Laboratory Mathematics and Solution Preparation PDF

Chapter 6 Laboratory Mathematics and Solution Preparation Preamble PowerPoints are a general overview and are provided to help students take notes over the video lecture ONLY. PowerPoints DO NOT cover the details needed for th...

Chapter 6 Laboratory Mathematics and Solution Preparation Preamble PowerPoints are a general overview and are provided to help students take notes over the video lecture ONLY. PowerPoints DO NOT cover the details needed for the Unit exam Each student is responsible for READING the TEXTBOOK for details to answer the UNIT OBJECTIVES Unit Objectives are your study guide (not this PowerPoint) Test questions cover the details of UNIT OBJECTIVES found only in your Textbook! 1  Chapter 7 Objectives 1. Calculate proportions and ratios 2. Describe the use of exponents 3. Define the terms density and specific gravity. 4. Calculate the requirements for solutions of a given volume and molarity 5. Describe the procedures for making a single dilution and a serial dilution 6. Differentiate the expressions of solution concentration weight per unit weight and weight per unit volume. 7. Prepare a percent solution Significant Figures  Rounding Off Numbers ◦ In the laboratory, test results can sometimes produce insignificant digits. ◦ It is necessary to round off the numbers to a significant value.  General rules to remember 1.If the digit next to the last one to be kept is 5, then the last digit should be increased by 1. 3.If the digit next to the last one is 5, the last digit reported is changed to the nearest even number. 2 Examples 7.342 g is rounded off to 7.34 g 3.659 g is rounded off to 3.66 g 13.5 mg is rounded off to 14 mg 14.5 mg is rounded off to 14 mg Exponents Used to indicate that a number must be multiplied by itself. Two types to consider: Positive Exponent (102) Negative Exponent (10-1) 3 Exponents Positive exponents Indicates the number of times the base is to be multiplied by itself. 102 = 10 x 10 = 100 105 = 10 x10 x 10 x 10 x 10 =100,000 Lab values expressed as exponents Hematology Cell Counts RBC Normal Range – Female: 4.8 x 1012 /L WBC Normal Range – 4.5 x 109 /L Exponents Negative Exponents Indicates the number of times the reciprocal of the base is to be multiplied by itself. Negative exponents represent a fraction 10-1 = 1/10 = 0.1 10-3 = 1/10 x 1/10 x 1/10 = 1/1000 = 0.001 4 Density and Specific Gravity Density = the amount of matter per unit volume of a substance. Example : Specific gravity = the weight of 1 mL of a solution compared to the weight of 1 mL of pure water at 4° C. Specific gravity × Percent assay = Grams of compound/mL Example: Concentrated HCl has a specific gravity of 1.25 g/mL and an assay value of 38%. What is the amount of HCl/mL? 1.25 g/mL × 0.38 = 0.475 g of HCl/mL Proportions and Ratios Proportions are a way of saying that two ratios are equal  a1 = a 2 b1 b2 A ratio is an amount of something compared to an amount of something else 5 Proportions and Ratios Example:  5 grams dissolved in 100 ml of solvent can be expressed as any of the following: Ratio is expressed as: (1) 5/100 (2) 5:100 (3) 5% (4) 5 divided by 100 or 0.05 Proportion is expressed as: 5:100 = 1:20 (expressing the ratios are equal) Expressions of Solution Concentration Solutions are made up of a mixture of substances. Usually two main parts: 1. Solute- the substance being dissolved 2. Solvent- the substance into which the solute is being dissolved Concentration of the solution is the amount of one substance relative to the amount of the other substances in the solution and is expressed in several ways with the most common being: Weight (Mass) per Unit Volume Most common way to express concentration – expressed as milligrams per milliliter (mg/mL) Volume per Unit Volume Liquid diluted with another liquid - expressed as milliliter per milliliter (mL/mL) Percent Parts per hundred parts 6 Expressions of Solution Concentration Molarity Gram-molecular mass (weight) of a compound per liter of solution. Formula: Molecular weight × Molarity = Grams/liter Osmolarity Number of osmoles of solute per liter of solution Osmole is the amount of a substance that will produce 1 mol of particles having osmotic activity. Osmole = 1 gmw (1 mol) of the substance divided by the # of particles formed by the dissociation of the molecules of the substance. Particles that do not ionize, 1 osm =1 mol Concentration – Percentages Percentages Percent solution usually means g or mL of solute per 100mL of solution Are not real numbers Representation of a whole Three ways to represent Percentage 75% Fraction75/100 Decimal 0.75 7 Concentration – Percentages A percent solution is determined in the same manner regardless of whether Weight/weight (w/w) Volume/volume (v/v) recommended that milliliters per liter or mL/L be used in the place of percent Weight/volume (w/v) (most common) Percent implies “parts per 100” represented as percent Concentration – Percentages To make up a 5% aqueous solution of hydrochloric acid (using 12 M HCl) multiply the total amount by the percent expressed as a decimal (5%) 5% = 5/100 = 0.050  Therefore 0.050 x 100 = 5 g of HCl and dilute to 100 mL will make a 5% solution or use 5 = x 100 100 X=5 8  Concentration – Percentages Most frequently used is w/v solution expressed as grams per 100 mL of diluent To make up 1000 mL of a 10% (w/v) solution of NaOH 0.10 x 1000 = 100g Measure out 100g of NaOH add to 1000 mL volumetric Class A flask and dilute to the calibration mark with Type II water and it will be a 10% solution of NaOH Concentration – Percentages Example of v/v Make up 50 mL of a 2% (v/v) concentration of HCL solution 0.02 x 50 = 1 mL Add 40 mL of water to a 50 mL Class A volumetric flask, and add 1 mL of HCL, mix and dilute up to the calibration mark with Type II water Remember – always add acid to water!! 9 Concentrations Can never increase a dilution – only decrease Prepare 60 mL of a 6% NaOH from 40% NaOH USE: Relating concentrations formula C1V1 = C2 V2 This formula is used if you want to change concentrations (40)(X) = (6)(60) X=9 Measure 9 ml of the 40% solution and QS to 60 mL and will have 60 mL of a 6% solution of NaOH Percentage Solutions % solutions may be made by weighing out a specific amount of solute for each 100 ml of solvent. This is w/v Example: saline 0.85% NaCl Thus 100 ml of saline contain 0.85 grams of NaCl 10 Problem Prepare 500 ml of saline. Saline is 0.85% NaCl, therefore every 100 ml of saline contains 0.85 grams of NaCl 0.85 x X = 4.25 g of NaCl and 100 500 dilute to 500 ml Weight out 4.25 g of NaCl, fill a 500 ml volumetric flask approximately half full of distilled water, add the 4.25 g of NaCl to the water and swirl gently to dissolve and then fill the flask to the marked line with distilled water. Problem Prepare one Liter of 2% acetic acid from concentrated glacial acetic acid 2% acetic acid contains 2ml in each 100 ml 2 = x 100 1000 X = 20 mL of glacial acetic acid Fill a one liter volumetric flask approximately half full of type I or II water. Add 20 ml of concentrated acetic acid and swirl to mix. Fill the flask to the line with type I or II water. 11 Concentrations If you want 1000 ml of a 4% solution how much solute do you need to make the solution. Use the a1/b1 = a2/b2 formula Use this formula when you want a different amount of the same concentration 4% solution is 4g/100mL _4_ = _X_ 100 1000 40g of solute and QS to 1000 diluent Concentration - Molarity Molarity = GMW per liter of solution Another way of expressing is Moles per Liter A Mole is the molecular weight of a compound in grams You must know the molecular formula – then you check the periodic table for the correct molecular weights and add them up 12 Concentration - Molarity Example NaCl = Na = 23 Cl = 35.5 58.5 GMW = 58.5 grams which would be 1 Mole A one molar solution of NaCl would contain 58.5 grams of NaCl in one liter of solution because MOLARITY EQUALS MOLES PER LITER Concentration - Molarity BaSO4 Barium Sulfate 1 Ba = 137 x 1 = 137 1S = 32 x 1 = 32 4O = 16 x 4 = 64 233 233 grams of BaSO4 would be 1 mole and a 1 Molar solution would contain 233 g / l liter 13 Concentration - Molarity Molarity is not always in whole units – can use formula: Grams = (M) (Molecular Wt.) (L) How many grams are there in 0.18 M solution of NaCl? g = (0.18) (58.5) (1) = 10.5 grams of NaCl How many grams in 250mL of a 3.16 M solution of BaSO4 g = (3.16)(233)(.25L) = 184.1 Concentration - Molarity What is the molarity of a solution containing 10 grams of NaCl per Liter? Grams = (M) (Molecular Wt.) (L) 10 = (M) (58.5g)(1) M = 0.17 So it is 0.17 M solution 14 Concentration - Molarity Now it’s your turn~ What is the M of a solution of NaOH that contains 16 grams? (NaOH has a molecular weight of 40) 16 = (M) (40) (1L) M = 0.4 Concentration - Normality Molarity does not give a basis for direct comparison of strength for all solutions 1 liter of a 1 M NaOH will exactly neutralize 1 L or 1 M HCl BUT It will only neutralize 0.5 L of a 1 M H2SO4 15 Concentration - Normality Normality (N) is expressed as the number of equivalent weights per liter (Eq/L) or milliequivalents per milliliter (mEq/mL) Equivalent weight is equal to gmw divided by the valence (V) Eq. Wt. = gmw/V Concentration - Normality Normality has often been used in acid-base calculations because an equivalent weight of a substance is also equal to its combining weight Another advantage in using eq.wt is that an eq.wt of one substance is equal to the eq.wt of any other chemical 16 Concentration - Normality Give equivalent weight, in grams, for each substance listed below: (Eq. Wt. = gmw/V)  NaCl (gmw = 58 g, valence = 1)  HCl (gmw = 36 g, valence = 1)  H2SO4 (gmw = 98g, valence = 2) Changing Molarity to Normality Use the formula M x V = N What is the molarity of a 2.5 N solution of HCl? (M) (1) = 2.5 N M= 2.5 17 Dilutions Diluting specimens and other substances are necessary in the lab Can never increase a dilution – can only decrease a dilution A dilution is PART TO TOTAL PART A ratio is PART TO PART Ratio 1:1 Dilution is 1/2 Prepare 150 ml of a 1/5 dilution of serum in saline Dilutions 1:5 dilution is 1 part of serum to 4 parts of saline. If 150 mL of the substance is desired: Use proportion formula 1 = X X = 30 5 150 Use 30 ml of serum and QS to 150 ml of saline 18 Dilutions What is dilution of 100 mEq Na solution from a 3000 mEq stock solution? 100 cancel the zeros 1 is the dilution 3000 30 If in the preceding example 150 ml of the 100 mEq sodium was required 1 = x x=5 30 150 5ml of stock solution and QS to 150 ml of diluent Dilutions A concentration of a solute can be determined by dividing the denominator into the nominator Determine the concentration of a 200 umol/ml hCG standard that was diluted 1/50 200 umol/mL x 1/50 or 4 umol/ml is the concentration of the final dilution 19  Dilutions Dilutions should always be written with a “1” on the left side If a dilution is 2/32 then reduce to 1/16 3/32 reduce to 1/10.7 Dilutions What is the dilution of the following? 18 oz. of saline and 2 oz. of serum. 10 grams of serum and enough saline to make 100 grams of final product. 0.1 quart of iodine solution into a volume of 5 quarts of water. 20 Dilution One dilution: the simplest of dilution procedures involves making a single dilution from a single substance Example: 1 mL of serum with 9 mL of saline 1 mL of serum + 9 mL of saline 10 mL Total volume 1:9 ratio 1/10 dilution Dilutions A. 100 uL of serum added to 900 uL of saline B. 20 uL of serum added to 180 uL of saline C. 1 mL of serum added to 9 mL of saline D. 2 mL of serum added to 18 mL of saline 1. What is the ratio for A – D? 2. What is the dilution for A - D? All of the above are a 1:9 ratio or a 1/10 dilution 21 Dilutions Dilute 3 mL of serum with 25 mL of saline 3 mL serum +25 mL saline 28 mL Total volume 3:25 ratio 3:28 dilution Can’t leave dilution as number larger than 1 so reduce 1:8.33 ratio 1:9.33 dilution Dilutions If you want all your answers to be per 100 mL, the dilution factor must be determined. Example: 5 mL of serum diluted to 25 mL 5/25 = 1/X 5X = 25 X=5 The dilution factor is 5. To convert to 100 mL use a1 = a2 1 = x = 20 mL b1 b2 5 100 5:25 will equal a 1/5 dilution or a ratio of 1:4 22 Dilution 0.5 mL of blood is diluted to 10 mL with saline Same proportion, different amount than 0.5 mL blood = 1mL blood 10 mL saline x mL X = 1 mL x 10 mL = 20 mL or 0.5/10 mL is the 0.5 same as a 1/20 dilution Dilution Find out how much blood you will need to make 1 mL solution of a 1:20 dilution Use same formula 1 mL of blood x mL blood = 0.05 mL 20 mL solution 1 mL solution 23 Dilutions Serial Dilutions Series of Dilutions used in the laboratory usually seen in serological testing. A general rule for calculating the concentrations of solutions is to multiply the original concentration by the first dilution (fraction) by the second dilution and so on until desired concentration is known. Serial Dilutions 24 Dilutions and Concentrations Tube 1 contains 1ml of undiluted serum = 1:1 dilution Tube 2 contains 1ml of undiluted serum + 1 ml of diluent= 1:2 dilution; concentration = ½ x 1 = 0.5 ml of serum Tube 3 contains 1ml of 1:2 diluted serum + 1ml diluent = 1:4 dilution; concentration = 1/4 x 1 = 0.25 ml of serum Tube 4 contains 1ml of 1:4 diluted serum + 1ml diluent = 1:8 dilution; concentration = 1/8 x 1 = 0.125 ml of serum Tube 5 contains 1ml of 1:8 diluted serum + 1ml diluent = 1/16 dilution; concentration = 1/16 x 1 = 0.06 ml of serum Dilutions Standard Solutions Working Standards These are prepared from the stock solution. Standards Used in Spectrophotometry Blank Solutions A blank solution contains reagents used in the procedure, but it does not contain the substance to be measured. 25 Team Work Work on Exercises in Teams: #1 – Dilution/Percentage Problems #2 - Molarity Problems Postamble READ the TEXTBOOK for the details to answer the UNIT OBJECTIVES. USE THE UNIT OBJECTIVES AS A STUDY GUIDE All test questions come from detailed material found in the TEXTBOOK (Not this PowerPoint) and relate back to the Unit Objectives 26