BUFFER SOLUTIONS

Buffer solutions CONTENTS What is a buffer solution? Uses of buffer solutions Acidic buffer solutions Alkaline buffer solutions Buffer solutions - ideal concentration Calculating the pH of a buffer solution Salt hydrolysis Check list

Before you start it would be helpful to… Buffer solutions Before you start it would be helpful to… know that weak acids and bases are only partly ionised in solution be able to calculate pH from hydrogen ion concentration be able to construct an equation for the dissociation constant of a weak acid

Buffer solutions - Brief introduction Definition “Solutions which resist changes in pH when small quantities of acid or alkali are added.” Acidic Buffer (pH < 7) made from a weak acid + its conjugate base ethanoic acid sodium ethanoate Alkaline Buffer (pH > 7) made from a weak base + its conjugate acid ammonia ammonium chloride Uses Standardising pH meters Buffering biological systems (eg in blood) Maintaining the pH of shampoos Maintaining the pH of hydroponic solutions

Buffer solutions - uses Definition “Solutions which resist changes in pH when small quantities of acid or alkali are added.” Biological Uses In biological systems (saliva, stomach, and blood) it is essential that the pH stays ‘constant’ in order for any processes to work properly. e.g. If the pH of blood varies by 0.5 it can lead to unconsciousness and coma Most enzymes work best at particular pH values. Other Uses Many household and cosmetic products need to control their pH values. Shampoo Buffer solutions counteract the alkalinity of the soap and prevent irritation Baby lotion Buffer solutions maintain a pH of about 6 to prevent bacteria multiplying Others Washing powder, eye drops, fizzy lemonade, hydroponic solutions

Acidic buffer solutions - action It is essential to have a weak acid for an equilibrium to be present so that ions can be removed and produced. The dissociation is small and there are few ions. CH3COOH(aq) CH3COO¯(aq) + H3O+(aq) relative concs. HIGH LOW LOW NB A strong acid can’t be used as it is fully dissociated and cannot remove H3O+(aq) HCl(aq) ——> Cl¯(aq) + H3O+(aq) Adding acid Any H3O+ is removed by reacting with CH3COO¯ ions to form CH3COOH via the equilibrium. Unfortunately, the concentration of CH3COO¯ is small and only a few H3O+ can be “mopped up”. A much larger concentration of CH3COO¯ is required. To build up the concentration of CH3COO¯ ions, sodium ethanoate is added.

Acidic buffer solutions - action It is essential to have a weak acid for an equilibrium to be present so that ions can be removed and produced. The dissociation is small and there are few ions. CH3COOH(aq) CH3COO¯(aq) + H3O+(aq) relative concs. HIGH LOW LOW NB A strong acid can’t be used as it is fully dissociated and cannot remove H+(aq) HCl(aq) ——> Cl¯(aq) + H3O+(aq) Adding alkali This adds OH¯ ions which react with H3O+ ions H3O+(aq) + OH¯(aq) 2H2O(l) Removal of H3O+ from the weak acid equilibrium means that, according to Le Chatelier’s Principle, more CH3COOH will dissociate to form ions to replace those being removed. As the added OH¯ ions remove the H3O+ from the weak acid system, the equilibrium moves to the right to produce more H3O+ ions. Obviously, there must be a large concentration of undissociated acid molecules to be available.

Alkaline buffer solutions - action Very similar but is based on the equilibrium surrounding a weak base; AMMONIA NH3(aq) + H2O(l) OH¯(aq) + NH4+(aq) relative concs. HIGH LOW LOW but one needs ; a large conc. of OH¯(aq) to react with any H3O+(aq) added a large conc of NH4+(aq) to react with any OH¯(aq) added There is enough NH3 to act as a source of OH¯ but one needs to increase the concentration of ammonium ions by adding an ammonium salt. Use AMMONIA (a weak base) + AMMONIUM CHLORIDE (one of its salts)

Buffer solutions - ideal concentration The concentration of a buffer solution is also important If the concentration is too low, there won’t be enough CH3COOH and CH3COO¯ to cope with the ions added. Summary For an acidic buffer solution one needs ... large [CH3COOH(aq)] - for dissociating into H3O+(aq) when alkali is added large [CH3COO¯(aq)] - for removing H3O+(aq) as it is added This situation can’t exist if only acid is present; a mixture of the acid and salt is used. The weak acid provides the equilibrium and the large CH3COOH(aq) concentration. The sodium salt provides the large CH3COO¯(aq) concentration. One uses a WEAK ACID + its SODIUM OR POTASSIUM SALT

Calculating the pH of an acidic buffer solution Calculate the pH of a buffer whose [HA] is 0.1 mol L-1 and [A¯] of 0.1 mol L-1.

Calculating the pH of an acidic buffer solution Calculate the pH of a buffer whose [HA] is 0.1 mol L-1 and [A¯] of 0.1 mol L-1. Ka = [H3O+(aq)] [A¯(aq)] [HA(aq)]

Calculating the pH of an acidic buffer solution Calculate the pH of a buffer whose [HA] is 0.1 mol L-1 and [A¯] of 0.1 mol L-1. Ka = [H3O+(aq)] [A¯(aq)] [HA(aq)] re-arrange [H3O+(aq)] = [HA(aq)] x Ka [A¯(aq)]

Calculating the pH of an acidic buffer solution Calculate the pH of a buffer whose [HA] is 0.1 mol L-1and [A¯] of 0.1 mol L-1. Ka = [H3O+(aq)] [A¯(aq)] [HA(aq)] re-arrange [H3O+(aq)] = [HA(aq)] x Ka [A¯(aq)] from information given [A¯] = 0.1 mol L-1 [HA] = 0.1 mol L-1

Calculating the pH of an acidic buffer solution Calculate the pH of a buffer whose [HA] is 0.1 mol L-1 and [A¯] of 0.1 mol L-1. Ka = [H3O+(aq)] [A¯(aq)] [HA(aq)] re-arrange [H3O+(aq)] = [HA(aq)] x Ka [A¯(aq)] from information given [A¯] = 0.1 mol L-1 [HA] = 0.1 mol L-1 If the Ka of the weak acid HA is 2 x 10-4 [H3O+(aq)] = 0.1 x 2 x 10-4 = 2 x 10-4 mol L-1 0.1

Calculating the pH of an acidic buffer solution Calculate the pH of a buffer whose [HA] is 0.1 mol L-1 and [A¯] of 0.1 mol L-1. Ka = [H3O+(aq)] [A¯(aq)] [HA(aq)] re-arrange [H3O+(aq)] = [HA(aq)] x Ka [A¯(aq)] from information given [A¯] = 0.1 mol L-1 [HA] = 0.1 mol L1 If the Ka of the weak acid HA is 2 x 10-4. [H3O +(aq)] = 0.1 x 2 x 10-4 = 2 x 10-4 mol L-1 0.1 pH = - log10 [H3O+(aq)] = 3.699

Calculating the pH of an acidic buffer solution Calculate the pH of the solution formed when 500mL of 0.1 mol L-1 of weak acid HX is mixed with 500mL of a 0.2 mol L-1 solution of its salt NaX. Ka = 4 x 10-5.

Calculating the pH of an acidic buffer solution Calculate the pH of the solution formed when 500mL of 0.1 mol L-1 of weak acid HX is mixed with 500mL of a 0.2 mol L-1 solution of its salt NaX. Ka = 4 x 10-5. Ka = [H3O+(aq)] [X¯(aq)] [HX(aq)]

Calculating the pH of an acidic buffer solution Calculate the pH of the solution formed when 500mL of 0.1 mol L-1 of weak acid HX is mixed with 500mL of a 0.2 mol L-1 solution of its salt NaX. Ka = 4 x 10-5. Ka = [H3O+(aq)] [X¯(aq)] [HX(aq)] re-arrange [H3O +(aq)] = [HX(aq)] Ka [X¯(aq)]

Calculating the pH of an acidic buffer solution Calculate the pH of the solution formed when 500mL of 0.1 mol L-1 of weak acid HX is mixed with 500mL of a 0.2 mol L-1 solution of its salt NaX. Ka = 4 x 10-5. Ka = [H3O+(aq)] [X¯(aq)] [HX(aq)] re-arrange [H3O+(aq)] = [HX(aq)] Ka [X¯(aq)] The solutions have been mixed; the volume is now 1 L therefore [HX] = 0.05 mol L-1 and [X¯] = 0.10 mol L-1

Calculating the pH of an acidic buffer solution Calculate the pH of the solution formed when 500mL of 0.1 mol L1 of weak acid HX is mixed with 500mL of a 0.2 mol L-1 solution of its salt NaX. Ka = 4 x 10-5. Ka = [H3O+(aq)] [X¯(aq)] [HX(aq)] re-arrange [H3O +(aq)] = [HX(aq)] Ka [X¯(aq)] The solutions have been mixed; the volume is now 1 L therefore [HX] = 0.05 mol L-1 and [X¯] = 0.10 mol L-1 Substituting [H3O+(aq)] = 0.05 x 4 x 10-5 = 2 x 10-5 mol L1 0.1 pH = - log10 [H3O+(aq)] = 4.70

SALT HYDROLYSIS Many salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates. There are four distinct systems. All dissociated ions are aqueous ions. When mixed, the ions of strong acids and bases remain apart. Ions of weak acids and bases associate.

SALT HYDROLYSIS Many salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates. There are four distinct systems. All dissociated ions are aqueous ions. When mixed, the ions of strong acids and bases remain apart. Ions of weak acids and bases associate. Salts of strong acids and strong bases - SODIUM CHLORIDE NaCl dissociates completely in water Na+ Cl¯ ——> Na+ + Cl¯ Water only ionises to a very small extent 2H2O OH¯ + H3O+ Na+ and OH¯ are ions of a strong base so remain apart H3O+and Cl¯ are ions of a strong acid so remain apart all the OH¯ and H3O+ ions remain in solution therefore [H3O +] = [OH¯] and the solution will be NEUTRAL

SALT HYDROLYSIS Many salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates. There are four distinct systems. All dissociated ions are aqueous ions. When mixed, the ions of strong acids and bases remain apart. Ions of weak acids and bases associate. Salts of strong acids and weak bases - AMMONIUM CHLORIDE NH4Cl dissociates completely in water NH4+ Cl¯ ——> NH4+ + Cl¯ Water only ionises to a very small extent 2H2O OH¯ + H3O+ NH3 + H2O Na+ and OH¯ are ions of a strong base so tend to be associated H3O+and Cl¯ are ions of a strong acid so remain apart all the H3O+ ions remain in solution therefore [H3O+] > [OH¯] and the solution will be ACIDIC

SALT HYDROLYSIS Many salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates. There are four distinct systems. All dissociated ions are aqueous ions. When mixed, the ions of strong acids and bases remain apart. Ions of weak acids and bases associate. Salts of weak acids and strong bases - SODIUM ETHANOATE CH3COONa dissociates completely in water CH3COO ¯ Na + ——> Na+ + CH3COO¯ Water only ionises to a very small extent 2H2O OH¯ + H3O+ CH3COOH Na+ and OH¯ are ions of a strong base so remain apart H3O+and CH3OO¯ are ions of a weak acid so tend to be associated all the OH¯ ions remain in solution therefore [OH¯] > [H3O+] and the solution will be ALKALINE

SALT HYDROLYSIS Many salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates. There are four distinct systems. All dissociated ions are aqueous ions. When mixed, the ions of strong acids and bases remain apart. Ions of weak acids and bases associate. Salts of weak acids and weak bases - AMMONIUM ETHANOATE CH3COONH4 dissociates completely in water CH3COO ¯ NH4 + ——> NH4+ + CH3COO¯ Water only ionises to a very small extent 2H2O OH¯ + H3O+ NH3 + H2O CH3COOH Na+ and OH¯ are ions of a weak base so tend to be associated H3O+and CH3OO¯ are ions of a weak acid so tend to be associated the solution might be alkaline or acidic i.e. APPROXIMATELY NEUTRAL

What should you be able to do? REVISION CHECK What should you be able to do? Recall the definition of a buffer solution Recall the difference between an acidic and an alkaline buffer solution Recall the uses of buffer solutions Understand the action of buffer solutions Calculate the pH of an acidic buffer solution Recall and understand the reactions due to salt hydrolysis CAN YOU DO ALL OF THESE? YES NO

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BUFFER SOLUTIONS THE END