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CHEM1222 – Materials in Biology Metals have unique properties including high electrical/thermal conductivity, malleability and ductility. This is because electrons are arranged around metal atoms in such a way that they are mobile, following an electron sea model (sea of electrons, free flowing, delocalised electrons). Electrons do not belong to any one metal atom. Kinetic energy is transferred as heat into electrons, which then transfer energy from one to the other, thus allowing heat to move very fast through metals. Ions in metals are not arranged randomly, but rather maintain a lattice structure (as in crystals). If lateral force is applied to a metal, it will simply deform (as opposed to shattering – crystals) because of the same sea of electrons concept – metals are always shielded by electrons preventing same charged ions from meeting (repulsing). Described as a regular array of metal atoms embedded in a sea of electrons. Metals exhibit a wide range of melting points since different metals have different charges and form different lattices. Alloys are a mixture of different metals and/or other impurities. Addition of these impurities can alter the properties of metals such as strength and malleability, e.g. addition of carbon into interstitial holes in pure iron lattice prevents adjacent atoms from sliding past each other and hardens iron into steel. Bronze is 88% copper, 12% tin. Steel is 98% iron, 2% carbon. Mercury is the only metal element that is liquid at room temperature (melting point -39o C) with a density of 13.5g/cm. It forms amalgams (alloys) with many different metals. (When in water, you displace your weight’s worth in water but since water is less dense you have to displace more water than your own volume. Since mercury is dense, you only need to displace a little bit and you end up floating on it easily). Mercury and its compounds are often highly toxic and are readily absorbed through globes and skin. 0.1 mL of mercury is considered fatal due to high vapour pressure. To treat mercury poisoning, administration of sulphur-containing compounds is generally used, as they bind tightly to mercury and then can be excreted. Fluorescent lights function by running an electric current through gas, exciting mercury vapour which produces UV light, then causing the phosphor coating on the inside of the tube to glow. It is very efficient (uses less energy à less greenhouse gas), but contains mercury so is bad for the environment when it is released in landfill. Alloys in biomedical applications need to be biocompatible, corrosion resistant, fracture tough, last a long time, wear resistant and match native bone in stiffness.

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CHEM1222 – Ionic Strength; Colligative Properties & Osmosis Ionic strength – take account of concentration (c) and charge (z) of ions in a system. The variable that enables us to achieve this is as follows: I is the ionic strength in (mol/L or mol/dm3 – decimetres cubed), ci is the molar concentration of ion i, zi is the charge number on the ion i. (Reminder – summation symbol = sum of…). Example question – Calculate the ionic strength for a 0.01M Na2SO4 solution. Na2SO4 dissociates into the ions Na+ and SO4

2-. Thus I = ½ (12*0.02 + 22*0.01) = ½ (0.02 + 0.04) = 0.03 mol/L Colligative – properties that depend on the ratio of particles. Osmosis, depression of freezing point, elevation of boiling point, depression of vapour pressure. Depression of vapour pressure In a closed container containing water, some molecules contain enough energy to enter a gaseous state. The vapour pressure is the pressure of the gaseous phase of water in equilibrium with the liquid phase = P0. When solute is added to an identical container (water + solute), the new vapour pressure = P, is smaller than P0. This is because water molecules must escape through the surface of the water in order to entire the gaseous phase. When solute particles are dissolved in water, it is more difficult for water to exit from the surface since solute particles could be hindering the molecules. On the other hand, when water undergoes condensation from gas à liquid, the solute particles do not hinder them resulting in a shift in equilibrium à less gaseous phase à less vapour pressure. A general formula that can describe this interaction is P = Xsolv*P0 (don’t need to remember this), where X is the fraction of solvent present (or something like that). The freezing point of a solution is lower than that of a pure solvent (no solute). Anti-freeze (ethylene glycol) and salt lowering the freezing point of water follows this mechanism. The phase diagram of water is described in the figure. Where P represents pressure and T represents temperature. When solvent is added to pure water, this lowers the vapour pressure, thus lowering the curve between gas and liquid (visualise). An arbitrary line can be drawn on the Y-axis representing 1 atm (atmospheric pressure). Drawing this line across the graph, we find that it intercepts with the line between liquid/gas twice. The first interception can be described as a phase change (boiling point) of water - 100oC. The lowered vapour pressure (caused by impurities/solute) produces a second interception, also representing the boiling point of water. The new boiling point of solution (water + solute) thus becomes higher than the normal boiling point of water – an elevation in b.p. The red T represents the triple point, in which all three states of water can co-exist at the same time. The addition of solute decreases the triple point, thus also affecting the line of phase change between liquid and solid, meaning that the temperature required to change water from a liquid to a solid has also been lowered – depression of f.p. Reminder – Molal/molality = mol/kg (of solvent) Molality is used in place of molarity since the volume can change when temperature changes, but mass stays the same. The colligative properties of solutions can be described by the following formulae: Change in boiling point elevation = Kb (molal boiling point elevation constant – dependent purely on solvent, not solute) * b (molality of solution) Change in freezing point depression = Kf (molal freezing point depression constant – only solvent, not solute) * b (molality of solution) Example question – New boiling point of solution (water) = 100oC + delta Tb (change in b.p.e).

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Ethylene glycol is used as antifreeze in car radiators. Mol. Weight = 62.07 g/mol Kf = 1.86 K/mol/kg for water What is the freezing point of water when 250g of ethylene glycol is added to 1.0L water? n(EG) = 250/62.07 = 4.03 mol Change in f.p = 1.86 K/mol/kg * 4.03 mol/kg = 7.49 K = 7.49oC Therefore, the new freezing point is 0 – 7.49 = -7.49oC A solution of 5.65g of an unknown compound is dissolved in 110g of benzene and the freezing point was measured as 4.39oC. The normal freezing point of benzene is 5.45oC. What is the mass of the compound? Kf(Benzene) = 5.07 K/mol/kg Change in f.p = 1.06 K = Kf * b = 5.07 * b b = 0.209 = mol/0.110 mol = 0.0230 molar mass (Mr) = mass (m)/mol = 246 g/mol Osmosis A semi-permeable membrane allows small molecules (such as water) to pass through, but not larger molecules (glucose). When a dilute solution is separated from a more concentrated solution by a semi-permeable membrane, osmosis (a passive process) will attempt to equalise the concentrations of the two solutions by allowing ONLY solvent through the membrane. Dialysis occurs when a dialysing membrane allows both water and small solute particles through. Osmosis is a net shift of ONLY solvent through an osmotic membrane. A dialysis membrane may allow small molecules/solvents through, but not red blood cells. Osmosis is more than a just a filter, it represents a net shift of solvent through a membrane when there are different concentrations on each side, from a side of lower concentration to the side of higher concentration. The process of osmosis comes back to the concept of vapour pressure – vapour pressure is greater in pure water than in that of a solution (Raoult’s Law). If a beaker of pure water and a beaker of salt solution were placed in a closed system saturated with water vapour, there will be a net increase of salt solution – since water condenses at the same rate in both systems, but water from the beaker containing pure water will evaporate at a greater rate than that containing salt solution. The process of osmosis itself can be described as spontaneous meaning, delta-G is negative (exothermic). Osmotic pressure is described as the pressure needed (applied to the more concentrated solution) to prevent any osmotic flow (process of osmosis) from occurring when one of the liquids is a pure solvent. Osmotic pressure is symbolised by a capital pi (II without the bottom horizontal). Reminder – Tonicity; hypertonic (higher conc. leads to shrunk RBC) and hypotonic (lower conc. leads to lysis RBC). Osmotic pressure can be calculated by the van’t Hoff equation: icRT Where i = the van’t Hoff factor, c = the concentration (also can be represented by n/v), R = the gas constant and T = temperature. Rewritten using c=n/v, osmotic pressure is very similar to the gas law: IIV = inRT The van’t Hoff factor is the ratio between the concentration of particles produced when the substance is dissolved, and the concentration of a substance is calculated from its formula. Most non-electrolytes (sucrose) dissolved in water have a van’t Hoff factor of 1. For NaCl, which dissociates into 2 particles (when dissociation is 100%, as seen in dilute solutions), the van’t Hoff factor is 2. *In very concentrated solutions (exam), 100% dissociation will not occur, and a table will be provided giving you the actual van’t Hoff factor to be used. Example Question – A very dilute solution of urea, 0.001M is separated from pure water by an osmotic membrane. What osmotic pressure (in atm) develops at 25oC? II = icRT = 1*0.001*0.0826*298.15 = 0.025 atm

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0.01M K2SO4 solution is separated from pure water by an osmotic membrane. What osmotic pressure (in Pa) develops at 25oC? II = icRT = 2.7*0.01*8.314*298.15 = 66.9 Pa

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CHEM1222 – Solubility Key ideas regarding solutions:

- Liquid solutions contain a liquid solvent and a solute. - Solute can be a gas, liquid or solid. - A precipitate in a saturated solution is in equilibrium with the solute. - Precipitation depends on the solute concentrations and the solubility product Ksp.

The solubility of gases increases with pressure; Henry’s law states that cgas = kH * pgas Where cgas is the concentration of the gas (M), pgas is the partial pressure of the gas above the solution (Pa) and kH is the Henry’s law constant unique to each gas (M/Pa). Since the partial pressure also depends on the temperature of the environment, we assume that T is constant for Henry’s law. The more gas present on top of the solution, the higher the partial pressure, and thus the greater the dissolvability of the gas. Solubility of gases decrease in water as T increases. On the other hand, most solids in aqueous solutions increase their solubility with increasing temperature. This is because a solid requires energy to break apart the crystal lattice structure, before forming new bonds with surrounding solvent. Solubility is the maximum amount of solute that dissolves completely in a given amount of solvent at a particular temperature, T. (mg/L, g/L, mol/L, etc.) A saturated solution is a solution in which no more solute will dissolve. Dissolution is the process of dissolving a solute to a solvent to give a homogenous solution. To determine solubility, a known volume of saturated solution is transferred to a beaker, the water is evaporated leaving a residue of the solid compound. The mass of the remaining compound can then be used to calculate solubility. The equilibrium expression is called the solubility product – Ksp, and does not include solids. For a balanced equation A2B3 à 2A + 3B, Ksp = [A]2[B]3 Example Question: One litre of a solution saturated at 25oC with calcium oxalate (Mr = 128 g/mol) is evaporated to dryness, giving 0.0061g residue of CaC2O4. Calculate Ksp for calcium oxalate at this temperature. n(CaC2O4) = 0.0061/128 = 0.0000477 mol c(CaC2O4) = 0.0000477 M = c(Ca) = c(C2O4) Ksp = 0.00004772 = 2.3*10-9 10L of a saturated solution of calcium fluoride, CaF2 was evaporated and found 4.28 mmol fluoride. What is the value of Ksp? n(F) = 0.00428 mol, c(F) = 0.00428/10 = 0.000428 M, c(Ca) = 1/2c(F) = 0.000214 M Ksp = 0.000214 * 0.0004282 = 3.9*10-11 Determine the solubility of barium sulphate in water at 25oC, Ksp = 1.1*10-10 BaSO4 à Ba2+ + SO4

2- The ICE table (initial, change, equilibrium) shows that: Where s is the change in concentration (solubility) of barium and sulphate ions. Ksp = s*s = 1.1*10-10, s = 1.0*10-5 M The Common Ion Effect A few drops of HCl is added to a saturated solution of NaCl. What happens? NaCl à Na + Cl Le Chatelier’s principle predicts that due to the addition of Cl ions, the equilibrium will shift to the left, causing precipitation of NaCl, ultimately lowering the solubility of NaCl in the solution. Example Question: Magnesium hydroxide, Mg(OH)2 is found in many commercial antacids. It’s Ksp at 25oC is 1.8*10-11. Calculate its molar solubility. Once again using the ICE box, we find that the change in hydroxide

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ions is twice as great as the change in magnesium ions (s vs. 2s). Therefore, Ksp = s*(2s)2 = 4s3 = 1.8*10-11 Solving for s gives a solubility of 1.65*10-4 M What is the molar solubility of calcium fluoride in 0.100M sodium fluoride solution at 25oC? Ksp for CaF2 = 3.4*10-11

Ca F ICE box à Initial 0 0.1 Change s 2s Equilm s 2s + 0.1 3.4*10-11 = s*(2s + 0.1)2 = 4s3 + 0.4s2 + 0.01s à no CAS no party… However, since the solubility s is so small compared to the initial concentration of 0.1M, we simplify 2s + 0.1 to just 0.1. à 3.4*10-11 = 0.01s, s = 3.4*10-9 M