12.1 Expressing Concentration The Amount of Solute in the solvent Dr. Fred Omega Garces Chemistry 201 Miramar College

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.

9 Types of Solution (derived from 3 phases) Solute Solvent Solution gas O2 in gas N2 Air gas CO2 in Liquid H2O Carbonated Water liquid H2O in Gas Air Fog liquid EtOH in Wine Liquid Hg in Solid Ag Dental-filling Solid NaCl in Brine Solid Ag in Solid Au 14 Karat gold

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?

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 (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.

Concentration Relationship moles mass  m %m M* } Solute } Solvent Molecular Weight Molc’ Wt MassSolution Density Solution VolSolution N Equivalence/mol * Volume of solution must be used and not just volume of solvent

Concentration Relationship * Volume of solution must be used and not just volume of solvent

Calculating molality (m): Example 3.5 g of CoCl2 (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?

Calculating molality (m): Example 3.5 g of CoCl2 (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 molality concentration 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

Concentration by Parts Solute (mass or volume) Solution (mass or volume) x multiplier % Concentration w/w = Wt Solute• g •100  % (pph) Wt Soln g w/v = Wt Solute• g • 100  % (pph) Vol Soln ml v/v = Vol Solute• ml • 100  % (pph) ppm & ppb (For dilute solution) m/m = mass Solute • g •106  ppm (ppm) mass Soln g v/v = Vol Solute• ml • 109  ppb (ppb) Vol Soln ml

Calculating % Concentration: Example 3.5 g of CoCl2 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 95 g Solution = 3.7% (m/m) x 100 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) ppm (v/v)= (0.05 mL/1000 mL)•106 = 50 ppm Answer 3.7% m/v CoCl2

% Concentration: % Mass Example Example#2: What is the v/v (%) & (ppm) concentration if one drop of alcohol (1/20 ml = 0.05 ml) is added to 1.00L? Anws: %(v/v) = (0.05mL/1000mL)•100 = 0.005 % (pph) ppm(v/v)= (0.05mL/1000mL)•106 = 50 ppm Example#3: The legal BAC in California cannot exceed .080% of a person’s blood by volume. How much alcohol (oz & mL) if a 160lb person has 4.7 L blood and has 0.12% BAC. 1 L = 1000ml, 29.57 mL = 1.0 oz 5.64 mL or 0.191 oz Example#1 3.5 g of CuSO4 is dissolved in 100mL solution. Assume the density of the solution is 1.0 g/mL, what is concentration of the solution in % mass? %m = 3.5 g CuSO4 100g Solution = 3.5% (m/m)

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 1mi3 volume ? Ans: ppt= (0.05mL/4.17•1015ml)•109 = 0.000000012ppb = 0.000012ppt *Quintillion (1 •1018) = 12 ppquintlllion

Expressing Normality Normality: Equivalent: 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+ H2SO4: 1 mol H2SO4 = 2 eq H+ H3PO4: 1 mol H3PO4 = 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 H2SO4 = 5mol H2SO4 • 2eq = 10.0 N H2SO 1 L 1mol

Interconverting Concentration: A Calculation Example A perchloric acid (HClO4 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? Answer 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 H2O molality = 1.11 m ® 9.95•10-2 mole ® 5.00 mol H2O cA = .0195

Interconverting Concentration: More Examples If sodium thiosulfate (Na2S2O3, 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 cK2CO3 = 0.0545 solution which contains exactly 750.5 g of water? MWK2CO3 138.21 g/mol

...and ever more Examples Example#8: 50.00ml of ethylene glycol ( = 1.114 g/mL; MW = 62.07 g/mol) is added to 1.000-L water ( = 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.

Misc Problems Answers Harris 7th ed p18 1. The density of 70.5 wt% aqueous perchloric acid, HClO4, 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 HClO4 are in 1.000L? 1180 g (c) How many moles of HClO4 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, C6H12O6) 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•102 m and and average depth of 10.0 m. (V = p r2 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 H2SO4 are required to react with 4.35 g of solid containing 23.2 m:m% Ba(NO3)2 if the reaction produces BaSO4 precipitate. 133, 1.29 mL

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