Chem 1102 Semester 1, 2011 ACIDS AND BASES

Chem 1102 Semester 1, 2011 ACIDS AND BASES

Acids and Bases Lecture 23: Weak Acids and Bases Calculations involving pk a and pk b Strong Acids and Bases Lecture 24: Polyprotic Acids Salts of Acids and Bases Buffers Indicators Titrations

Polyprotic Acids Monoprotic acids: donate only a single proton to water Diprotic: can donate two protons. Polyprotic: can donate more than two protons. H 3 PO 4(aq) #H + (aq) + H 2 PO 4 (aq) #pk a1 = 2.2 # H 2 PO 4 (aq) #H + (aq) + HPO 4 2 (aq) #pk a2 = 7.2 # HPO 4 2 (aq) #H + (aq) + PO 4 3 (aq) # #pk a3 = 12.4 # removing more protons is harder: increasing pk a = decreasing K a : K a1 > K a2 > K a3 # reason: harder to remove +ve charge against increasing ve charge.# large difference in pk a values: only need to consider one equilibrium at a time (simplifies maths).#

pk a1 = 2.2 pk a2 = 7.2 pk a3 = 12.4 pk a1 =3.13, pk a2 = 4.76, pk a3 =6.40.

Hydrolysis of Ions: Examples Is a solution of NaCN acidic or basic? Does a solution of NH 4 Cl have ph > 7 or < 7?

Salt of a Weak Acid & Weak Base What is the ph of ammonium acetate at 25 C? 2 possible reactions are: CH 3 COO (aq) + H 2 O(l) CH 3 COOH(aq) + OH (aq) K b = 10 9.24 Anions of weak acids hydrolyse OH Basic effect NH 4+ (aq) + H 2 O(l) NH 3 (aq) + H 3 O + (aq) K a = 10 9.24 =5.5 x 10-10 Cations of weak bases hydrolyse H 3 O + Acid effect In this case K a = K b, so salt is neutral. Also: salts of strong acids & strong bases are always neutral!

If K a = K b, then ph = 7 and solu5on is neutral If K a > K b, then ph < 7 and solu5on is acidic (A salt formed between a strong acid and a weak base is an acid salt, e.g. NH 4 Cl) If K a < K b, then ph > 7 and solu5on is basic (A salt formed between a weak acid and a strong base is a basic salt, e.g. NaCH 3 COO) Ions of Neutral Salts Cations Na + K + Rb + Cs + Mg 2+ Ca 2+ Sr 2+ Ba 2+ Anions Cl - Br - I -, ClO 4 - BrO 4 - ClO 3 - NO 3 - Acidic Ions NH 4 + Al 3+ Pb 2+ Sn 2+ Transition metal ions HSO 4 - H 2 PO 4 - Basic Ions F - C 2 H 3 O 2 - NO 2 - HCO 3 - CN - CO 3 2- S 2- SO 4 2- HPO 4 2- PO 4 3-

Another Example Rank the following 1.0 M solutions in order of decreasing ph. HNO 3 NaCN KOH HCl CH 3 COOH

The Common Ion Effect If you add the salt of a weak acid to a solution of the same acid then the equilibrium will shift towards neutral ph.# CH 3 COOH (aq) + H 2 O (l) CH 3 COO (aq) + H 3 O+ (aq) Addition of CH 3 COO Na + will raise [CH 3 COO ]. From Le Châtelier s principle, the first equilibrium will shift to the left to remove CH 3 COO and therefore decrease [H 3 O + ].!The same holds true for a base: If you add the salt of a weak base to a solution of the same base then the equilibrium will shift towards neutral ph.#

Buffers A solution containing both:# a weak acid + its salt# OR# a weak base + its salt# withstands ph changes when (limited) amounts of acid or base are added.# Reason: Le Châtelierʼs principle.#

ph Change and Buffers Consider change in ph of pure water (ph = 7) if we add an equal amount of 10 3 M HCl: ##[H 3 O + ] = 5 x 10 4 M (can neglect amount already present in water), so ph drops from 7 to 3.3! ## # # A whopping change! What about buffer solutions?

ph Change and Buffers The addition of small amounts of acid or base to a buffer solution will result in a negligible change to the ph. Why? Buffer after addition of H 3 O + # Buffer with equal concentrations of conjugate base and acid # Buffer after addition of OH -# CH 3 COO -# CH 3 COOH # H 3 O +# CH 3 COO -# CH 3 COOH # OH -# CH 3 COO -# CH 3 COOH # H 3 O + + CH 3 COO - H 2 O + CH 3 COOH# CH 3 COOH + OH - H 2 O + CH 3 COO -#

Henderson-Hasselbalch Equation For a buffer solution, which contains similar concentrations of a conjugate acid/base pair of a weak acid:# Since the dissociation of HA or protonation of A doesnʼt lead to a significant change in the concentrations of these species:# Rearranging:#

Buffer Preparation and Capacity Buffer Preparation If the ph of a required buffer = pk a of available acid then use equimolar amounts of acid and conjugate base. If the required ph differs from the pk a then use the Henderson- Hasselbalch equation to work out ratios of acid and conjugate base.

Buffer Preparation and Capacity Buffer capacity is a measure of the ability of a buffer to resist ph change and depends on both the absolute and relative component concentrations. Buffer range is the ph range over which the buffer acts effectively. Buffer capacity is related to the amount of strong acid or base that can be added without causing significant ph change. Depends on amount of acid & conjugate base in solution: - Highest capacity when [HA] and [A ] are large. - Highest capacity when [HA] [A ] (most effective buffers have acid/ base ratio less than 10 and more than 0.1 ph range is ±1).

Buffer Example In the H 3 PO 4 / NaH 2 PO 4 / Na 2 HPO 4 / Na 3 PO 4 system, how could you make up a buffer with a ph of 7.40? (K a1 = 7.2 x 10 3, K a2 = 6.3 x 10 8, K a3 = 4.2 x 10 13 ) To make up a buffer, we need ph ~ pk a pk a1 = 2.14 pk a2 = 7.20 pk a3 = 12.38 must use mixture of H 2 PO 4 and HPO 2 4 Simply use the Henderson-Hasselbalch equation...

Buffer Example cont d ph = pk a2 + log 10 [HPO 4 2 ] [H 2 PO 4 ] original amounts 7.40 = 7.20 + log 10 [HPO 4 2 ] [H 2 PO 4 ] [HPO 4 2 ] [H 2 PO 4 ] = 10 0.20 = 1.58 So the required ratio of Na 2 HPO 4 to NaH 2 PO 4 = 1.58:1

Buffers in Natural Systems Biological systems, e.g. blood, contain buffers: ph control essential because biochemical reactions are very sensitive to ph.# Human blood is slightly basic, ph 7.39 7.45.# In a healthy person, blood ph is never more than 0.2 ph units from its average value.# ph < 7.2, acidosis ; ph > 7.6, alkalosis.# Death if ph < 6.8 or > 7.8.#

Buffer System in Blood Extracellular buffer (outside cell)# H + (aq) + HCO 3 (aq) H 2 CO 3(aq) H 2 CO 3(aq)# H 2 O (l) + CO 2(g)# Removal of CO 2 shifts equilibria to right, reducing [H 3 O + ], i.e. raising the ph.# The ph can be reduced by:# H 2 CO 3(aq) + H 2 O (l) HCO 3 (aq) + H 3 O + (aq)#

Acid-Base Titrations Equivalence Point, is the point at which the number of moles of added base = original number of moles of acid.# Strong acid/strong base ph = 7# Weak acid/strong base ph > 7# Strong acid/weak base ph < 7! End Point:#!When a colour change in the indicator is observed.! Always choose an indicator that changes colour close to the equivalence point#

Indicators weak acid base# Each form has a different colour # The ph at which acid base depends on the pk a of the indicator.

Titrations: Strong Acid / Strong Base Equivalence point at ph 7#

Titrations: Weak Acid / Strong Base Equivalence point ph > 7# (ph value depends on starting# concentrations)# ph change is more gradual#

Titrations: Weak Base / Strong Acid Equivalence point ph < 7# ph value depends on starting # concentrations#

What is the ph of a 0.10 M solu5on of NH 4 Br, if the pk b of NH 3 is 4.74?