6 Buffer solutions

Buffer Solutions A Buffer solution is one where the pH does not change significantly if small amounts of acid or alkali are added to it. An acidic buffer solution is made from a weak acid and a salt of that weak acid ( made from reacting the weak acid with a strong base). Example : ethanoic acid and sodium ethanoate CH3CO2H (aq) and CH3CO2 – Na+ A basic buffer solution is made from a weak base and a salt of that weak base ( made from reacting the weak base with a strong acid). Example :ammonia and ammonium chloride NH3 and NH4+ClHow Buffer solutions work In an ethanoic acid buffer CH3CO2H (aq) CH3CO2 – (aq) + H+ (aq) Acid conjugate base In a buffer solution there is a much higher concentration of the salt CH3CO2 – ion than in the pure acid. If small amounts of acid is added to the buffer: Then the above equilibrium will shift to the left removing nearly all the H+ ions added, CH3CO2 – (aq) + H+ (aq)  CH3CO2H (aq) As there is a large concentration of the salt ion in the buffer the ratio [CH3CO2H]/ [CH3CO2 -] stays almost constant, so the pH stays fairly constant. If small amounts of alkali is added to the buffer. The OHions will react with H+ ions to form water. H+ + OH –  H2O The Equilibrium will then shift to the right to produce more H+ ions. CH3CO2H (aq) CH3CO2 – (aq) + H+ (aq) Some ethanoic acid molecules are changed to ethanoate ions but as there is a large concentration of the salt ion in the buffer the ratio [CH3CO2H]/ [CH3CO2 -] stays almost constant, so the pH stays fairly constant. Learn these explanations carefully and be able to write the equilibrium to illustrate your answer. The buffer contains a reservoir of HA and A – ions [CH3CO2H (aq)] [CH3CO2 – (aq) ]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 Calculating the pH of Buffer Solutions [H+ (aq)][A- (aq)] [HA (aq)] Ka= We still use the weak acids dissociation expression But here we assume the [A-] concentration is due to the added salt only [HA(aq)] [A- (aq) ] [H+ (aq) Normally we ] = Ka rearrange to We also assume the Initial concentration of the acid has remained constant, because amount that has dissociated or reacted is small. The salt content can be added in several ways: a salt solution could be added to the acid or some solid salt added. A buffer can also be made by partially neutralising a weak acid with alkali and therefore producing salt.Example 12: making a buffer by adding a salt solution What would be the pH of a buffer made from 45cm3 of 0.1M ethanoic acid and 50cm3 of 0.15 M sodium ethanoate (Ka = 1.7 x 10-5 ) ? Work out the moles of both solutions Moles ethanoic = conc x vol = 0.1 x 0.045 = 0.0045mol Moles sodium ethanoate = conc x vol = 0.15 x 0.050 = 0.0075 [HA(aq)] [A- (aq) ] [H+ (aq)] = Ka We can enter moles of acid and salt straight into the equation as they both have the same new final volume 0.0045 0.0075 [H+ (aq)] = 1.7 x 10-5 x [H+ (aq)] = 1.02x 10-5 pH = – log [H+ ] = -log 1.02x 10-5 = 4.99 Example 13 : making a buffer by adding a solid salt A buffer solution is made by adding 1.1g of sodium ethanoate into 100 cm3 of 0.4M ethanoic acid. What is its pH? Ka =1.7 x10-5 Work out the moles of both solutions Moles ethanoic = conc x vol = 0.4 x 0.1 = 0.04mol Moles sodium ethanoate = mass/Mr= 1.1/82 = 0.0134 [HA(aq)] [A- (aq) ] [H+ (aq)] = Ka We can enter moles of acid and salt straight into the equation as they both have the same new final volume 0.04 0.0134 [H+ (aq)] = 1.7 x 10-5 x [H+ (aq)] = 5.07x 10-5 pH = – log [H+ ] = -log 5.07x 10-5 = 4.29 N Goalby chemrevise.org If a buffer is made by adding sodium hydroxide to partially neutralise a weak acid then follow the method below Example 14 55cm3 of 0.5M CH3CO2H is reacted with 25cm3 of 0.35M NaOH. What will be the pH of the resulting buffer solution? 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. Calculating change in pH of buffer on addition of small amount of acid or alkali If a small amount of alkali is added to a buffer then the moles of the buffer acid would reduce by the number of moles of alkali added and the moles of salt would increase by the same amount so a new calculation of pH can be done with the new values. CH3CO2H (aq) +OH-  CH3CO2 – (aq) + H2O (l) If a small amount of acid is added to a buffer then the moles of the buffer salt would reduce by the number of moles of acid added and the moles of buffer acid would increase by the same amount so a new calculation of pH can be done with the new values. CH3CO2 – (aq) + H +  CH3CO2H (aq) Example 15: 0.005 mol of NaOH is added to 500cm3 of a buffer where the concentration of ethanoic acid is 0.200 mol dm-3 and the concentration of sodium ethanoate is 0.250 mol dm-3 . (Ka = 1.7 x 10-5 ) Calculate the pH of the buffer solution after the NaOH has been added. Work out the moles of acid and salt in the initial buffer solution Moles ethanoic acid= conc x vol = 0.200 x 0.500 = 0.100mol Moles sodium ethanoate = conc x vol = 0.25 x 0.500 = 0.125mol Work out the moles of acid and salt in buffer after the addition of 0.005mol NaOH Moles ethanoic acid = 0.100 – 0.005 = 0.095 mol Moles sodium ethanoate = 0.125 +0.005 = 0.130 mol [CH3COOH (aq)] [CH3COO- (aq) ] [H+ (aq)] = Ka We can enter moles of acid and salt straight into the equation as they both have the same new final volume 0.095 0.130 [H+ (aq)] = 1.7 x 10-5 x [H+ (aq)] = 1.24x 10-5 pH = – log [H+ ] = -log 1.24x 10-5 = 4.91