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Introduction to Biochemistry, Water, Biophyscial Chemistry, Ph

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Introduction to Biochemistry, Water, Biophyscial Chemistry, Ph Solutions Disperse Systems, Protective Colloids, Molar and Equivalent Concentrations, Percentage Solutions. Disperse System • The term "Disperse System" refers to a system in which one substance (the dispersed phase) is distributed, in discrete units, throughout a second substance (the continuous phase or vehicle). • There are 3 types of disperse system depending on the size of disperse phase particles. ▫ True Solutions ▫ Collaidal Solutions/Dispersion ▫ Suspensions/Dispersion Disperse Systems True Solution Colloidal Solution Suspensions It has particles It has particles It has particles <1 nm. between 1 and 100 nm. >100 nm . Homogeneous mixture. Heterogeneous mixture. Heterogeneous mixture. Behaves like a third unit other than soluble and solvent. Low viscosity. Viscosity is high. Viscosity is veryhigh. Osmotic pressure is high. Osmotic pressure is low. Does not show osmotic pressure. When the light passes, the definition of When light passes, it gets foggy. This is the light is illuminated called tyndall effect. Particles can not be seen even with the Particles can be seen with an electron Particles can be seen with light most powerful optical and electronic microscope. microscopy or even with the eye. systems Slow Brownian motion is seen in the Particles make molecular movements. Brownian motion is seen in particles. particles. The particles are separated by filtration Filtering the particles through the filter The particles is separated by filtration through filter paper, separated by paper is not possible by dialysis. through filter paper or by dialysis. dialysis. Brownian Motion Tyndall Effect Disperse Systems Important terms of the colloidal state Colloidal solution is given to the particles of solute in the Disperse Phase solution. Dispersion Medium The solvent in the colloidal solution is given to the liquid. Colloid particles do not have affinity to the molecules of solvent Suspensoid (Lyophobic system) Mostly complexes of inorganic particles which do not have affinity to the solvent. In the case of Suspension, the dispersion medium is called Hydrophobic hydrophobic. a mixture of two or more immiscible liquids, one liquid (the Emulsoid (Lyophilic System) dispersed phase) is dispersed in the other (the continuous phase). If the dispersion medium is water in emulsified , such system is Hydrophilic called hydrophilic. Disperse Systems Viscosity states of emulsions SOL GEL The viscosity is low, Viscosity increased, Close to the true solution, Taking a jellylike shape, is called a colloidal system that It is called colloids which must can flow from one vessel to be pressurized to provide another. fluidity. Protective Colloids • Emulsoids are much more durable than suspensoids. • If a small amount of an emulsoid is added to the suspension, the suspension is more stable. • The emulsoid forms a protective layer around the suspended particles and gives the emulsoid most of its own strength. The emulsions used in this way are called protective colloids. Protective Colloids • Except for globulins, various proteins have protective effect. • Many water-insoluble substances in blood plasma can be transported without collapse by the protective colloids in the plasma. • Lipids dissolve as colloids under the influence of proteins. Protective Colloids • Insoluble substances, such as calcium phosphate and uric acid, are brought to the extraction door by the action of protective colloids in the urine without causing undue and oversaturated solubilizing. • It has been reported that the reduction of the protective colloids contained in the urine may make urinary stones possible. Mole Unit • Mole = 6,02 x 1023 Avogadro number (Amedeo AVOGADRO) • 1 mole = 6.02 x 1023 = grams (atomic mass) • 1 mole = 6.02 x 1023 = grams (molecular mass) • 56Fe, 27Al 1 mole 0.5 mole 3 moles 5 moles Mole Unit 1. How much grams should we weigh to prepare 2 moles of NaOH? ( Na=23 g/mol, O=16 g/mol, H=1 g/mol) 2. How many moles KCl in 7.45 g? (K=39.0 g/mol; Cl=35.5 g/mol) 3. What is the mole number of 19.6 g H2SO4? (H=1 g/mol, S=32 g/mol, O=16 g/mol) 4. How many moles of Sn atom that contains 3.01x1023 particles? Molar Concentration • The most common measure of concentration in the laboratory is molarity - the number of moles of solute per liter of solution. • Symbol = M or mol/L • Preparing Molar Concentration from solid chemicals. ▫ NaOH MW= 40 g • Preparing Molar Concentration from acids. ▫ H2SO4 MW= 98 g Density= 1.84 g/ml % 98 Molecular Weight (g) x wanted M x Desired Volume (mL)) M = Density x Percentage x 1000 Molal Solutions (Molality) • A molal solution is a solution that contains 1 molecular weight (mole number) of solute in a kilogram of solvent. A solution of concentration 1 mol/kg is also sometimes denoted as 1 molal. • Symbol= m veya mol/kg • Preparing Molal Concentration from solid chemicals. ▫ What is the molality of a solution of 10 g NaOH in 500 g water? (MW= 40 g/mol) mole (solute) Molarity = L (solution) ! mole (solute) Molality = Kg (solvent) Molarity & Molality Problems • We prepared a 500 ml of solution with using 23.4 g NaCl. So, What is the molarity of the solution? (NaCl MW=58.5 g/mol) • Prepare 0.5 M 400 mL HCl solution (MW=36.5 g/mol - Density=1.2 g/mL - 36.5%). • 32.5 g of NaF is dissolved in 425g of water. Calculate the molality of the solution (NaF MW=42 g/mol). • 30.8 g of KOH is dissolved in 1100 g of water. Calculate the molality of the solution (KOH MW=56 g/mol). Molarity & Molality Problems • How many grams of KNO3 should be added to 250 g water to prepare a 0.200 m KNO3 solution? (KNO3: 101.1 g / mol) • How many mL of HNO3 solution should be used to prepare an 100 mL of 100 mmol/L HNO3? (MW = 63.01 g / mol - Density = 1.51 g / ml - 70%) • How many grams of CaCO3 is required to prepare a 500 g of 0.5 mol/kg CaCO3 solution? (CaCO3: 100.08 g / mol) • A solution of 74.5 g of CaCl2 in 560 g of water was prepared. The density of the solution is 1.15 g/mL. Calculate the molality and molarity of this solution? (CaCl2 MW = 110.98 g/mol) Osmole Unit • Osmole = Molecular Mass: Osmotically active particle number • NaCl dissociates to Na and Cl ions in aqueous solution. Since each molecule forms two osmotically active particles, 1 osmole grams of NaCl 58.5/2 = 29.25 grams. So, to prepare 1 osmolar solution of NaCl, we should weigh 29.25 g NaCl. • 1 osmolar glucose is 180 grams. Because glucose is dissolved in the molecular state and 1 active particles. • Ionic - Nonionic compound difference is important. Osmolar Solutions (Osmolarity-Osmotic Concentration) • It is the measure of solute concentration, defined as the number of osmoles (Osm) of solute per litre (L) of solution. • Symbol= Osm, Osm/L veya osmol/L • Osmolarity can be used to predict whether water will pass from one side of a semipermeable membrane to the other (also referred to as water retention). • How much grams should we weigh to prepare 2 Osm of 4 L NaCl? (NaCl MW=58 g/mol) • Plazma Osmolarity (Osm) = 2 Na + Glucose + Ure (all in mmol/L) Osmolality Solutions (Osmolality) • Osmolality is the number of osmoles of solute in a kilogram of solvent. • Symbol= Osmol/kg osmole (solute) Osmolarity = L (Solution) ! osmole (solute) Osmolality = Kg (Solvent) Equivalent Concentration (Normality) • Normality is another way of expressing the concentration of a solution. It is based on an alternate chemical unit of mass called the equivalent weight. • The normality of a solution is the concentration expressed as the number of equivalent weights (equivalents) of solute per liter of solution. ▫ The equivalent concentration or normality of a solution is defined as the molar concentration divided by an equivalence factor. • Equilibrated grams are found by dividing the molecular weight of the substance to be prepared by the solubility value by the valency of the substance to be prepared. Equivalent Concentration (Normality) • A normal solution contains one equivalent of solute per liter of solution. • For acid-base reactions, an equivalent is the amount of a reactant that can produce or consume one mole of hydrogen ions (using the Brønsted-Lowry definition). • So, for example, a mole of HCl or NaOH is one equivalent, but a mole of H₂SO₄ or Ca(OH)₂ is two equivalent. Equivalent Concentration (Normality) Equivalent grams of the solute Normality = L (Solution) Equivalent Concentration (Normality) • Symbol= N/eq/L/Val/L • Preparing normal solution from solid chemicals. ▫ NaOH MW= 40 g • Preparing normal solutions from acids ▫ H2SO4 MW= 98 g Dansite= 1.84 g/ml % 98 Molecular Weight (g) x Desired M x Desired Volume (mL) N = Density x Percentage x Equivalence Factor x 1000 Normal Solutions (Normality) • A 250 ml of solution was prepared with using 18.5 g Ca(OH)2. What is the normality of the solution? [Ca(OH)2: 74 g mol)] • How many grams of NaOH is required to prepare a 250 mL of 0,2 N NaOH solution? (NaOH= 40 g/mol) • A 0.1 L solution of 4.9 g of H2SO4 was prepared. Calculate the Normality of this solution? (H2SO4 MW=98 g/mol) Percentage Solutions • A percentage solution is an amount or volume of chemical or compound per 100 mL of a solution. It is a relative expression of solute to solvent: X amount/100 ml = X%. • Weight/Volume ▫ 250 ml % 10 NaOH Preparation • Volume/Volume ▫ Preparation of 100 ml of ethyl alcohol of 40% by using 96% ethyl alcohol • Weight/Weight ▫ % 5 NaOH Preparation Dilution • Dilution is the process of decreasing the concentration of a solute in a solution, usually simply by mixing with more solvent like adding more water to a solution.
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