Solution-Properties-and-Laboratory-Mathematics_compressed.pdf

Solution Properties and Laboratory Mathematics Grams Molecular Weight (GMW)- summation of the atomic weight of each element present in the solution. A. Terminologies...

Solution Properties and Laboratory Mathematics Grams Molecular Weight (GMW)- summation of the atomic weight of each element present in the solution. A. Terminologies II. MOLARITY (M) Solute- any substance dissolved in liquid - Number of moles per 1 liter of solution (mol/L) Solvent- liquid (water- universal solvent) - 1 mole of the substance equals its gram molecular Biologic solutes (analytes)- substance weight (GMW) dissolved in biologic fluids. (ex. - A molar solution is also defined as the gram macromolecules, micromolecules, molecular weight of a substance dissolved to a final metabolic waste products, vitamins) volume of 1L of solution. - These solutes in the body fluids are the substance of interest in clinical FORMULA: chemistry assay. When performing a test, analytes are the target substance to be analyzed. (Kaya ANALYtes). Biologic fluids- the liquid in which analytes are dissolved. Body fluids are made up of water, that’s why they are the solvent. (ex. blood- plasma, if blood is clotted- serum, urine, synovial fluids, CSF, effusion, amniotic fluid) Solution- end result of solute plus solvent. SOLUTION PROPERTIES A. Concentration- expression of analyte concentration: NOTE: solute must be in grams and the solution must be in I. Percent solution- the amount of solute per 100 liters. total units of solution. Percent means parts per 100, represented by percent symbol (%). EXAMPLE PROBLEMS: Molarity 1. How many grams are needed to make 1L of a 2M Three expressions used in the laboratory: solution of HCl? MW of HCl= 36.5g 1. % Weight per weight (w/w) 2. % Volume per weight (v/w) 3. % Weight per volume (w/v) 2. A liter of solution contains 24g of NaOH, what is the molarity? MW of NaOH= 40g Grams of solute= % w/v x mL of solution/100 3 ways to solve a problem: 1. identify the given 2. identify the unknown 3. identify the formula to be used in the calculation NOTE: convert if the given is not in grams or mL  3. Make up 250ml of a 4.8M solution of HCl? 2. If a solution contains 111g CaCl2 per liter, what is the MW of HCl= 36.5g normality? MW of CaCl2 = 111g; Valence Ca = +2 III. NORMALITY (N) 3. 3. To make 500 mL of 3N NaSO4 how much substance - Number of Gram equivalent weight per 1 liter of must be weighted? solution. MW of Na2SO4 =142 g; Valence = 2 - An equivalent weight is equal to the molecular weight divided by its valence. - Valence: number of electrons exchanged in oxidation reduction reactions. FORMULA: CONVERSION: a. Normality to Molarity Formula: Normality/Valence = M Example: NOTE: grams unit for solute, liters unit for volume. If not in grams or liters, convert first. EXAMPLE PROBLEMS: Normality 1. What is the normality of a 500mL solution that contains 7g of H2S04? GMW of H2S04= 98g, Valence H (2) = 1x2 = 2 b. Molarity to Normality Formula: Molarity x Valence = N Example:  IV. MOLALITY (m) - Amount of solute per 1kg of solvent B. Colligative Properties- behavior of particles or solutes - Expressed as mol/kg in a solution. FORMULA: FOUR PROPERTIES- only applicable based on a relative number or specific kind of molecule present. 1. Vapor Pressure - pressure at which the liquid solvent is in equilibrium with the water vapors. - The rate of condensation and vaporization are equal. 2. Freezing Point - Vapor pressure of a solid or liquid is equal - FP of water: 0 3. Boiling Point - The vapor pressure of solvent reaches one atmosphere. EXAMPLE PROBLEM: Molality 4. Osmotic Pressure 1. A solution contains 15.6 g of NaCl dissolved in 500g of - Pressure that composes osmosis when a solvent water. Determine molal concentration. flows through a semipermeable membrane. MW of NaCl = 58.5 g ANY SUBSTANCE DISSOLVE IN SOLVENT WILL: 1. Decrease the freezing point by 1.858°∁ - The more dissolved solutes in the solution, the lower freezing point it will be. 2. Increase the boiling point by 0.52°∁ - The more dissolved solutes in the solution, higher boiling point level it will be. 3. Decrease vapor pressure (Dew point) by 0.3 mmHg 4. Increase the osmotic pressure by 17,000 mmHg/Torr C. OSMOTIC PRESSURE D. REDOX POTENTIAL (Oxidation-Reduction Potential) EXTENT OF SATURATION: - Measure of the ability of a solution to accept or a. Dilute- little quantity of solute in a solution. Amount donate electrons. of solvent is greater that makes the solution less - LEORA- Loss of Electrons Oxidation Reducing Agent concentrated. - GEROA- Gain of Electrons Reduction Oxidation Agent b. Concentrated- large quantity of solute in a solution. c. Saturated- there is excess in undissolved solute E. CONDUCTIVITY particles in the solution. - Measure how well electricity passes through a d. Supersaturated- there are greater undissolved solute solution. particles present in the solution. Considered as the - Good conductor- electricity is able to pass through a most thermodynamically unstable solution. solution. - Bad conductor- electricity is unable to pass through a Two ways remove the excess material out of the solution supersaturated solution: - Depends mainly on the charges of ions present in the solution 1. Add crystal solutes - Unit: OHM-1 (Mho) 2. Do mechanical agitation When these are applied, the supersaturated solution results in crystallization of the excess material present in the solution. F. RESISTIVITY OTHER LABORATORY MATHEMATICS AND CALCULATIONS - Resistance of substance to the passage of electrical I. MILLIEQUAVALENT current - Use for expressing electrolyte concentrations - High resistivity able to resist the passage of an Electrolytes: electric current - Low resistivity freely allows the passage of electrical Cations- sodium, potassium, magnesium, current calcium - Unit: OHMS Anions- chloride, bicarbonate phosphate - Used to assess water purity. When a suspended or - A milliequivalent is the EW expressed in milligrams undissolved particle substance is present in water, - mg/dL → mEq/L this substance can actually block or prevent the FORMULA: passage of electric current. - The higher the impurity level in water, the higher the resistivity of the solution/water - If water is pure resistivity is 0 means all electric current passed through - Impurities in water block the passage of electric current in water, then resistivity increases. EXAMPLE PROBLEMS: Milliequivalent 1. If a solution contains 350 mg/dL Na+, how many G. pH and Buffers mEq/L of Na+ does it contain? - In CC, we’re after in pH plasma MW of Na= 23, Valence of Na= 1 - Normal arterial plasma pH: 7.35 to 7.45pH - Th pH of the solution would determine its acidity or basicity. - The pH of a solution is inversely proportional to the hydrogen ion concentration. The lower the pH, the higher the hydrogen ion concentration (vice versa) Acidic solutions- higher HIC Basic solutions (high pH)- lower HIC - pH plasma must be maintained because pH can directly affect the metabolic activity of the cells and could also affect the physiology of various tissues and 2. A solution containing 12 mg/dL Ca++ contains cells. how many mEq/L Calcium? - Buffer is needed to minimize changes in the hydrogen MW of Ca= 40, Valenc of Ca= 2 ion concentration. Buffers: weak acids or bases and their related salts EXAMPLE OF BUFFERS: 1. Bicarbonate carbonic acid buffer system - Chemical formula: Bicarbonate: HCO3- II. MILLIMOLES Carbonic acid: H2CO3- - Molecular weight expressed in milligrams - The major buffer system in our body. FORMULA: 2. Hemoglobin - Specifically, the deoxygenated or reduced HGB 3. Plasma proteins EXAMPLE PROBLEMS: Millimoles IV. DILUTION - Ratio of the volume of substance to be diluted to the 1. Convert a 3 mg/dL magnesium to mmol/L. final volume. MW of Mg= 24.31 - In CC, dilution is more used than ratio FORMULA: 2. Convert 8.2 mg/dL calcium to millimoles per liter. EXAMPLE PROBLEM: Dilution MW of Ca= 40 1. Calculate the dilution using 50 uL of blood and 950 uL of diluting fluid. III. RATIO - Volume of solute per volume of solvent FORMULA: 2. Calculate the dilution using 0.5 mL of urine and 8.5 mL of isotonic saline. EXAMPLE PROBLEMS: Ratio 3. Calculate the dilution of 0.1 mL serum in 0.9 mL water.  V. SERIAL DILUTION - Multiple progressive ranging from more concentrated solutions to less concentrated solutions. - Dilution increases = concentration decreases - Used in measuring titer of antibodies in the serum Titer- reciprocal of the highest dilution showing a positive reaction. Used in reporting the antibody level in the serum. EXAMPLE PROBLEM: Serial Dilution 1. A serum sample was diluted 1:10, then 1:10, then 1:2. What is the final dilution?