12.1 Expressing Concentration The Amount of Solute in the solvent Dr. Fred Omega Garces Chemistry 201 Miramar College 1 Expressing Concentration
Components of Solution Mixtures: Variable components, retains properties of its component. Homogeneous systems: Solutions Solution - Homogeneous mixture of two or more substances Components of solution Solute - Substance being dissolve Solvent - Substance in which solute is dissolved in. If solvent is water, then solution is considered aqueous. 2 Expressing Concentration
9 Types of Solution (derived from 3 phases) Solute Solvent Solution gas O 2 in gas N 2 Air gas CO 2 in Liquid H 2 O Carbonated Water liquid H 2 O in Gas Air Fog liquid EtOH in Liquid H 2 O Wine Liquid Hg in Solid Ag Dental-filling Solid NaCl in Liquid H 2 O Brine Solid Ag in Solid Au 14 Karat gold 3 Expressing Concentration
Solution & Concentration Concentration - Proportion of substance in mixture or amount of solute in given amount of solvent. Most common type: Molarity (M) Molarity - moles solute / Liter solution Example: What mass (g) of copper(ii)sulfate pentahydrate (250. g/mol) is needed to make 1.00-L of 0.10M solution? MW CuSO4 5H 2 O = 249.6 g mol Mass CuSO4 5H 2 O = 0.10 mol L 249.6 g 1 mol 1 L Mass CuSO4 5H 2 O = 24.96 or 25.0 g 4 Expressing Concentration
Expressing Concentration 5 ways of expressing concentration- Molarity (M) - moles solute / Liter solution Molality * (m) - moles solute / Kg solvent Conc. by parts (% m)- (solute [mass] / solution [mass]) * 100 w/v [mass solute (g) / volume solution (ml)] * 100 v/v [vol solute (ml) / vol solution (ml)] * 100 mole fraction ( χ A) - moles solute / Total moles solution Normality (N) - Number of equivalent / Liter solution Note that molality is the only concentration unit in which denominator contains only solvent information rather than solution. It is also a concentration unit together with mole fraction that are temperature independent. 6 Expressing Concentration
Concentration Relationship! Molecular Weight moles mass } Solute moles Molc Wt mass } Solvent χ Mass Solution m %m Density Solution M* Vol Solution Equivalence/mol N * Volume of solution must be used and not just volume of solvent 7 Expressing Concentration
Concentration Relationship! * Volume of solution must be used and not just volume of solvent 8 Expressing Concentration
Calculating molality (m): Example Example # 1 3.5 g of CoCl 2 (129.93 g/mol) is dissolved in water to produce 100-ml solution. Assume the density of the solution is 0.95g/mL, what is molal concentration of the solution? 9 Expressing Concentration
Calculating molality (m): Example Example # 1 3.5 g of CoCl 2 (129.93 g/mol) is dissolved in water to produce 100-ml solution. Assume the density of the solution is 0.95 g/ml, what is concentration the molality of the solution? mol CoCl2 = 3.5 g CoCl2 mol CoCl2 129.93 g = 0.0269 mol = 0.027 mol mass H2O = (100ml 0.95 g) - 3.5 g 1 ml = 91.5 g m = 3.5 g CoCl2 mol CoCl2 1 1 129.93 g 0.0915kg = 0.294 m = 0.29 molal 10 Expressing Concentration
% Concentration Concentration by Parts Solute (mass or volume) Solution (mass or volume) x multiplier w/w = Wt Solute g 100 g % (pph) Wt Soln g w/v = Wt Solute g 100 g % (pph) Vol Soln ml v/v = Vol Solute ml 100 g % (pph) Vol Soln ml ppm & ppb (For dilute solution) m/m = mass Solute g 10 6 g ppm (ppm) mass Soln g v/v = Vol Solute ml 10 9 g ppb (ppb) Vol Soln ml 11 Expressing Concentration
Calculating % Concentration: Example Example # 2 3.5 g of CoCl 2 is dissolved in a solvent to form a 100-ml solution. Assume the density of the solution is 0.95 g/ml, what is concentration of the solution in % mass? %m = 3.5 g CoCl2 x 100 95 g Solution = 3.7% (m/m) 3.7% m/v CoCl 2 Example #3 What is the v/v concentration if one drop of alcohol (1/20 ml = 0.050 ml) is added to 1.00L? % or ppm Ans: %(v/v) = (0.05mL/1000mL) 100 = 0.005 % (pph) Answer ppm (v/v)= (0.05 ml/1000 ml) 10 6 = 50 ppm 12 Expressing Concentration
Shark Sense pp(?): Making Chem Relevant Example#4 A shark can smell blood in water from several miles away. What is the concentration of 1 drop blood in 1mi 3 volume? Ans: ppt= (0.05mL/4.17 10 15 ml) 10 9 = 0.000000012ppb = 0.000012ppt *Quintillion (1 10 18 ) = 12 ppquintlllion 14 Expressing Concentration
Normality: Expressing Normality The number of equivalents of acid or base solute in one litter of solution. Equivalent: The number of specie (acid or base) that gives one mole of charge. Note: HCl : 1mol HCl = 1eq H + H 2 SO 4 : 1 mol H 2 SO 4 = 2 eq H + H 3 PO 4 : 1 mol H 3 PO 4 = 3 eq H + Mg +2 : 1 mol Mg +2 = 2 eq charge NaOH: 1 mol NaOH = 1 eq OH - Therefore: 5.0 M HCl = 5mol HCl 1eq = 5.0 N HCl 1 L 1mol and 5.0 M H 2 SO 4 = 5mol H 2 SO 4 2eq = 10.0 N H 2 SO 1 L 1mol 15 Expressing Concentration
Interconverting Concentration: A Calculation Example Example#5: A perchloric acid (HClO 4 MWt = 100.5 g/mol) solution is 10.0 % m/m. The density of the solution is 1.060 g/cc. What is the Molarity, molality, mole fraction and Normality of the solution? 10.00 g 9.95 10-2 mole 100 g solution 94.34 cc Molarity = 1.05 M 9.95 10-2 mole 0.090 Kg H 2 O Answer molality = 1.11 m 9.95 10-2 mole 5.00 mol H 2 O χa =.0195 16 Expressing Concentration
MW Na 2 S 2 O 3 = 158.11 g/mol vol = 800.0 ml H 2 O 1ml m = 0.0245 mol/kg solvent mol mol Na 2 S 2 O 3 = 0.0245 kg H 2 O 0.800 kg H O = 0.0196 mol Na S O 2 2 2 3 = 1 mol = 3.099 g Na 2 S 2 O 3 = 4.41 10-4 χ K2 = 0.0545 mass H CO 2 O = 750.5 g H 2 O 3 MW K2 = 138.21 g/mol CO 3 mol H 2 O = 750.5 g H 2 O 1 mol 18.0 g = 41.69 mol H O 2 χ K2 = 0.0545 = CO 3 x x + 41.69 mol H 2 O 0.0545 (x + 41.69 mol H 2 O) = x 0.0545x + 2.27235 = x 2.272235 =.9455 x x = 2.4032 mass K2 = 2.4032mol 138.21 g = 332.15 g CO 3 1 mol χ K2 CO 3= % m= 2.4032 mol 41.69 mol H 2 O + 2.4032 mol = 0.0545 332.15 g 332.15 g + 750.5 g H 2 O 100 = 30.68% Interconverting Concentration: More Examples Example#6: If sodium thiosulfate (Na 2 S 2 O 3, MW = 158.11 g/mol) solution (0.0245 molal) contains 800. ml water, what is the mole fraction and the mass of the solute for this solution? Example#7: What is the % mass and the molality of a χk 2 CO 3 = 0.0545 solution which contains exactly 750.5 g of water? MW K 2CO3 138.21 g/mol mol H 2 O = 800.0 ml 1g 1 mol 18.0 g = 44.44 mol H O 2 massna 2 S 2 O 3 = 0.0196 mol Na 2 S 2 O 3 158.11 g mol fraction χ Na2 S 2 O 3 0.0196 mol Na 2 S 2 O 3 0.0196 mol Na 2 S 2 O 3 + 44.44 mol H 2 O = 0.000441 18 Expressing Concentration
...and ever more Examples Example#8: 50.00ml of ethylene glycol (ρ = 1.114 g/ml; MW = 62.07 g/mol) is added to 1.000-L water (ρ = 1.00 g/ml) at 20 C. Answer the following questions and assume additive volumes. i) What is the density of the mixture ii) Calculate the % mass of the ethylene glycol in the solution. iii) Calculate the molarity and molality of ethylene glycol in the solution. Mass H 2 O = 1000 g vol = 50.0 ml ethylene Glycol 50.00 ml 1.114 g ml D = vol = 1000 ml H 2 O = 55.70 g mass H2O + mass ethylene glycol vol H2O +vol ethylene glycol D = 1055.70 g 1050.0 ml = 1.0054 = 1.005 g 19 Expressing Concentration ml % m = 55.70 g 1055.7 g mol glycol, 55.70 g 0.89737 mol mol H 2 O, 1000 g 55.56 mol molality = Molarity = 100 = 5.276 %.89737 mol 1.00kg.89737 mol 1.050 L = 0.8974 m = 0.8546 M
Misc Problems Answers Harris 7 th ed p18 1. The density of 70.5 wt% aqueous perchloric acid, HClO 4, is 1.67 g/ml. 1.20 (a) How many grams of solution are in 1.000 L 1670 g (b) How many grams of HClO 4 are in 1.000L? 1180 g (c) How many moles of HClO 4 in 1.000L? 11.7 mol 2. An aqueous solution containing 20.0% wt% KI has a density of 1.168 g/ml. Find the molality, mole fraction, and molarity of the KI solution. 1.21 1.51 m 3. The concentration of sugar (glucose, C 6 H 12 O 6 ) in human blood ranges from about 80mg/100mL before meal to 120mg/100mL after eating. Find the molarity before and after eating. 1.22 4.4e-3M, 6.7e-3M 4. It is recommended that drinking water contain 1.6 ppm fluoride (F-) for preventing of tooth decay. Consider a reservoir with a diameter of 4.50 10 2 m and and average depth of 10.0 m. (V = π r 2 h) How many grams of fluoride should be added to give 1.6 ppm? How many grams of sodium fluoride, NaF contains this much fluoride? 1.25, 2.5e6 g F-, 5.6e6 g NaF 5. How many ml of 3.00 M H 2 SO 4 are required to react with 4.35 g of solid containing 23.2 m:m% Ba(NO 3 ) 2 if the reaction produces BaSO 4 precipitate. 133, 1.29 ml 20 Expressing Concentration
Solution at a Glance Solutions can be describe by the following: Solvent The component of a solution present in the greatest quantity Solute The component of solution present in the lesser quantity Solution A homogeneous mixture of two or more substances in which each substance retains its chemical identity Concentration of a Solution The amount of solute in a specific amount of solution. Molarity (M) moles of solute Liters of solution 22 Expressing Concentration