Lecture 4 Acid Base Titration PDF

 Lecture 4  by Dr/Roshdy saraya Analytical chemistry qualitative quantitative  Quantitative analysis Sample technique concentr...

 Lecture 4  by Dr/Roshdy saraya Analytical chemistry qualitative quantitative  Quantitative analysis Sample technique concentration 1-According to Sample concentration 2- According to technique: A-macro for 100 mg or more quantities A-volumetric Depends on the controlled addition of a reagent B-semi-micro for 10 – 100mg have known conc. to sample until the reaction C-micro less than 1mg complete B-gravimetric By forming ppt. and weight final product C-physieochemical or instrumental Depending on physical and chemical properties  Types of reactions used in volumetric analysis 1- Ionic combination reaction A- Neutralization reaction(acid base titration) a- formation of water H++ OH-→H2O b- displacment reaction i- formation of weak acid Salt of weak acid + strong acid → weak acid + salt of strong acid ii- formation of weak base Salt of weak base + Strong base →Weak base + salt of strong base B- formation of weakly ionizable salt(precipitation). C-Complex formation reaction(complexmetry) 2- Electron transfer reactions(redox)  Requirements of titrimetric reaction 1- Rapid 2- End point easily detected 3- Simple 4- Single 5- Standard titrant is available  Standard solutions They are solution of known concentrations thy are classified according to the type of concentration to Methods for expressing the standard solution concentration: 1. Weight to weight expressions; A) Weight% B) Mole fraction C) Molality. m 2. Weight to volume expressions; D) Molarity. M E) Normality. N Standard solutions Primary standard Secondary standard  standard solutions Primary standard chemicals: Primary standard chemicals are substances of definite known composition and high purity. Secondary standard chemicals: Secondary standard chemicals are substances which may be used for standardization and whose content have been found by comparison against primary standard. Primary standard chemicals 1. high purity and known composition 2. They must be easily tested for impurity. 3. They must be stable, (not absorbing water or CO2, not be efflorescent or volatile and withstand drying at 110-120º C) 4. must react with other substances quantitatively according to a balanced chemical equation. 5. They must be readily soluble in the solvent. 6. They should have high equivalent weight to minimize error due to incomplete transfer during weighing. 7. Inexpensive. Example: Na2CO3 Primary standard prepared by direct method HOW?  Secondary standard Are substance which don't fulfill the previous requirements of the primary standard It must be standardized against primary standard Example: HCl & NaOH Secondary standard prepared by in direct method HOW? Indirect method If the solute is secondary standard, it is used to prepare solution of approximate concentration. must be standardized against primary solution f = volume of exact standard / volume of approximate standard Where f is standardization factor  Calculation of the concentration of a sample during volumetric analysis 1. Find a relation between the analyzed sample and standard solution (balanced chemical equation) Na2CO3 + 2 HCI → H2O + CO2 + 2 NaCl 2. Find a relation between the strength of standard and the analyzed sample. 1ml 0.1 N HCI = 1 ml 0.1 N Na2CO3 3. Calculation of equivalent factor (F) 1 ml 0.1 N HCI = 0.0053 gm. Na2CO3 𝑴.𝑾𝒕 𝒐𝒇 𝒂𝒏𝒂𝒍𝒚𝒕𝒆 ×𝑵𝒐𝒓𝒎𝒂𝒍𝒊𝒕𝒚 𝒐𝒇 𝒕𝒊𝒕𝒓𝒂𝒏𝒕 Each 1ml of titrant= 𝒗𝒂𝒍𝒆𝒏𝒄𝒆 ×𝟏𝟎𝟎𝟎 4. calculate concentration of solution sample  Acid base titration by Dr/Roshdy saraya Acid base titration aqueous Non aqueous  Acid base titration in aqueous medium  IONIC THEORY OF SOLUTION Classification of aqueous solutions: 1. Electrolytes. 2. Non-electrolytes. Arrhenius "Ionization" / “Dissociation” For an electrolyte, there is a state of equilibrium between the non-dissociated molecules and the ions; the process being a reversible one. NaCl ⇌ Na+ + Cl- 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑒𝑑 α= 𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑏𝑒𝑓𝑜𝑟𝑒 𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑜𝑛 Law of mass action "The rate of a chemical reaction is proportional to the active masses of the reacting substances."  A+B ⇌ C+ D Vf= [A].[B]. Kf Vb = [C]. [D]. Kb At equillibrium Vf = Vb [A].[B]. Kf = [C]. [D]. Kb 𝑲= 𝑪 [𝑫] 𝑨 [𝑩]. K is called “Equilibrium constant” The Dissociation of Water H2O ⇌ H+ + OH– Kw = [H+] x [OH −] =10-14 “Ionic product of water” For pure water; [H+] = [OH–] = 10–7 “Neutral" Hydrogen ion exponent (pH): pH = –log [H+] *pH can be expressed by positive numbers (0 to14) pOH = –log [OH–] pKw = –log Kw pKw = pH + pOH = 14  A) Strong acids, having a high degree of dissociation B) Weak acids, which are feebly dissociated. Monobasic acids, which dissociate in one stage, Polybasic acids dissociate in consecutive stages. Acids and bases HCl ⇌ H+ + Cl- H2SO4⇌ H+ + HSO4– primary HSO4–⇌ H+ + SO42– Secondary H3PO4⇌ H+ + H2P O4− primary H2PO4−⇌ H+ + HPO42− Secondary HPO42−⇌ H+ + PO43− tertiary Acidity constant: HA ⇌ H+ + A− K = [H+] [A-] / [HA] K is called Dissociation, Ionization or Acidity constant. The stronger the acid, the larger its acidity constant. H2SO4 ⇌ H+ + HSO4− very large=1 HSO4− ⇌ H+ + SO42− 1.3 × 10−2  pH Calculations Strong acids and bases Are considered completely ionized in water ❑ Example 1: Calculate the [H+] and pH of 0.009 N hydrochloric acid? [H+] = 0.009 N pH = – log (0.009) = 2.05 ❑ Example 2: Calculate the pH values of a solution of sodium hydroxide whose [OH–] is 1.05 x 10–3? pOH = – (log 1.05 x 10–3) = 2.98 pH = 14 – 2.98 = 11.02 Example 3: Calculate the hydrogen ion concentration of a solution of pH 5.3? pH = – log [H+] [H+] = antilog-pH [H+] = 5.01 x 10–6 M Example 4: Calculate the hydroxyl ion concentration of a solution of pH 10.75 ? pOH = 14 – 10.75 = 3.25 [OH–] = the antilog of –3.25 [OH–] = 5.62 x 10–4 M  Weak acids and bases A. pH of solution of weak acids HA⇌ H+ + A – K = [H+] [A-] / [HA] Since [H+] = [A–] and if Ca is the total acid concentration, then [HA] =Ca – [H+] K = [H+]2 / Ca - [H+] [H+] is very small compared to Ca and can be neglected K = [H+]2 / Ca [H+]= √K.Ca pH = ½ (pKa + pCa) ❑ Example 1: Calculate the pH and [H+] of 0.10 N acetic acid (pKa= 4.76)? pH = ½ (pKa + pCa) = ½ x 4.76 + ½ (– log 0.1) = 2.38 + 0.5 = 2.88 ❑ Example 2: Calculate the pH and [H+] of a 0.0045 M solution of phenobarbital (pKa= 7.41)? PCa = – log (0.0045) = 2.35 pH = ½ (7.41 + 2.35) = 4.88 [H+] = antilog – 4.88 = 10-4.88 = 1.32 x 10–5 M B. pH of solution of weak bases pH =pKw - ½ (pKb + pCb) ✓ Example 3: Calculate the pH, and the [H+] of 0.13 N ammonia solution (pKb = 4.76)? pH = 14 – ½ (4.76) – ½ (– log 0.13) = 14 – ½ (4.76 + 0.89) = 11.18 log [H+] = 0.82x10– 12 [H+] = antilog 0.82 x 10–12 = 6.6 x 10–12 M ✓ Example 4: Calculate the pH and the [H+] of a 0.0037M solution of cocaine base (pKb = 5.59)? pH = 14.00 – ½ (5.59 + 2.43) = 9.99 [H+] = antilog of – 9.99 = 1.03 x 10–10  pH of salts (Hydrolysis) A. Salts of strong acids and strong bases Aqueous solution of strong salts is neutral KCl  K+ + Cl- K+ + OH- ⇌ KOH, and Cl- + H+ ⇌ HCl B. Salts of weak acids and strong bases CH3COONa  CH3COO− + Na+ H2O + CH3COO− ⇌ CH3COOH + OH− HAc is slightly ionized according to its acid constant Ka. This requires that some of the water molecules ionise to maintain Kw, constant; produces more OH− A solution of salt of a weak acid and a strong base is alkaline in reaction pH = ½ (pKw – pCs + pKa) The degree of hydrolysis, h, of a salt, is the fraction of the salt hydrolyzed when equilibrium is established h = y / Cs %h = y / Cs x 100  Example: Calculate the pH of a 0.165 M solution of sodium sulphathiazole (pKa = 7.12)? pH = ½ (pKw – pCs + pKa) pCs = – log 1.65 x 10–1 = 0.78 pH = ½ (14.00 – 0.78 + 7.12) = 10.17 Example: Calculate the pH, the [H+], the [OH–], and the degree of hydrolysis of a 0.1 M solution acetate (pKa = 4.76)? pH = ½ (pKw – pCs + pKa) pH = ½ (14.00 – 1.00 + 4.76) = 8.88 [H+] = 1.3 x 10–9 M [OH-] = 10–14/ 1.3x10–9 = 7.5x10–6 M h = 7.5 x 10–6/ 0.1 = 7.5 x 10–5 %h = 7.5 x 10–6 /0.1 x 100 = 0.007% [OH–] is 75 fold greater in this salt solution than in pure water C. Salts of weak bases and strong acids: NH4Cl  Cl− + NH4+ NH4++ H2O ⇌ NH4OH + H+ pH = ½ (pKw + pCs – pKb) ❑ Example1: Calculate the pH of a 0.025 M solution of ephedrine sulphate (pKb = 4.64)? pH = ½ (pKw + pCs – pKb) pCs = – log 2.5 x 10–2 = 1.58 pH = ½ (14.00 – 1.58 – 4.64) = 5.47 ❑ Example 2: Calculate the pH, the [H+] and the degree of hydrolysis of a 0.05 N solution of ammonium chloride? pH = ½ (pKw + pCs – pKb) pH = ½ (14.00 + 1.30 – 4.76) = 5.27 [H+] = 5.4 x 10–6 M h = 5.4 x 10–6 / 0.05 = 1.1 x 10–4 %h = 100 x (1.1 x 10–4) = 0.011 % D. Salts of weak acids and weak bases: Ex. Ammonium acetate, undergo hydrolysis in aqueous solutions. If the dissociation of the acid and the base are not widely different, the hydroxyl and hydrogen ions will be produced in approximately equal amounts pH = ½ pKw + ½ pKa – ½pKb ❑Calculate the pH, the [H+] of a 0.05 N solution of ammonium acetate? pH = ½ pKw + ½ pKa – ½pKb pH= ½ (14.00 + ½ 4.76 – ½ 4.76) pH= 7 BUFFER SOLUTIONS  BU FFER SOLU TION S Buffers are mixtures of compounds in solution which resist changes in pH caused by addition of small amounts of acid or base; or upon dilution. Types of buffers: 1. Acidic buffers: prepared by dissolving a weak acid [HA] and its salt [A–] in water. Upon addition of acids… H+ + A– ⇌ HA Addition of bases… OH– + HA ⇌ A– + H2O 2. Basic buffers: prepared from a weak base [BOH] and its salt [B+]. Upon addition of acids… H+ + BOH ⇌ B+ + H2O Addition of bases… OH– + B+ ⇌ BOH  The Henderson-Hasselblach equation For acidic buffer: Whatever the relative proportions of the acid and its salt in the solution, their concentrations must always be related to acidity constant. Ka = [H+] [A-] / [HA] By taking logarithm of both sides -pka = -pH + log [A-] / [HA] pH = pKa + log [salt] / [Acid] ❑ Example: Calculate the pH of a solution containing 0.1N acetic acid and 0.1N sodium acetate? pH = 4.76 + log 0.1/0.1 = 4.76 Compare to pH of 0.1N acetic acid!! pH = ½ (pKa + pCa) = ½ (4.76 + 1) = 2.88 ❑ Example : Calculate the pH of the solution produced by adding 10.0 ml of 1N HCl to 1 liter of solution which is 0.1N in acetic acid and 0.1N in sodium acetate (Ka= 1.82 x 10-5)? Neglecting the volume change from 1000 to 1010 ml H+ + CH3COO− ⇌ CH3COOH [CH3COO−] = 0.1 – 0.01 = 0.09 [CH3COOH]= 0.1 + 0.01 = 0.11 pH = pKa + log [salt] / [Acid] pH = 4.74 + log0.09/0.11 = 4.74 – 0.09 = 4.65 ❑ Example :Calculate the pH of the solution produced by adding 20.0 ml of 2N NaOH to 1 liter of solution which is 0.1N in acetic acid and 0.1N in sodium acetate (Ka= 1.82 x 10-5)? Neglecting the volume change from 1000 to 1020 ml NaOH + CH3COOH ⇌ CH3COONa + H2O [CH3COOH] = 0.1 – (20×2/1000) = [CH3COOH] = 0.1 – 0.04 = 0.06 [CH3COO−]= 0.1 + (20×2/1000)= [CH3COO−]= 0.1+ 0.04 = 0.14 pH = pKa + log [salt] / [Acid] pH = 4.74 + log 0.14/0.06 = 4.74 + 0.37= 5.11  For basic buffers: pH = pKw – pKb – log [B-]/ [BOH] pH = pKw – pKb – log [Salt] / [Base] ❑ Example: Calculate the pH of a solution containing 0.07N ammonia, and 0.28N ammonium chloride? pH = 14.00 – 4.75 – log0.28/0.07 = 8.64 Compare to pH of 0.07N ammonia!! pH =pKw - ½ (pKb + pCb) pH = 14.00 - ½(4.76 + 1.16) = 11.04 ❑ Example : Calculate the pH of the solution produced by adding 5.0 ml of 2N HCl to 500 ml of solution which is 0.5N ammonia, and 0.5N ammonium chloride (Kb= 4.75)? HCl + NH4OH ⇌ NH4Cl + H2O NH4OH= 0.5- (5×2/500)= 0.5-0.02= 0.48 NH4Cl= 0.5 + (5×2/500)= 0.52 pH = pKw – pKb – log [Salt] / [Base] pH = 14 – 4.75 – log [0.52] / [0.48]=14 – 4.75 -2.03= 7.22  Properties of buffer mixtures Effect of temperature: No effect on acidic buffers, but affect most basic buffers. Why? Kw changes significantly with temperature. Effect of dilution: Dilution should have no effect on pH. Why? pH depends on the ratio of the molar concentrations of the two compounds in a buffer solution. This ratio remains constant on dilution  The number gram equivalents of strong acid or strong base required to change the pH of one liter of buffer solution by one unit. Buffer 1. The higher the concentration Capacity of the two components of the buffer solution, the higher is the buffer capacity. 2. Buffer capacity is a maximum when the two components are present in equal concentrations,  Example: Calculate the buffering capacity of a solution containing 0.1 g equivalent of sodium acetate and acetic acid? PKa = 4.76 pH = 4.76 + log 0.1/0.1 = 4.76 To increase (or decrease) the pH by one unit, the ratio salt/acid will have to alter by a factor of 10: 5.76 = 4.76 + log [salt] / [Acid] If log [salt] / [Acid] = 1; therefore [salt] / [Acid] = 10 Suppose that the amount of base added (X gram equivalents). So; [salt] = (0.1+X) and [acid ] = (0.1-X) (0.1+X) / (0.1-X) = 10 X=0.082 The buffering capacity of the buffer mixture is therefore (0.082 gram equivalent). i.e. 0.082 gm eq. of strong base is required to increase pH of the buffer by 1 unit Thank you