Presentation on theme: "Solution properties pH pKa Weak acid, base Buffered solution Acid Base"— Presentation transcript:
1 Solution properties pH pKa Weak acid, base Buffered solution Acid Base Ionization of waterpH scalesignificancepKaWeak acid, baseHenderson-Haslebach equationBuffered solutionDefinitionActionBuffer capacityDepend on ratio salt/acidDepend on concentration salt + acidDepend on amount of added a, bMaximum buffer capacityPreparationexamples
2 Acid a substance which donates a proton (or hydrogen ion) the addition of an acid to water will increase hydrogen ion concentration (more than 10-7 mol/l)
3 Base a substance that accepts protons the addition of a base will decrease the concentration of hydrogen ions.
4 dissociation of water The dissociation of water can be represented by: H2O H+ + OH-In pure water the concentrations of H+ and OH- ions are equal and at 25°C both have the values of 1 x 10-7 mol/l.
5 dissociation of waterThe hydrogen ion concentration range is from 1 mol/l for a strong acid down to 1 x mol/l for a strong base.To avoid the use of such low values pH has been introduced as a more convenient measure of hydrogen ion concentration.
7 pH the negative logarithm of the hydrogen Ion concentration [H+] pH = -log10 [H+]
8 pH pH of a neutral solution like pure water is 7, why? because the conc. of H+ ions (and OH- ) ions = 1 x 10-7 mol/l
9 pH pHs of acidic solutions will be less than 7 pHs of alkaline solutions will be greater than 7
10 pH has several important applications in pharmaceutical practice. Affect the solubilities of drugs that are weak acids or basesAffect the stabilities of many drugsAffect the ease of absorption of drugs from the GIT
11 Solution properties Acid Base Ionization of water pH pKa significancepKaHenderson-Haslebach equationBuffer solutionDefinitionActionBuffer capacityPreparationexamples
12 Dissociation (or ionization) constants and pKa: In solutions of weak acids or weak bases equilibria exist between undissociated molecules and their ions.
13 Dissociation (or ionization) constants and pKa: In a solution of a weakly acidic drug HA the equilibrium may be represented by:HA H+ + A-
14 Dissociation (or ionization) constants and pKa: The protonation of a weakly basic drug B can be represented by :B + H BH+
15 Dissociation (or ionization) constants and pKa: Such equilibrium do not occur in solutions of strong acids or bases in water, why?because they are completely ionized.
16 The ionization constant (dissociation constant) Ka of a weak acid can be obtained by applying the Law of Mass Action:[H+] [A-]Ka =[HA]pKa = the negative logarithm of KapH = the negative logarithm of the hydrogen ion conc. [H+]pKa = pH + log[A-]Henderson-Hasselbalch equation:A general equation that is applicable to any acidic drug with one ionizable group where:Cu = conc. of the unionized ; Ci = conc. of the ionized speciesCuCi
17 The ionization constant (dissociation constant) Ka of a protonated weak base is given by [H+] [B]Ka =[BH+]Taking the negative log of this equation:pKa = pH + log[B]Henderson-Hasselbalch equation:A general equation that is applicable to any weak basic drug with one ionizable group where:Ci = conc. of protonated ; Cu = conc. of the unionized speciesCiCu
18 The degree of ionization of a drug in a solution can be calculated from the Henderson-Hasselbalch equations for weak acids and bases if the pKa of the drug and the pH of the solution are knownSuch calculations are useful in determining the degree of ionization of drugs in various parts of the GIT and in the plasmaExamples1. The pKa of the weakly acidic drug sulphapyridine is about 8.0 and if the pH of the intestinal contents is 5.0 then the ratio of unionized:ionized drug is given by:Culog = pKa - pH = 8 – 5 = 3CiCu:Ci = antilog 3 = 103 : 1
19 The pKa value of aspirin, which is a weak acid, is about 3 The pKa value of aspirin, which is a weak acid, is about 3.5, and the pH of the gastric contents is 2.0Culog = pKa - pH = = 1.5Cithe ratio of the conc. of unionized acetylsalicyclic acid to acetylsalicylate anion is given by:Cu:Ci = antilog 1.5 = :1The pH of plasma is 7.4 so that the ratio of unionized:ionized aspirin in this medium is given by:log = pKa - pH = 3.5 – 7.4 = -3.9Cu:Ci = antilog -3.9 = x 10-4: 1
20 Solution properties Acid Base Ionization of water pH pKa significancepKaHenderson-Haslebach equationBuffer solutionDefinitionActionBuffer capacityPreparationexamples
21 Buffered SolutionsA buffered solution or buffer is a solution that resists a change in pH upon addition of small amounts of strong acid or strong base.
22 Composition and Action of Buffered Solutions A buffer consists of a mixture of a weak acid (HX) and its conjugate base (X–).HX(aq) ↔ H+(aq) + X–(aq)Thus a buffer contains both:An acidic species (to neutralize OH–) andA basic species (to neutralize H+).
24 Buffer CapacityA buffer counteracts the change in pH of a solution upon the addition of a strong acid, a strong base, or other agents that tend to alter the hydrogen ion concentration.Buffer capacity β: buffer efficiency, buffer index or buffer valueIs the resistance of a buffer to pH changesupon the addition of a strong acid or base.Definition:It can be defined as being equal to the amount of strong acid or strong base , expressed as moles of H +or OH- ions, required to change the pH of one litre of the buffer by one pH unite.Maximum buffer capacity (βmax) obtain when ratio of acid to salt = 1 i.e. pKa = pH
25 DefinitionThe ratio of increase of strong base (or acid) to small change in pH brought about by this addition.β = Δ BΔ pHΔB = the small increment in gram equiv./liter of strong baseadded to the buffer solution to produce a pH changeΔ pH = pH change
26 The buffer capacity of the solution has a value of 1: when the addition of 1 gram equiv. of strong base (or acid) to 1 liter of the buffer solution results in a change of 1 pH unit.Ex:Acetate buffer contains: 0.1 mole each of acetic acid & sodium acetatein 1 liter of solution.a) 0.01 mole portions of NaOH is addedb) The conc. of Na acetate (the [salt] in buffer equation) ↑ by 0.01 mol/l & the conc. of acetic acid [acid] ↓ by 0.01 mol/lbecause each increment of base converts 0.01 mole of acetic acid into 0.01 mole of sodium acetate according to the reaction.HAc NaOH NaAc + H2O(0.1 – 0.01) (0.01) ( )
27 The changes in concentration of the salt and the acid by the addition of a base are represented in the buffer equation by using the modified formpH = pKa + log [salt ] + [base][acid] - [base]Before the addition of the first portion of NaOH, the pH of the buffer solution ispH = log ( ) = 4.76( 0.1 – 0 )
28 pH of solutionMoles of NaOH added4.764.850.014.940.025.030.035.130.045.240.055.360.06BufferCapacity, β0.110.100.090.080.07
29 The buffer capacity is not a fixed value for a given buffer system, but depends on the amount of base added.With the addition of more NaOH, the buffer capacity decreases rapidly, and, when sufficient base has been added the acid convert completely into sodium ions and acetate ionsThe buffer has it’s greatest capacity before any base is added where [salt] / [acid] = 1, and according to equation, pH = pKa.The buffer capacity is influenced by an increase in the total conc. of the buffer constituents since a greater conc. of salt and acid provides a greater alkaline and acid reserve.
30 C = The total buffer concentration (the sum of the molar Van Slyke developed a more exact equation for calculation of buffer capacity βKa [H3O+] 2.3 C = β[H3O+])2+ Ka)C = The total buffer concentration (the sum of the molarconcentrations of the acid and the salt).Ka = dissociation constantH3O+ = hydrogen ion concentrationThe equation permits the calculation of the buffer capacity at any hydrogen ion concentration, i.e. when no acid or base has been added to the buffert
31 If hydrogen ion concentration is 1.75 x 10-5, pH = 4.76 Example:If hydrogen ion concentration is 1.75 x 10-5, pH = 4.76what is the capacity of the buffer containing 0.10 mole of each of acetic acid and sodium acetate per liter of solution ?The total concentration , C = [acid] + [salt], is 0.20 mol/l and the dissociation constant Ka is 1.75 x 10-5Ka [H3O+] 2.3 C = β(Ka + [H3O+])2β = 2.3 X 0.20 X (1.75x10-5) X (1.75 X 10-5) = 0.115[(1.75x10-5) +(1.75 X 10-5)]2ion
32 Prepare a buffer solution of pH 5 having a capacity of 0.02. 1-One chooses a weak acid having a pKa close to the pH desired.Acetic acid, pKa= 4.76 is suitable in this case
33 2-The ratio of salt and acid required to produce a pH of 5 was found to be [salt] / [acid] = 1.74/ 1
34 3] The buffer capacity equation is used to obtain the total buffer concentration , C = [salt ] + [acid]Ka[H3O+] 2.3 C = β[H3O+])2+ Ka)0.02 = 2.3 X C X (1.75x10-5) X (1X 10-5)[(1.75x10-5) + (1 X 10-5)]2C= 3.75 x 10-2 mol/lFrom (b): [salt] = 1.74 x [acid], andfrom (c): C = 1.74 x [acid] + [acid] = 3.75 X 10-2 mol/l[acid] = 1.37 x 10-2 mol/l[salt] = 1.74 X [acid] = 2.38 X 10-2 mol/l
35 The influence of concentration on buffer capacity . The buffer capacity is affected not only by the (salt)/(acid) ratio but also by the total concentrations of acid and salt . As shown previously in the table.when 0.01 mole of base was added to a 0.1 molar acetate buffer, the pH increased from 4.76 to 4.85 or a ∆ pH of 0.09If the concentration of acetic acid and sodium acetate is raised to 1 molar ,the pH of the original buffer solution remains at about 4.76 , but now upon the addition of 0.01 mole of base it becomes 4.77 and ∆ pH of only 0.01Therefore, an increase in the concentration of the buffer components results in a greater buffer capacity or efficiency.In summary , the buffer capacity depends on(a) the value of the ratio salt/acid , β increase as the valueapproaches unity.(b) the magnitude of the individual concentrations of thebuffer components, β increase as C increased
36 Maximum buffer capacity . The maximum buffer capacity occurs when pH = pKa orwhen (H3O+) = Kamax = C (H3O+) 2 = β2 (H3O+) 2β max = CWhere C is the total buffer concentrationExample:What is the maximum buffer capacity of an acetate buffer with a total concentration of 0.20 mol/l?= = 0.01
37 Solution properties pH pKa Weak acid, base Buffered solution Acid Base Ionization of waterpH scalesignificancepKaWeak acid, baseHenderson-Haslebach equationBuffered solutionDefinitionActionBuffer capacityDepend on ratio salt/acidDepend on concentration salt + acidDepend on amount of added a, bMaximum buffer capacityPreparationexamples
39 I think you are referring to buffers I think you are referring to buffers. At the point of half neutralisation, the acid and its conjugate base are at equal concentrations Ka= [H+][A-]/[HA] But when [HA]=[A-] half way through the reacion, this cancels down to Ka=[H+] => pKa=pH
40 Isoelectric pointThe isoelectric point (pI), sometimes abbreviated to IEP, is the pH at which a particular molecule or surface carries no net electrical charge.Amphoteric molecules called zwitterions contain both positive and negative charges depending on the functional groups present in the molecule. The net charge on the molecule is affected by pH of their surrounding environment and can become more positively or negatively charged due to the loss or gain of protons (H+). The pI is pH value at which the molecule carries no electrical charge or the negative and positive charges are equal.The pI value can affect the solubility of a molecule at a given pH. Such molecules have minimum solubility in water or salt solutions at the pH which corresponds to their pI and often precipitate out of solution.