Lec2ChemicalConcepts (1) (1) PDF

Important Chemical Concepts and a Basic Approach to Chemical Equilibrium Solutions and Their Concentrations. Molar Concentration Ex. 1. Calculate the molar concentration of ethanol in an aqueous solution that contains 2.30g of C2H5OH in 3.50L of solution. Because M =moles of solute/L of soluti...

Important Chemical Concepts and a Basic Approach to Chemical Equilibrium Solutions and Their Concentrations. Molar Concentration Ex. 1. Calculate the molar concentration of ethanol in an aqueous solution that contains 2.30g of C2H5OH in 3.50L of solution. Because M =moles of solute/L of solution, both the no. of moles of solute and the volume of solution will be needed. The volume (V) is given as 3.50 L, so all we need to do is convert the mass of ethanol to the corresponding moles. nC2H5OH = amountC2H5OH = 2.30g C2H5OH x 1mol C2H5OH 46.07 gC2H5OH = 0.04992 mol C2H5OH To obtain molar Conc. , we divide the no. of moles of ethanol by the volume. C C2H5OH = n C2H5OH = 0.04992 mol C2H5OH V 3.50 L = 0.0143mol C H OH/L or 2 5 0.0143 M Analytical Molarity Gives the total no. of moles of a solute in 1 L of the solution, or the total no. of millimoles in 1 mL. Analytical molarity specifies a recipe by which the solution can be prepared. Ex. A sulfuric acid solution that has an analytical conc. Of 1.0M can be prepared by dissolving 1 mol,or 98 g of H2SO4 in water and diluting to exactly 1.0 L. Equilibrium Molarity (Species Molarity) Expresses the molar conc. of a particular species in a solution at equilibrium. Ex. The species molarity of H2SO4 in a solution with an analytical concentration of 1.0 M is 0.0M because the sulfuric acid is entirely dissociated into a mixture of H3O+, HSO4- and SO42- ions; there are essentially no H2SO4 molecules in the solution. The equilibrium concentration and thus the species molarities of these three ions are 1.01, 0.99. and 0.01M respectively. Equilibrium M are usually symbolized by placing square brackets around the chemical formula for the specie; [H2SO4] = 0.00 M [H3O+] = 1.01 M [HSO4-] = 0.99 M [SO4 -2] = 0.01 M Some chemist distinguish species and analytical concentrations in different way. They use molar conc. for species conc. and formal concentration (F) for analytical conc. For analytical concentration. Formal conc. of H2SO4 is 1.0F whereas its equilibrium conc. is 0.0M The analytical molarity of H2SO4 is given by C H2SO4 = [SO4 -2] + [HSO4-] because these are the only two sulfate- containing species in the solution. Ex. Calc. the analytical and equilibrium molar of the solute species in an aqueous solution that contains 285 mg of trichloroacetic acid. Cl3CCOOH( 163.4 g/mol). In 10.0 mL Trichloroacetic acid is 73% ionized in water. Calculate the no. of moles of Cl3CCOOH. Which we designate as HA and divide the volume of the solution. amount HA = 285mgHA x 1g HA x 1 mol HA 1000mgHA 163.4 g HA = The molar analytical concentration, CHA is 1.744 x 10 -3 mol HA CHA = 1.744 x 10 -3 mol HA x 1000 mL = 10.0 mL 1L 0.174 mol HA/l or 0.174 M In this soln., 73% dissociates. Giving H+ and A- HA H+ + A- The species molarity is then 27% of CHA [HA] = CHA x (100 – 73)/100 = 0.174 x 0.27 = 0.047M The species molarity of A- is equalto 73% of the analytical conc. of HA [A-] = 73 mol A- x 0.174 mol HA = 0.127 M 100mol HA L Because one mole H+ is formed for each mol A- ,we can also write [H+] = [A-] = 0.127 M Ex. Describe the preparation of 2.00 L of 0.108 M BaCl2 from BaCl2. 2H2O (244g/mol). To determine the no. of grams of solute to be dissolved and diluted to 2.00 L, we note that 1 mol of the dihydrate yields 1 mol BaCl2. 2.00 L x 0.108 mol BaCl2.H20 = 0.216 mol BaCl2.H20 L Mass of BaCl2.H20 0.216 mol BaCl2.H20 x 244.3 g BaCl2.H20 = 52.8 g BaCl2.H20 mol BaCl2.H20 Dissolve 52.8 g of BaCl2.H20 in water and dilute to 2.00 L Describe the preparation of 500mL M Cl- solution from solid BaCl2.2H20 (244 g/mol) Mass BaCl2.2H20 = 0.0740 mol Cl- x 0.500 L x 1 mol BaCl2. 2H20 x 244.3g BaCl2.2H20 L 2 mol Cl- mol BaCl2.2H2O = 4.52 g BaCl2.2H2O Dissolve 4.52g of BaCl2.2H2O in water and dilute to 0.500L or 500 mL. Activity 4 : Molarity (Formality/Species) Using the template, solve for the following problems. 1. Calc. the analytical and equilibrium molar of the solute species in an aqueous solution that contains 58 g of trichloroacetic acid. Cl3CCOOH( 163.4 g/mol). In 10.0 mL Trichloroacetic acid is 835% ionized in water. 2. Describe the preparation of 3.5.00 L of 0.035 M BaCl2 from BaCl2. 2H2O (244g/mol). 3. Describe the preparation of 1000mL M Cl- solution from solid BaCl2.2H20 (244 g/mol) 4. Calculate the molar concentration of ethanol in an aqueous solution that contains 5.20g of C2H5OH in 7.00L of solution.