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Analytical Chemistry (Volummetric analysis) Lecture 2 Prepared by Dr.Ahmed Bahgat 1.Which of the following is a 2.Calculate the mole weak acid? fraction of C2H5OH in a (a) H2SO4 solution that contains 46 (b) HClO3...

Analytical Chemistry (Volummetric analysis) Lecture 2 Prepared by Dr.Ahmed Bahgat 1.Which of the following is a 2.Calculate the mole weak acid? fraction of C2H5OH in a (a) H2SO4 solution that contains 46 (b) HClO3 grams of ethanol, C2H5OH, (c) HF and 64 grams of methanol, (d) HCl CH3OH. (a) 1/3 (b)0.42 (c)½ (d)2/3 3.A 5.2 molal aqueous 4.The molal solution is solution of methyl alcohol, that contain 1 mole of CH3OH is supplied. What is solute per the mole fraction of methyl (a)1 kg of solvent alcohol in the solution? (b)1 kg of solution (a)0.050 (c)1 L of solvent (b)1.100 (d)1 L of solution (c)0.190 (d)0.086 5.Which of the following 6. A 2 M H2SO4 solution will represent ppm have a (a) µg/L Normality of (b) mg/L (a) 1 (c) g/L (b) 2 General problems 1. A solution of NaCl was prepared by dissolving 5g of it in 500 ml of distilled water. Express the concentration by g/l, ppm and ppb units. 2. Express the concentration of 200 g solution containing 25g of Na2SO4 by 24 (w/w%). 3. Calculate the percentage by weight, for solution prepared by dissolving 5g AgNO3 in 100 ml of water (density=1g/ml). 4. Calculate the percentage by volume for a solution prepared by adding 50 ml methanol to 200 ml of water. 5. Calculate the molarity of solution prepared by dissolving 4g NaOH in 500ml of water. 6. Calculate the normality of HCl contain 4.01 g/l. 7. Calculate the molality of sodium hydroxide solution contain 8g NaOH in 500 g of water. 8. A solution contains 5 mole of ethyl alcohol and 6 moles of water. Acids and Bases and Their Reactions Definitions -Arrhenius definition for Acids and Bases Acids are H+ donors Bases are OH- donors Broadened Definition for Acids and Bases Acids increase H+ concentration or [H+] increases Bases increase H+ acceptor Arrhenius 1903 Nobel Prize  ARRHENIUS THEORY for ACIDS and BASES - Arrhenius acid: is a substance that increases the concentration of the hydronium ion, H3O+, when dissolved in water. - HCl H O + Cl 3 + - - Arrhenius base: is a substance that dissociates into hydroxide ions OH- and cations in solution. NaOH Na+ + OH- Problems with Arrhenius Theory * H3 O+ Hydronium ion rather than H+ * OH(H2O)3- presents in solution, not OH- * Other substances also have acidic or basic properties Brønsted-Lowry theory of acids and bases: A Brønsted-Lowry acid (or simply Brønsted acid): is a species that donates a protons HA  H+ + A- Brønsted-Lowry base: is the substance that can accept a proton. B + H+ BH+ Brønsted-Lowry acid-base theory has several advantages over Arrhenius theory: ex acetic acids -CH3COOH acts as an Arrhenius acid because it acts as a source of H3O+ when dissolved in water, and it acts as a Brønsted acid by donating a proton to water. -In the second example CH3COOH undergoes the same transformation, in this case donating a proton to ammonia (NH3), but cannot be described using the Arrhenius definition of an acid because the reaction does not produce hydronium Lewis-Lowry theory of acids and bases: Lewis acid: a chemical compound which have tendency to accept an electron pair or more from a donor compound. Lewis base: a chemical compound which have tendency to donate an electron pair to an acceptor compound. Trimethylborane (Me3B) The Leveling Effect: The level of acidity governed by acid strength resulted from the dissociation of acids in solvents. -If HB > HS+, so HB will transfer proton to the solvent, the equilibrium will lie toward the right. HB + S HS + + B- (equilibrium sifted to right) -If HB >>>> HS+ (equilibrium sifted to right, 100% dissociation of HB) The effect of Solvent on the dissociation of acids Aqueous solvent Less basic solvent (H2O) ( Acetic acid) All acids are equally strong Acids are not leveled and HClO4 HNO3 or HCl or HClO4 Is stronger than the other. Acid strength: The general reaction that occurs when an acid is dissolved in water can be best represented as HA(aq) + H2O H3O+(aq) + A-(aq) Acid Base Conjugate Conjugate acid base Ka = [ H3O+][A-] = [H+][A-] [HA] [HA] In this reaction, water as a polar molecule pulls the protons from the acid thus it acts as a base. Note that, the conjugate base is everything that remains of the acid molecule after a proton is lost , while a conjugate acid is formed when the proton is transferred to the base. Acid Strength: The strength of an acid is defined by the equilibrium position of its dissociation reaction: HA(aq) + H2O → H3O+(aq) + A-(aq) A strong acid is one for which this equilibrium lies far to the right. This means that almost all the original HA is dissociated (ionized). E.g., Hydrochloric acid (HCl) is a strong acid HCl (aq) → H+ (aq) + Cl- (aq) (reaction essentially complete)  Strong Acids A strong acid: is one that dissociates completely in water to produce H+(aq). Weak acids : A weak acid is one for which the equilibrium lies far to the left. A weak acid dissociates only to a very small extent. A weak acid has a strong conjugate base. Examples: phosphoric acid (H3PO4), acetic acid (CH3COOH), benzoic acid (C6H5COOH).  Graphic representations of strong and weak acid equilibria Strong Acid: 100% A strong acid: equilibrium Dissociation lies far to the right into ions HA + H2O →H3O+ + A– A weak acid: equilibrium lies far Weak Acid: to the left Very little HA + H2O ← H3O+ + A – Dissociation A weak acid yields a relatively into ions strong conjugate base The Calculation of pH for Strong Acids pH = -log [H3O+] = -log [H+] Calculate the pH of 0.10M solution of HCl. HCl → H+ + Cl- 0.1M 0.1 0.1 pH = -log[0.1] = 1 pH + pOH = pKw = 14 [OH-] = Kw / [H+] = 10-14/10-1 = 10-13 Calculate the pH of 0.10M solution of H2SO4. H2SO4 → 2H+ + SO42- 0.1M 2×0.1 0.1 pH = -log[0.2 ] = ---  Calculate the pH of Strong Acid-Base Solutions EXAMPLE 7: -Calculate the pH (at 25oC) of an aqueous solution that has an OH- (aq) concentration of 1.2 x 10- 6 M (i.e., mol/liter). Solution - The concentration of H+ (aq) is [H+][OH-] = KW [H+] = KW/[OH-] = 10-14/1.2x10-6 = 8.3 x 10-9 pH = -log[8.3 x 10-9] = 8.1  Calculating pH of weak acid solutions Weak acids dissociates partially (