotic acid and strong alkali 1. Calculate the initial number of moles of acid and OH¯ ions in the solutions 2. As H+ and OH¯ ions react in a 1:1 ratio, calculate unreacted moles of excess acid unreacted CH3COOH 20cm3 of 0.1M NaOH 25cm3 of 0.1M CH3COOH 5.0 x 10-4 moles 2.0 x 10-3 moles 2.5 x 10-3 moles The reaction taking place is CH3COOH + NaOH CH3COONa + H2O 2.0 x 10-3 moles of H+ will react with the same number of H+; this leaves 2.5 x 10-3 - 2.0 x 10-3 = 5.0 x 10-4 moles of CH3COOH in excess

  37. WORKED EXAMPLE pH of mixtures EXCESS Weak monoprotic acid and strong alkali 1. Calculate the initial number of moles of acid and OH¯ ions in the solutions 2. As H+ and OH¯ ions react in a 1:1 ratio, calculate unreacted moles of excess acid 3. Calculate moles of salt anion formed; 1 mol of anion is formed for every H+ removed CH3COONa produced 20cm3 of 0.1M NaOH 25cm3 of 0.1M CH3COOH 2.0 x 10-3 moles 2.0 x 10-3 moles 2.5 x 10-3 moles The reaction taking place is CH3COOH + NaOH CH3COONa + H2O 2.0 x 10-3 moles of H+ will produce the same number of CH3COONa this produces 2.0 x 10-3 moles of the anion CH3COO

  38. WORKED EXAMPLE pH of mixtures EXCESS Weak monoprotic acid and strong alkali 1. Calculate the initial number of moles of acid and OH¯ ions in the solutions 2. As H+ and OH¯ ions react in a 1:1 ratio, calculate unreacted moles of excess acid 3. Calculate moles of salt anion formed; 1 mol of anion is formed for every H+ removed 4. Obtain the value of Ka for the weak acid and substitute the other values Substitute the number of moles of anion produced here... it will be the same as the number of moles of H+ used up Ka = [H+(aq)] [CH3COO¯(aq)]mol dm-3 [CH3COOH(aq)] Substitute the Ka value Substitute the number of moles of unreacted acid here

  39. WORKED EXAMPLE pH of mixtures EXCESS Weak monoprotic acid and strong alkali 1. Calculate the initial number of moles of acid and OH¯ ions in the solutions 2. As H+ and OH¯ ions react in a 1:1 ratio, calculate unreacted moles of excess acid 3. Calculate moles of salt anion formed; 1 mol of anion is formed for every H+ removed 4. Obtain the value of Ka for the weak acid and substitute the other values Substitute the number of moles of anion produced here... it will be the same as the number of moles of H+ used up 1.7 x 10-5 = [H+(aq)] x (2 x 10-3)mol dm-3 (5 x 10-4) Substitute the Ka value Substitute the number of moles of unreacted acid here

  40. [H+(aq)] = 1.7 x 10-5x5 x 10-4 mol dm-3 2 x 10-3 = 4.25 x 10-6mol dm-3 WORKED EXAMPLE pH of mixtures EXCESS Weak monoprotic acid and strong alkali 1. Calculate the initial number of moles of acid and OH¯ ions in the solutions 2. As H+ and OH¯ ions react in a 1:1 ratio, calculate unreacted moles of excess acid 3. Calculate moles of salt anion formed; 1 mol of anion is formed for every H+ removed 4. Obtain the value of Ka for the weak acid and substitute the other values 5. Re-arrange the expression and calculate the value of [H+]

  41. [H+(aq)] = 1.7 x 10-5x5 x 10-4 mol dm-3 2 x 10-3 = 4.25 x 10-6mol dm-3 pH = - log10[H+(aq)] = 5.37 WORKED EXAMPLE pH of mixtures EXCESS Weak monoprotic acid and strong alkali 1. Calculate the initial number of moles of acid and OH¯ ions in the solutions 2. As H+ and OH¯ ions react in a 1:1 ratio, calculate unreacted moles of excess acid 3. Calculate moles of salt anion formed; 1 mol of anion is formed for every H+ removed 4. Obtain the value of Ka for the weak acid and substitute the other values 5. Re-arrange the expression and calculate the value of [H+] 6. Convert concentration to pH usingpH = - log[H+]

  42. 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 acids and bases

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