Basics of Analytical Chemistry PDF

Basics of Analytical Chemistry (chm201, chm219B) Dr Ibrahem M A Hasan  Examples Example 1: How many mL of 0.2 N hydrochloric acid are required to neutralize 25.0 mL, 0.1 N sodium hydroxide? Solution: HCl NaOH VA x NA = VB x NB...

Basics of Analytical Chemistry (chm201, chm219B) Dr Ibrahem M A Hasan  Examples Example 1: How many mL of 0.2 N hydrochloric acid are required to neutralize 25.0 mL, 0.1 N sodium hydroxide? Solution: HCl NaOH VA x NA = VB x NB VA x 0.2 = 25.0 x 0.1 Example 2: 25 mL of a ferrous sulfate solution react completely with 30 ml of 0.125 M KMnO4. Calculate the strength of the ferrous sulfate solution in g/L. A molar solution of FeSO4 contains 1 mol/L or 151.90 (MW) g/L. Solution: FeSO4 KMnO4 m (g) 25 x MA = 30 x 0.125 M n (mol) (g/mol) MA = 30 x 0.125/25 = 0.150 M Conc (g/L) = M x molar mass = 0.150 mol/L x 151.90 g/mol = 22.78 g/L Example 3: What volume of 0.127 N reagent is required for the preparation of 1000 mL of 0.1 N solution. Solution: (before dilution) VA x NA = VB x NB (after dilution) VA x 0.127 = 1000 x 0.1 VA = (1000 mL x 0.1 N)/0.127 N = 787.4 mL 2  The Equipment Volumetric analysis involves a few pieces of equipment: Pipette: for measuring accurate and precise volumes of solutions Burette: for pouring measured volumes of solutions Conical flask: for mixing two solutions Wash bottle: these contain distilled water for cleaning equipment Funnel: for transfer of liquids without spilling Volumetric flask: a flask used to make up accurate volumes for solutions of known concentration  Requirements of primary standard substances 1. It must be easy to obtain, to purify, to dry (preferably at 110-120oC) and preserve in pure state. 2. It should not be altered in air during weighing (it should not be hygroscopic, oxidized by air, affected by carbon dioxide). 3. It should be capable of being tested for impurities. 4. It should have a high equivalent weight so that the weighing errors may be negligible. 5. It should be readily soluble under the conditions in which it is employed. Examples: Sodium carbonate, borax, potassium hydrogen phthalate, and sodium chloride. 4 Preparation of standard solutions Weigh out an equivalent weight, or a definite fraction or multiple of pure reagent. Dissolve in the least volume of solvent, usually water. Transfer the solution into a volumetric flask. Complete the volume till the mark with solvent and shake well. 1.0 M NaCl solution Add water to Weigh out dissolve the 1 mole (58.45 g) NaCl, then of NaCl and add add water to Swirl to mix. it to a 1.0 L the mark. volumetric flask. Step 3 5 Step 1 Step 2  Theory of acid-base titration General conception of acids and bases 1. Arrhenius theory Arrhenius acid: solution contains an excess of H+ ions. Arrhenius base: solution contains an excess of OH- ions. H+ + OH− → H2O 6  2. Bronsted-Lowry theory Bronsted-Lowry acid: is a proton (H+) donor. Bronsted-Lowry base: is a proton (H+) acceptor. 7  proton proton acceptor Bronsted-Lowry Acid Bronsted-Lowry donor Base HCl(g) + H2O(l) → H3O+(aq) + Cl-(aq) 8 ahydrogen hydronium ionion ishydrogen ion does formed not existcombines with water in water 9  Conjugate acid-base pairs differ by a proton. When an acid donates a proton it becomes the conjugate base. HCl(g) → Cl-(aq) acid base 10  Conjugate acid-base pairs differ by a proton. When a base accepts a proton it becomes the conjugate acid. H2O (l) → H3O+(aq) base acid 11  Conjugate acid-base pairs differ by a proton. HCl(g) + H2O (l) → Cl-(aq) + H3O+(aq) acid base base acid acid + base conjugate base + conjugate acid 12  3. Lewis theory Lewis acid: an electron-pair acceptor. Lewis base: an electron-pair donor. 14  Electron Pair Lewis Acid Acceptor Electron pair donated to H+ Lewis Base Electron Pair Donor 15  Electron Pair Lewis Acid Acceptor Electron pair donated to B Lewis Base Electron Pair Donor 16  Equivalent weight Equivalent weight of an acid: is its weight which contains one replaceable hydrogen (proton); i.e.; 1.008 g of hydrogen (= MW/basicity). The equivalent weight of a monobasic acid (HCl, HBr, HI) is the same as its molecular weight (MW). A normal soln of a monobasic acid will therefore contain 1 MW (or 1 mole) in a liter of soln. The equivalent weight of a dibasic acid (H2SO4) is 1/2 its MW. The equivalent weight of a tribasic acid (H3PO4) is 1/3 its MW. 17  Equivalent weight The equivalent weight of a base: is its weight which contains one replaceable hydroxyl group; i.e.; 17.008 g of ionizable Hydroxyl (= MW/acidity). 17.008 g of hydroxyl are equivalent to 1.008 g of hydrogen. The equivalent weights of sodium hydroxide and calcium hydroxide are 1 mole and 1/2 mole, respectively. 18  Equivalent weight Equivalent weight of a compound: Equivalent weight of NaHCO3 = (1x23 + 1x1 + 1x12 + 3X16)/1 = 84. Salts of strong bases and weak acids possess alkaline reactions in aqueous solution because of hydrolysis. Sodium carbonate, with methyl orange as indicator, reacts with 2 moles of HCl to form 2 moles of NaCl, hence its equivalent weight is 1/2 MW. Na2CO3 + 2HCl → 2NaCl + CO2 + H2O Borax, under similar conditions, also reacts with 2 moles of HCl and its equivalent weight is 1/2 MW. Na2B4O7 + 2HCl + 5H2O → 2NaCl + 4H3BO3 19 Hydrogen-ion concentration and the pH scale Self ionization of water Water ionizes according to the equation H 2 O + H 2O OH - + H 3O+ Acid Base Conjugate Conjugate base of H 2O Acid of H 2O [H3O+][OH-] The equilibrium expression is Keq = [H2O]2 The concentration of water is constant since the degree of self ionization is slight Kw = Keq[H2O]2 = [H3O+][OH-] In pure water: [H3O+] = [OH-] = 1.0 x 10-7 Thus, -14 Kw = 1.0 x 10  pH and pOH Because hydronium ion concentrations for most solutions are numbers with negative exponents, these concentrations are more commonly expressed as pH pX = -log X pH = -log[H+] pOH = -log[OH-]  Kw = 1.0 x 10-14 Kw = [H+] [OH-] = 1x10-14 -log Kw = -log [H+] + -log [OH-] = -log (1x10-14) -14 since pKw = -logKw = -log 1.0 x 10 pH + pOH = 14 Examples: (1) The [H3O+] of a certain liquid detergent is 1.4 x 10-9 M. What is its pH. (2) The pH of black coffee is 5.3. What is its [H3O+]. (3) Calculate the pH of a 0.01 N solution of acetic acid (the degree of dissociation is 12.5 %). Solution: (1) pH = -log[H+] pH = -log[1.4 x 10-9] = 8.85 (2) [H3O+] = antilog-5.3 = 5 x 10-6 OR [H3O+] =10-pH (3) [H3O+] = 0.125 x 0.01 = 1.25 x 10-3 pH = -log[H3O+] = -log(1.25 x 10-3) = 2.903 Calculating [H+] or pH from ionization constants Henderson-Hasselbalch equation For a weak acid: HA(aq) H+(aq) + A-(aq) [ H + ][ A − ] Ka = [ HA ] [H+] = Ka([HA]/[A–]) pH = pKa - log([HA]/[A–]) pH = pKa + log([A–]/[HA]) pH = pKa + log([salt]/[acid]) Solutions with the same [A–]/[HA] ratio have same pH. Calculating of [H+] or pH from ionization constants During acid-base titration: HA(aq) + B-(aq) ⇌ H+(aq) + AB(aq) Acid base salt Ka = [H+][AB]/[HA] [H+] = Ka ([HA]/[AB]) = Ka ([acid not yet titrated]/[Salt already formed]) pH = pKa - log([HA]/[AB]) Concentrations are in moles/L; M. The molar concentration of the base, is equal to the moles of base added divided by the volume in liters. This is identical with the salt concentration since the base is converted quantitatively into the corresponding salt prior to the end point. 26