Complete table Calculate Kc with units if any A + B = 2C component A B C Initial amount (mol) 0.4 0.2 Equilibrium amount (mol) 0.25

pH for strong and weak acids L.O.: Calculate pH of strong and weak acids.

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 Leave out

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 Leave out

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 Leave out

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

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 From chemfact 25

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

Exercise

Finding the pH of weak acids What is a “weak acid”? What is a “weak base”? Calculate the pH of a weak acid. 4.3 pH questions

whether the acids and alkalis are STRONG or WEAK 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