Concentration of Solutions - Comprehensive Notes

Concentration of Solutions - Solutions can be described qualitatively or quantitatively depending on solute relative to solvent. - Qualitative descriptions: dilute (small amount of solute) vs concentrated (large amount of solute). - Examples: a solution with 5 g of salt in 100 mL of water is considered diluted; 40 g of salt in the same volume is concentrated. - In industries and laboratories, quantitative measures of concentration are necessary. - Common quantitative expressions of concentration include: percent by mass, mole fraction, molarity, molality, percent by volume, and ppm. - General definitions and relationships - Mass-based concentration reference: - Mass of solution = mass of solute + mass of solvent. - \%\text{ by mass} = m{\text{solute}} / m{\text{solution}} \times 100, with m{\text{solution}} = m{\text{solute}} + m{\text{solvent}}. - Percent by Mass - Definition: the ratio of the mass of solute to the mass of the solution, expressed as a percentage. - Key equation: - \%\text{ by mass} = m{\text{solute}} / m{\text{solution}} \times 100. - Example 1 (from transcript): Brine solution - Given: mass of solute (NaCl) = 0.892 g; mass of solvent (H2O) = 54.6 g. - Mass of solution: m{\text{solution}} = 0.892 + 54.6 = 55.492\text{ g}. - \%\text{ by mass} = 0.892 / 55.492 \times 100 = 1.61\%. - Example 2: Benzoic acid in soda - Given: percent by mass = 0.50%; mass of solution = 150 g. - Mass of solute: m{\text{solute}} = 0.50 / 100 \times 150 = 0.75\text{ g}. - Percent by Volume - Definition: the ratio of the volume of solute (liquid) to the volume of the solution, multiplied by 100. - Equation: - \%\text{ by volume} = V{\text{solute}} / V{\text{solution}} \times 100, with V{\text{solution}} = V{\text{solute}} + V{\text{solvent}}. - Example 1: Acetic acid in water - Given: V{\text{solute}} = 3.5\text{ mL}; V{\text{solvent}} = 100\text{ mL}. - \%\text{ by volume} = 3.5 / (3.5+100) \times 100 = 3.38\% \approx 3.4\%. - Example 2: Isopropanol in water - Given: V{\text{solute}} = 25\text{ mL}; V{\text{solvent}} = 45\text{ mL}. - \%\text{ by volume} = 25 / (25+45) \times 100 = 25 / 70 \times 100 = 35.71\%. - Mole Fraction - Definition: mole fraction of a component A is the ratio of moles of A to the total moles in the solution. - Equations: - X{\text{A}} = n{\text{A}} / (n{\text{A}} + n{\text{B}}), \quad X{\text{B}} = n{\text{B}} / (n{\text{A}} + n{\text{B}}), - and X{\text{A}} + X{\text{B}} = 1. - Example 1: 1.00 mol HCl and 8.00 mol water - X{\text{HCl}} = 1.00 / (1.00 + 8.00) = 1 / 9 \approx 0.11, - X{\text{H2O}} = 8.00 / (1.00 + 8.00) = 8 / 9 \approx 0.889. - Example 2: Maple syrup with sucrose and water - Given: m{\text{solu}} = 10.0 g sucrose (C12H22O11), 50.0 g water (H2O). - Molar masses: C12H22O11 = 342 g/mol; H2O = 18 g/mol. - Moles: n{\text{C12H22O11}} = 10.0 / 342 = 0.0292\ \text{mol}, n{\text{H2O}} = 50.0 / 18 = 2.78\ \text{mol}. - Total moles = 2.81\ \text{mol}. - X{\text{C12H22O11}} = 0.0292 / 2.81 = 0.0104, - X{\text{H2O}} = 2.78 / 2.81 = 0.9896. - Molarity (M) - Definition: number of moles of solute per liter of solution. - Equation: M = n{\text{solute}} / V{\text{solution}}\text{ (L)}, with unit: mol L^{-1}. - Example 1: 684 g sucrose (C12H22O11) dissolved to make exactly 1.00 L solution - Molar mass of sucrose = 342 g/mol; n{\text{solute}} = 684 / 342 = 2.00\ \text{mol}. - M = 2.00 / 1.00 = 2.00\ \text{M}. - Example 2: Benzene in 250 mL bottle with concentration 0.000151 M - Given: M = 0.000151\ \text{mol L}}^{-1}, V = 250\text{ mL} = 0.250\text{ L}. - Moles of benzene: n = M \times V = 0.000151 \times 0.250 = 3.78 \times 10^{-5}\ \text{mol}. - Molar mass of benzene (C6H6) = 78 g/mol; mass = n \times M{\text{m}} = 3.78\times10^{-5} \times 78 = 2.94\times10^{-3}\ \text{g} = 2.94\ \text{mg}. - Molality (m) - Definition: number of moles of solute per kilogram of solvent. - Equation: m = n{\text{solute}} / m{\text{solvent}}\text{ (kg)}, with unit: mol kg^{-1}. - Example 1: 2.00 g aspartame in 250 g H2O - Molar mass of aspartame (C14H18N2O5) = 294 g/mol; n{\text{solute}} = 2.00 / 294 = 0.00680\ \text{mol}. - Mass of solvent kg: 0.250\ \text{kg}. - Molality: m = 0.00680 / 0.250 = 0.0272\ \text{mol kg}}^{-1}. - Example 2: Theobromine in 375 g water to make 3.75 m solution - Molar mass of C7H8N4O2 = 180 g/mol. - Required moles: n = m \times m{\text{solvent}} = 3.75 \times 0.375 = 1.406\ \text{mol}. - Mass needed: 1.406 \times 180 = 253.1\ \text{g} \approx 254\ \text{g}. - Parts per Million (ppm) - Definition: ppm = (mass of solute in mg) / (volume of solution in L). - Equation: \text{ppm} = \text{mass of solute (mg)} / V{\text{solution}}\text{(L)}. - Example: Brine solution - Given: mass of solute = 2.5 g; volume of solution = 125 mL = 0.125 L. - Mass of solute in mg: 2.5\text{ g} \times 1000 = 2500\text{ mg}. - \text{ppm} = 2500 / 0.125 = 2.0 \times 10^{4}\ \text{ppm}. - Quick practice problems and prompts (from page 8) - Problem a: Percent by mass of a silver ring - Ring mass = 12.0 g; pure silver mass = 11.1 g. - \%\text{ by mass} = 11.1 / 12.0 \times 100 = 92.5\%. - Problem b: Rubbing alcohol (C3H7OH) 70% solution in a 500.0 mL bottle - Assuming percent by volume, volume of pure C3H7OH = 0.70 \times 500 mL = 350 mL. - Problem c: Mole fraction of solute in a solution with 3.50 g sucrose (M = 342 g/mol) and 100.0 g water (M = 18 g/mol) - Moles: n{\text{sol}} = 3.50 / 342 = 0.010235\ \text{mol}, n{\text{solvent}} = 100.0 / 18 = 5.5556\ \text{mol}. - Total = 5.5660 mol. - X{\text{sol}} = 0.010235 / 5.5660 \approx 0.00184, X{\text{solvent}} = 5.5556 / 5.5660 \approx 0.99816. - Problem d: Mass of sugar, C6H12O6, to make 8.75 m solution in 450.0 g water - Mass solvent in kg: 0.450 kg; moles needed: n = m \times \text{kg solvent} = 8.75 \times 0.450 = 3.9375\text{ mol}. - Molar mass C6H12O6 = 180 g/mol; mass solute = 3.9375 \times 180 = 708.75\text{ g}. - Problem e: Volume of a 0.25 mol/L NaCl solution that contains 5.8 g NaCl - Molar mass NaCl = 58.44 g/mol; moles = n = 5.8 / 58.44 = 0.0993\ \text{mol}. - Volume: V = n / M = 0.0993 / 0.25 = 0.397\ \text{L} \approx 398\ \text{mL}. - Problem f: Importance of knowing the concentration of a solution - Concentration knowledge is essential for calculating reaction stoichiometry, predicting reaction rates, ensuring safety and efficacy, quality control in industries, and communicating solution compositions clearly. - Practical notes and assumptions - Some expressions assume densities near 1 g/mL for aqueous solutions when converting volumes to masses; real values may vary slightly with temperature and composition. - The volume-based expressions (percent by volume) use volumes that may not be strictly additive in all cases, but are commonly used approximations for dilute solutions. - Quick reference: important equations (LaTeX) - \%\text{ by mass} = m{\text{solute}} / m{\text{solution}} \times 100,\quad m{\text{solution}} = m{\text{solute}} + m{\text{solvent}}. - \%\text{ by volume} = V{\text{solute}} / V{\text{solution}} \times 100,\quad V{\text{solution}} = V{\text{solute}} + V{\text{solvent}}. - X{\text{A}} = n{\text{A}} / (n{\text{A}} + n{\text{B}}),\quad X{\text{B}} = n{\text{B}} / (n{\text{A}} + n{\text{B}}),\quad X{\text{A}} + X{\text{B}} = 1. - M = n{\text{solute}} / V{\text{solution}}\text{(L)}, with unit: mol L^{-1}. - m = n{\text{solute}} / m{\text{solvent}}\text{(kg)}, with unit: mol kg^{-1}. - \text{ppm} = \text{mass of solute (mg)} / V{\text{solution}}\text{(L)}, where V{\text{solution}} is in liters. - Summary of key takeaways - Concentration can be