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.

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 = 1 x 10 -14 mol 2 dm -6 (at 25°C)

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 10 -2 mol dm -3 pH = - log [H + ] = 1.7 WORKED EXAMPLE WORKED EXAMPLE

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 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) K w = [H + ][OH¯] = 1 x 10 -14 mol 2 dm -6 therefore [H + ] = K w / [OH¯] = 1 x 10 -13 mol dm -3 pH = - log [H + ] = 13 WORKED EXAMPLE WORKED EXAMPLE

Calculate the pH of 0.01 M HCl 0.1 M HNO 3 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

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

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

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

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

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

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 = -log [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

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

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

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

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

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 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 WORKED EXAMPLE WORKED EXAMPLE

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

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 -3 6. 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 10 -14 at 25°C therefore [H + ] = K w / [OH¯] then pH = - log[H + ]

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

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 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 WORKED EXAMPLE WORKED EXAMPLE

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 10 -3 moles 2.0 x 10 -3 moles WORKED EXAMPLE WORKED EXAMPLE

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 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 5.0 x 10 -4 moles of OH¯ UNREACTED 25cm 3 of 0.1M NaOH 20cm 3 of 0.1M HCl 2.5 x 10 -3 moles 2.0 x 10 -3 moles WORKED EXAMPLE WORKED EXAMPLE

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 25 + 20 = 45cm 3 WORKED EXAMPLE WORKED EXAMPLE

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 25 + 20 = 45cm 3 there are 1000 cm 3 in 1 dm 3 volume = 45/1000 = 0.045dm 3 WORKED EXAMPLE WORKED EXAMPLE

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 10 -4 / 0.045 = 1.11 x 10 -2 mol dm -3

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 6. 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 10 -14 mol 2 dm -6 (at 25°C) WORKED EXAMPLE WORKED EXAMPLE [OH¯]= 5.0 x 10 -4 / 0.045 = 1.11 x 10 -2 mol dm -3 [H + ] = K w / [OH¯] = 9.00 x 10 -13 mol dm -3 pH = - log[H + ] = 12.05

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 -3 6. Convert concentration to pH either using K w = [H + ] [OH¯] = 1 x 10 -14 at 25°C therefore [H + ] = K w / [OH¯] then pH = - log[H + ] or pOH = - log[OH¯] and pH + pOH = 14

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 10 -3 moles 2.2 x 10 -3 moles 22cm 3 of 0.1M CH 3 COOH 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 WORKED EXAMPLE WORKED EXAMPLE

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 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 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 10 -4 moles of OH¯ UNREACTED 25cm 3 of 0.1M NaOH 2.5 x 10 -3 moles 2.2 x 10 -3 moles 22cm 3 of 0.1M CH 3 COOH WORKED EXAMPLE WORKED EXAMPLE

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 25 + 22 = 47cm 3 WORKED EXAMPLE WORKED EXAMPLE

the volume of the solution is 25 + 22 = 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

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 25 + 22 = 47cm 3 there are 1000 cm 3 in 1 dm 3 volume = 47/1000 = 0.047dm 3 [OH¯]= 3.0 x 10 -4 / 0.047 = 6.38 x 10 -3 mol dm -3 WORKED EXAMPLE WORKED EXAMPLE

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 6. As the excess is OH¯ usepOH = - log[OH¯] then pH + pOH = 14 or K w = [H + ][OH¯] so [H + ] = K w / [OH¯] [OH¯]= 3x 10 -4 / 0.045 = 6.38 x 10 -3 mol dm -3 [H + ] = K w / [OH¯] = 1.57 x 10 -12 mol dm -3 pH = - log[H + ] = 11.8 pH of mixtures Weak acid and EXCESS strong alkali WORKED EXAMPLE WORKED EXAMPLE

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

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 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 WORKED EXAMPLE 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 5.0 x 10 -4 moles 20cm 3 of 0.1M NaOH 2.0 x 10 -3 moles 2.5 x 10 -3 moles unreacted CH 3 COOH 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 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

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 10 -3 moles of H + will produce the same number of CH 3 COONa this produces 2.0 x 10 -3 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 10 -3 moles 20cm 3 of 0.1M NaOH 2.0 x 10 -3 moles 2.5 x 10 -3 moles CH 3 COONa produced WORKED EXAMPLE WORKED EXAMPLE 25cm 3 of 0.1M CH 3 COOH

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

1.7 x 10 -5 = [H + (aq) ] x (2 x 10 -3 ) mol dm -3 (5 x 10 -4 ) 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

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 10 -5 x 5 x 10 -4 mol dm -3 2 x 10 -3 = 4.25 x 10 -6 mol dm -3 WORKED EXAMPLE 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 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 10 -5 x 5 x 10 -4 mol dm -3 2 x 10 -3 = 4.25 x 10 -6 mol dm -3 pH = - log 10 [H + (aq) ] = 5.37 WORKED EXAMPLE WORKED EXAMPLE

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

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.

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Presentation on theme: "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."— Presentation transcript:

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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.

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Presentation on theme: "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."— Presentation transcript:

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