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PH calculations A guide for A level students KNOCKHARDY PUBLISHING.

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1 pH calculations A guide for A level students KNOCKHARDY PUBLISHING

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 Calculating the pH of mixtures - weak acid and excess strong alkali Calculating the pH of mixtures - strong alkali and excess weak acid Check list 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 HCl H + + Cl¯ 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 HCl H + + Cl¯ 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 WEAK ACID and EXCESS STRONG BASE STRONG BASE and EXCESS WEAK ACID

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 HCl 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 HCl 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 HCl 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 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 pH of mixtures Strong acids and alkalis (either in excess) The reaction taking place is… HCl + NaOH NaCl + 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 HCl 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 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 pH of mixtures Strong acids and alkalis (either in excess) The reaction taking place is… HCl + NaOH NaCl + 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 HCl 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 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 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 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 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 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 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 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 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 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 dm 3 (divide cm 3 by 1000) 5. Divide moles by volume to find concentration of excess OH¯ in mol dm Convert concentration to pH either using K w = [H + ] [OH¯] = 1 x at 25°C therefore [H + ] = K w / [OH¯] then pH = - log[H + ] or pOH = - log[OH¯] and pH + pOH = 14

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

31 The reaction taking place is CH 3 COOH + NaOH CH 3 COONa + H 2 O or in its ionic form H + + OH¯ H 2 O (1:1 molar ratio) 2.2 x moles of H + will react with the same number of moles of OH¯ this leaves 2.5 x x = 3.0 x moles of OH¯ in excess pH of mixtures Weak acid and EXCESS strong alkali Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH and 22cm 3 of 0.1M CH 3 COOH 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¯ 3.0 x moles of OH¯ UNREACTED 25cm 3 of 0.1M NaOH 2.5 x moles 2.2 x moles 22cm 3 of 0.1M CH 3 COOH WORKED EXAMPLE WORKED EXAMPLE

32 pH of mixtures Weak acid and EXCESS strong alkali Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH and 22cm 3 of 0.1M CH 3 COOH 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 = 47cm 3 WORKED EXAMPLE WORKED EXAMPLE

33 the volume of the solution is = 47cm 3 there are 1000 cm 3 in 1 dm 3 volume = 47/1000 = 0.047dm 3 pH of mixtures Weak acid and EXCESS strong alkali Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH and 22cm 3 of 0.1M CH 3 COOH 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 dm 3 (divide cm 3 by 1000) WORKED EXAMPLE WORKED EXAMPLE

34 pH of mixtures Weak acid and EXCESS strong alkali Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH and 22cm 3 of 0.1M CH 3 COOH 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 dm 3 (divide cm 3 by 1000) 5. Divide moles by volume to find concentration of excess ion in mol dm -3 the volume of the solution is = 47cm 3 there are 1000 cm 3 in 1 dm 3 volume = 47/1000 = 0.047dm 3 [OH¯]= 3.0 x / = 6.38 x mol dm -3 WORKED EXAMPLE WORKED EXAMPLE

35 Calculate the pH of a mixture of 25cm 3 of 0.1M NaOH and 22cm 3 of 0.1M CH 3 COOH 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 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¯] [OH¯]= 3x / = 6.38 x mol dm -3 [H + ] = K w / [OH¯] = 1.57 x mol dm -3 pH = - log[H + ] = 11.8 pH of mixtures Weak acid and EXCESS strong alkali WORKED EXAMPLE WORKED EXAMPLE

36 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 K a 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 20cm 3 of 0.1M NaOH to 25cm 3 of 0.1M CH 3 COOH

37 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 20cm 3 of 0.1M NaOH 25cm 3 of 0.1M CH 3 COOH 2.0 x moles 2.5 x moles moles of OH ¯ = 0.1 x 20/1000 = 2.0 x moles of H + = 25 x 20/1000 = 2.5 x WORKED EXAMPLE WORKED EXAMPLE

38 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 5.0 x moles 20cm 3 of 0.1M NaOH 2.0 x moles 2.5 x moles unreacted CH 3 COOH 2.0 x moles of H + will react with the same number of H + ; this leaves 2.5 x x = 5.0 x moles of CH 3 COOH in excess The reaction taking place is CH 3 COOH + NaOH CH 3 COONa + H 2 O WORKED EXAMPLE WORKED EXAMPLE 25cm 3 of 0.1M CH 3 COOH

39 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 2.0 x moles of H + will produce the same number of CH 3 COONa this produces 2.0 x moles of the anion CH 3 COO  The reaction taking place is CH 3 COOH + NaOH CH 3 COONa + H 2 O 2.0 x moles 20cm 3 of 0.1M NaOH 2.0 x moles 2.5 x moles CH 3 COONa produced WORKED EXAMPLE WORKED EXAMPLE 25cm 3 of 0.1M CH 3 COOH

40 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 K a for the weak acid and substitute the other values K a = [H + (aq) ] [CH 3 COO¯ (aq) ] mol dm -3 [CH 3 COOH (aq) ] Substitute the number of moles of unreacted acid here Substitute the number of moles of anion produced here... it will be the same as the number of moles of H + used up Substitute the K a value WORKED EXAMPLE WORKED EXAMPLE

41 1.7 x = [H + (aq) ] x (2 x ) mol dm -3 (5 x ) 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 K a for the weak acid and substitute the other values Substitute the number of moles of unreacted acid here Substitute the number of moles of anion produced here... it will be the same as the number of moles of H + used up Substitute the K a value WORKED EXAMPLE WORKED EXAMPLE

42 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 K a 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 x 5 x mol dm -3 2 x = 4.25 x mol dm -3 WORKED EXAMPLE WORKED EXAMPLE

43 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 K a 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 x 5 x mol dm -3 2 x = 4.25 x mol dm -3 pH = - log 10 [H + (aq) ] = 5.37 WORKED EXAMPLE WORKED EXAMPLE

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

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

46 WELL DONE! Try some past paper questions

47 pH calculations THE END © 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


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