4 pH calculations involving neutralisation

pH Calculations involving Neutralisation Reactions These can be quite complex calculations working out the pH of a partially neutralised acid or the pH of the solution if too much alkali has been added and has gone past neutralisation. The method differs if the acid is strong or weak for the partially neutralised case. Strong Acid and Strong Base Neutralisations Work out moles of original acid Work out moles of base added Work out which one is in excess Work out new concentration of excess H+ ions [H+ ] = moles excess H+ total volume (dm3 ) pH = – log [H+ ] Total volume = vol of acid + vol of base added If excess acid Work out new concentration of excess OHions [OH-] = moles excess OH- total volume (dm3 ) [H+ ] = Kw /[OH– ] pH = – log [H+ ] Total volume = vol of acid + vol of base added If excess alkali Example 8 45cm3 of 1M HCl is reacted with 30cm3 of 0.65M NaOH. What will be the pH of the resulting mixture? Moles HCl = conc x vol = 1 x 0.045 = 0.045mol Moles NaOH = conc x vol = 0.65 x 0.030 = 0.0195 HCl + NaOH  NaCl + H2O Moles of HCl in excess = 0.045-0.0195 = 0.0255 (as 1:1 ratio) [H+ ] = moles excess H+ total volume (dm3 ) = 0.0255/ 0.075 = 0.34M pH = – log [H+ ] = -log 0.34 = 0.47 Example 7 15cm3 of 0.5M HCl is reacted with 35cm3 of 0.55M NaOH. What will be the pH of the resulting mixture? = 0.01175/ 0.05 = 0.235M [H+ ] = Kw /[OH– ] = 1×10-14 / 0.235 = 4.25×10-14 pH = – log [H+ ] = -log 4.25×10-14 = 13.37 Moles HCl = conc x vol = 0.5 x 0.015 = 0.0075mol Moles NaOH = conc x vol = 0.55 x 0.035 = 0.01925 HCl + NaOH  NaCl + H2O Moles of NaOH in excess = 0.01925 – 0.0075= 0.01175 (as 1:1 ratio) [OH-] = moles excess OH- total volume (dm3 ) Weak Acid and Strong Base Neutralisations Work out moles of original acid Work out moles of base added Work out which one is in excess Work out new concentration of excess HA [HA] = initial moles HA – moles OH- total volume (dm3 ) Work out concentration of salt formed [A-] [A-] = moles OH- added total volume (dm3 ) Rearrange ka = [H+ ] [A-] to get [H+ ] [HA] pH = – log [H+ ] If excess acid If excess alkali use the same method with excess alkali and strong acid above Example 9 55cm3 of 0.5M CH3CO2H is reacted with 25cm3 of 0.35M NaOH. What will be the pH of the resulting mixture? Moles CH3CO2H = conc x vol =0.5x 0.055 = 0.0275mol Moles NaOH = conc x vol = 0.35 x 0.025 = 0.00875 Moles of CH3CO2H in excess = 0.0275-0.00875 = 0.01875 (as 1:1 ratio) [CH3CO2H ] = moles excess CH3CO2H total volume (dm3 ) = 0.01875/ 0.08 = 0.234M CH3CO2H+ NaOH  CH3CO2Na + H2O pH = – log [H+ ] = -log 3.64 x 10-5 = 4.44 [CH3CO2 – ] = moles OH- added total volume (dm3 ) = 0.00875/ 0.08 = 0.109M ka = [H+ ] [CH3CO2 – ] [ CH3CO2H ] ka is 1.7 x 10-5 mol dm-3 [H+ ] = ka x[ CH3CO2H ] / [CH3CO2 – ] = 1.7 x 10-5 x 0.234 / 0.109 = 3.64 x 10-5. Working out pH of a weak acid at half equivalence When a weak acid has been reacted with exactly half the neutralisation volume of alkali, the above calculation can be simplified considerably. At half neutralisation we can make the assumption that [HA] = [A-] ka = [H+ ] [CH3CO2 – ] [ CH3CO2H ] So [H+ (aq)] = ka And pH = pka Example 10 What is the pH of the resulting solution when 25cm3 of 0.1M NaOH is added to 50cm3 of 0.1M CH3COOH (ka 1.7 x 10-5 ) From the volumes and concentrations spot it is half neutralisation (or calculate) pH = pka = -log (1.7 x 10-5 ) = 4.77 Diluting an acid or alkali pH of diluted strong acid [H+ ] = [H+ ]old x old volume new volume pH = – log [H+ ] pH of diluted base [OH– ] = [OH– ]old x old volume new volume [H+ ] = Kw [OH– ] pH = – log [H+ ] Example 11 Calculate the new pH when 50.0 cm3 of 0.150 mol dm-3 HCl is mixed with 500 cm3 of water. [H+ ] = [H+ ]old x old volume new volume 0.05 0.55 [H+ (aq)] = 0.150 x [H+ (aq)] = 0.0136 pH = – log [H+ ] = -log 0.0136 = 1.87