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

pH calculations www.argonet.co.uk/users/hoptonj/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.argonet.co.uk/users/hoptonj/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 Calculating the pH of mixtures - weak acid and excess strong alkali Calculating the pH of mixtures - strong alkali and excess weak acid Check list

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 WEAK ACID and EXCESS STRONG BASE STRONG BASE and EXCESS WEAK ACID

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)

Weak acid and EXCESS strong alkali 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

Weak acid and EXCESS strong alkali pH of mixtures Weak acid and EXCESS strong alkali WORKED EXAMPLE 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 2.5 x 10-3 moles 2.2 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

Weak acid and EXCESS strong alkali pH of mixtures Weak acid and EXCESS strong alkali WORKED EXAMPLE 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 2.5 x 10-3 moles 2.2 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

Weak acid and EXCESS strong alkali pH of mixtures Weak acid and EXCESS strong alkali WORKED EXAMPLE 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

Weak acid and EXCESS strong alkali pH of mixtures Weak acid and EXCESS strong alkali WORKED EXAMPLE 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

Weak acid and EXCESS strong alkali pH of mixtures Weak acid and EXCESS strong alkali WORKED EXAMPLE 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

Weak acid and EXCESS strong alkali pH of mixtures Weak acid and EXCESS strong alkali WORKED EXAMPLE 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

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 using pH = - 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

EXCESS Weak monoprotic acid and strong alkali pH of mixtures EXCESS Weak monoprotic acid and strong alkali WORKED EXAMPLE 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

EXCESS Weak monoprotic acid and strong alkali pH of mixtures EXCESS Weak monoprotic acid and strong alkali WORKED EXAMPLE 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 2.0 x 10-3 moles 5.0 x 10-4 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

EXCESS Weak monoprotic acid and strong alkali pH of mixtures EXCESS Weak monoprotic acid and strong alkali WORKED EXAMPLE 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

EXCESS Weak monoprotic acid and strong alkali pH of mixtures EXCESS Weak monoprotic acid and strong alkali WORKED EXAMPLE 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

EXCESS Weak monoprotic acid and strong alkali pH of mixtures EXCESS Weak monoprotic acid and strong alkali WORKED EXAMPLE 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

EXCESS Weak monoprotic acid and strong alkali pH of mixtures EXCESS Weak monoprotic acid and strong alkali WORKED EXAMPLE 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+] [H+(aq)] = 1.7 x 10-5 x 5 x 10-4 mol dm-3 2 x 10-3 = 4.25 x 10-6 mol dm-3

EXCESS Weak monoprotic acid and strong alkali pH of mixtures EXCESS Weak monoprotic acid and strong alkali WORKED EXAMPLE 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 using pH = - log[H+] [H+(aq)] = 1.7 x 10-5 x 5 x 10-4 mol dm-3 2 x 10-3 = 4.25 x 10-6 mol dm-3 pH = - log10[H+(aq)] = 5.37

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 acids and bases CAN YOU DO ALL OF THESE? YES NO

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