MEDL 210 Lab MATH 01_May 2019 PDF

Laboratory Math 1 Rationale In the clinical laboratory mathematic calculations of concentration and dilution are often performed. They are required in many instanced including preparation of reagents, controls, and standards as well as interpreting patient results. It is necessary for all laborato...

Laboratory Math 1 Rationale In the clinical laboratory mathematic calculations of concentration and dilution are often performed. They are required in many instanced including preparation of reagents, controls, and standards as well as interpreting patient results. It is necessary for all laboratory workers to have an adequate understanding of laboratory mathematics. Learning Outcome Apply mathematic principles used in the preparations of solutions and samples in the medical laboratory. Introduction In the medical laboratory many types of solutions are used and measured. Solutions are generally thought of as having a dispersed phase, or solute, and a dispersing phase, or solvent. The amount of solute relative to solvent is often measured and expressed as concentration. There are many ways concentration can be expressed. This module will define the expressions of concentration relevant to the medical laboratory, as well as demonstrate the calculation and manipulation of these concentrations. Dilutions of solutions and patient samples are routinely performed, so calculations involving single and serial dilutions will also be discussed. **Note on rounding off numbers:** If you have reason to suspect that a measurement such as 2.57 cm is reliable or accurate to the first decimal place only, you should round it off to that digit that is reliable -- so 2.57 cm would become 2.6 cm. Rules for "rounding off" numbers from experimental data: 1. 2. 3. **In the clinical laboratory, the number of decimals for results to be reported typically matches the associated reference (normal) range.** **Expressions of concentration in the medical laboratory** **Activity:** 1. Complete Exercise One and check your answers. **MOLARITY (M)** The **molarity** of a solution is a ratio of the number of moles of solute per litre of solution. Molarity = A solution containing 1 mole of KCl in 1 L of solution has a molarity of 1 and is said to be a 1 M (1 Molar) solution. To find the molarity of a solution, we need to know the solution volume in litres and the number of moles of solute present. An alternative to knowing the number of moles of solute is to know the number of grams of solute present and the molecular weight of the solute i.e. the number of grams per mole. **Examples of Molarity Problems** 1\. What is the molarity of a solution of NaCl when 58.5 g of NaCl are dissolved in 0.50 L of solution? 0.50 L. Substituting the known quantities into the equation for molarity results in: Molarity = Another way to solve this problem is by using ratios. Start with what you know on the left and setup ratios until you end up with the solution on the right. Make sure that all units cancel out from the numerator to the solve. This method can be used for most concentration problems. The full calculation in ratio form with units canceling is: 2\. 1.5 moles of NaOH are dissolved in 500 mL of H~2~O. What is the molarity? e.g. moles of solute = 1.5 moles of NaOH or Now, substituting the known quantities in the equation for molarity: Molarity = The full calculation in ratio form with units canceling is: 3\. How many grams of NaOH are present in 250 mL of a 0.15 M NaOH solution? In order to solve this problem, we must first convert mL to L. 1.5 g NaOH Another problem in solution preparation is when the volume of solution must be calculated from the number of grams of solute given and the molarity of the solution is specified. 4\. How many litres of 0.75 M NaOH can be prepared from 20 g of NaOH. 0.67 L Osmolarity (Osm) The **osmolarity** of a solution is a ratio of the number of osmoles of solute per litre of solution. The unit of volume, litres, is the one used in the ratio for molarity. The difference is that the amount of solute is expressed in osmoles, in osmolarity, and as moles, in molarity. To convert osmolarity to molarity or vice versa, it first must be determined what an osmole is, and then how it relates to a mole. An **osmole,** of an ionic solute, is an ion formed upon dissociation of the solute. The number of osmoles formed on dissociation is the same as the number of moles of ions that a mole of solute will form when it dissolves. ex. NaCl Na ^+1^ + Cl^-1^ 1 mole 1 osmole + 1 osmole = 2 osmoles 1 mole of NaCl 2 osmoles = 2 moles of ions in solution To calculate the osmolarity of a 1 molar solution of NaOH: 1. First, determine the number of osmoles or moles of ions that NaOH will form on dissociation. 2. Secondly, use the ratio of the number of osmoles per mole to change the molarity to osmolarity. **Examples of Osmolarity Problems** 1\. Calculate the osmolarity of a 0.30 molar solution of NaOH. 2\. How many grams of CaCl~2~ would you need to prepare 500 mL of a solution with an osmolarity of 0.50? **Weight per Volume - % (W/V)** Concentration is expressed in terms of grams of solute per 100 mL of solution. \%(W/V) = Therefore, a 5% (W/V) solution of NaCl would contain 5 grams of NaCl per 100mL of solution. Example problem: 1\. How would you prepare 0.35 L of 6.5% (w/v) NaOH? **Percentage by Volume - % (V/V)** Concentration is expressed in terms of mL of liquid solute per 100mL of solution. (This measurement of concentration is used for liquid in liquid solutions.) \%(V/V) = Therefore, a solution is 10%(V/V) when it contains 10 mL of a liquid solute, (the substance in the lesser amount), per 100 mL of solution. To solve percentage by volume problems the formula C~1~V~1~=C~2~V~2~ is often used. Available values for the volume and concentration of the original solution are substituted for V~1~ and C~1~ respectively, and the values for the second solution are substituted into V~2~ and C~2~. The missing value is then solved for. Example problem: 1\. How many mL of 95% ethanol would you need to prepare 500 mL of 70% ethanol? C~1~ × V~1~ = C~2~ × V~2~ 95% × V~1~ = 70% × 500 mL 95% × V~1~ = 350 mL V~1~ = 368 mL of 95% ethanol 2\. If you added water to 50 mL of 25% isopropanol to a total volume of 250 mL what would the concentration of the new solution be? C~1~ × V~1~ = C~2~ × V~2~ 25%× 50 mL = C~2~ × 250 mL 12.5 mL = C~2~ × 250 mL 0.05 = C~2~ C~2~ = 5% Molality (m) and Osmolality (osm) **Molality and osmolality are both methods of expressing concentration of solutions based on weight.** **Molality** A 1 molal (1 m) solution contains 1 mole of solute plus 1000 g of solvent -- or a 1 molal solution contains 1 mole of solute per kilogram of solvent. **Osmolality** The osmolality of a solution is the sum of the moles of all dissolved ions and undissociated molecules per kilogram of solvent. **\ Examples of Molality and Osmolality Problems** 1\. How would you prepare 1.00 kg of 0.200 molal Na~2~CO~3~? 2\. Calculate the osmolality of 35 g NaCl dissolved in 1500 g of H~2~O. **Molarity and pH** One can determine the pH of a solution of a strong acid from the molarity (concentration) of the hydrogen ions. The pH of a solution is the negative log of the H+ concentration. So the pH of a 0.01 mol/L HCl solution is 2 (-log of 0.01). One can also determine the pOH of a solution of a strong base from the molarity (concentration) of the hydroxide ions. The pOH of a solution is the negative log of the OH concentration. So the pOH of a 0.02 mol/L NaOH solution is 1.7 (-log of 0.02). As pH and the pOH always add up to 14, one can determine either the pH or pOH of a solution if the other value is available (pH + pOH =14). For example, if the pH of a solution is 2, then the pOH would be 12 (14-2=12). If the pOH of a solution is 1.7, then the pH will be 12.3 (14-1.7=12.3). **Exercise One** 1. How many moles of: a. silver nitrate are in 78.20 g? b. NaCl are in 92 mg? 2. How would you prepare: a. 500 mL of 0.050 M sodium sulfate? b. 200 mL of 0.40 M cupric sulfate pentahydrate (CuSO~4~ \* 5 H~2~O)? 3. Calculate the osmolarity of the following solutions: a. 2.5 M H~2~SO~4~ b. 1.5 M HCl c. 0.30 M NaOH 4. How would you prepare: a. b. 5. How would you prepare 500 mL of 5% (W/V) KCl? 6. How would you prepare 250 mL of 0.85% (W/V) NaCl? 7. A solution of ammonium chloride is 0.50 M. a. What is the concentration expressed as %(W/V)? b. What is the concentration expressed as osmolarity? 8. How would you prepare: a. 500 mL of 28 %(V/V) aqueous solution of methanol from 95 % methanol? b. 250 mL of 85 % (V/V) ethanol from absolute ethanol? 9. How would you prepare a 0.50 molal solution of CaCl~2~ containing 0.500 kg solvent? 10. How would you prepare approximately 1 kg: a. a 0.500 osmolal solution of NaCl? b. a 2.00 osmolal solution of NaCl? 11. Calculate the pH of a 0.06 mol/L solution of HCl. 12. What is the pH of solution when one gram of NaOH is added to 500 mL of water? Perform the calculations necessary when making dilutions of reagents and solutions. **Activity:** 1. Complete Exercise Two and check your answers. Dilution of Reagents and Solutions **Dilution** is the act of making a weaker solution from a more concentrated one. When using dilutions one of the main problems encountered is confusion between ratios and dilutions. A **ratio** is an amount of one substance relative to an amount of another. A ratio is usually expressed using a ratio sign (:), as in 1:4, unless it is being used in a mathematic calculation in which case it may be expressed as a common fraction (1/4), a decimal (0.25), or using a division sign (1÷4). When working with ratios it is important that you are certain of what is being compared. For example, there are many ways a ratio could be used to describe a tube containing 1 mL of ethanol and 4 mL of water. All of the following are correct. - Ethanol to water ratio is 1:4 - Ethanol to total volume ratio is 1:5 - Water to total volume ratio is 4:5 - Water to ethanol ratio is 4:1 - Total volume to ethanol ratio is 5:1 **When discussing dilutions the ratio used *should* express the number of parts of a substance being diluted to the *total number of parts in the final product***. This is not always the case so it is important to be clear before beginning any procedure involving dilution. To avoid confusion the common fraction is most often used to express a dilution. Using the example above the dilution should be expressed as 1/5 of ethanol (or "one in five"). Note that in most cases, a 1/5 dilution may also be written as 1:5. This represents 1 part to five parts total (parts:total). This is not always the case so it is important to be clear before beginning any procedure involving dilution. Dilutions are used a number of ways in the clinical laboratory. To determine volumes in single dilutions a proportion procedure can be used. Example problems: How would you make 500 mL of a ¼ dilution of solution A in water? 4x = 500 mL x = 125 mL of solution A  to 500mL How much of a 1/3 dilution of solution B could you make if you had only 20 mL of concentrated solution B? 3× 20 mL=1x X=60mL of the diluted solution could be prepared. When calculating concentration where a dilution has been made the following formula can be used: **C~(original)~ × d~(dilution)~ = C~(final)~** or **C~o~ × d = C~f~** The original concentration multiplied by the dilution made to it is equal to the final concentration. Example problems: **What is the concentration of the new solution if a 5 M NaCl solution is diluted 1/10?** **5M × 1/10 = 0.5M NaCl** **This formula can also be manipulated to determine the concentration of the original solution.** **\ After diluting a solution 1/20 its concentration is 12 mmol/L. What was the original concentration of this solution?** At times it may be necessary to make many different dilution of a given solution. This may occur when making working standards from a stock standard. Example problem: Your lab has saline stock standard with a concentration of100 mmol/L. Your procedure requires working standards of 10 mmol/L, 20 mmol/L, and 50 mmol/L. What dilution of the stock standard would be required to achieve these concentrations? 100 mmol/L × d = 10 mmol/L d = 10mmol/L ÷100mmol/L d = 10/100 or **1/10** 100 mmol/L × d = 20 mmol/L d = 20mmol/L ÷100mmol/L d = 20/100 or **1/5** 100 mmol/L × d = 50 mmol/L d = 50mmol/L ÷100mmol/L d = 50/100 or **1/2** **1/10, 1/5, and 1/2 dilutions of the stock standard would be requires to achieve the desired working standard concentrations.** **When a very large dilution is required it may be prudent to make multiple dilutions.** **Example problem:** **Your procedure requires that you make a 1/10 000 dilution of a reagent. Rather than making one huge dilution of something like 1 mL in 10 000 mL, you could perform two dilutions.** **First make a 1/100 dilution of the original reagent by diluting 1 mL of the reagent up to 100 mL with diluent, then take 1mL of the diluted reagent and dilute it up to100mL, making a second 1/100 dilution.** **By multiplying the two dilutions together you will get the total dilution of** **1/10 000.** **1/100 × 1/100 = 1/10 000** **Exercise Two** 1. a. What is the new concentration? b. If this solution is re diluted 1/5, what is the final concentration? c. What is the final concentration in mg/dL? MW urea = 60 g/mole 2. a. Two dilutions are prepared from this 'intermediate' standard: i. 3. **Perform the calculations necessary when making either single or serial dilutions of patient sample.** **Activity:** 1. Complete Exercise Three and check your answers. Single Dilutions When performing laboratory tests dilution of patient samples may be required. This can be part of the standard procedure or due to special circumstances such as an unusually high amount of analyte present, or a sample issues like lipemia. After diluting and testing a sample it is then necessary to calculate the original concentration of the analyte present in the patient sample to report to the physician. Example problem: A patient sample was diluted 1/5 with serum diluent and then tested for sodium. The sodium concentration of the diluted sample was determined to be 28 mmol/L. What is the concentration of sodium in the original patient sample? You can use the formula **or** C~o~= C~f~ × the reciprocal of d. C~o~ = 28 mmol/L × 5 C~o~ = 140 mmol/L The concentration of sodium in the patient sample is 140 mmol/L. Serial Dilutions A **serial dilution** is the systematic re dilution of a solution a number of times. As it is usually performed, the procedure is only semi-quantitative and is used in the clinical laboratory to observe a reaction at different concentrations. It is often used in serological procedures to estimate to what dilution a serological reaction is positive. Example: 2.0 mL of diluent is measured into each of 10 test tubes 1.0 mL of serum is added to the first tube and mixed with the diluent. The dilution in this tube is 1/3 (or 1:3 parts:total), meaning in 3 mL of solution, there is 1.0 mL of serum. Then, to continue the serial dilution, 1 mL of the diluted serum is removed by pipet and mixed with the 2.0 mL of diluent in the next tube. This gives another 1/3 dilution. This procedure of removing a measured amount of serum dilution and mixing it with the diluent in the next tube is repeated throughout the remaining tubes. The resulting dilutions of serum would be: ------ ---------------------------- -------------------- Tube **Dilution** **Total Dilution** 1 1/3 = 1/3 2 1/3 X 1/3 = 1/9 3 1/3 X 1/3 X 1/3 = 1/27 4 1/3 X 1/3 X 1/3 X1/3 = 1/81 5 1/3 X 1/3 X 1/3 X1/3 X 1/3 = 1/243 ------ ---------------------------- -------------------- The procedure of serial dilution is inexact and not comparable to dilutions made with volumetric equipment using quantitative technique. For this reason, it is never used in the preparation of reagents or standards. To Determine the Dilution in Any Tube of a Serial Dilution 1. Determine the dilution for the first tube. It may be different from the others. 2. Determine the dilution for successive tubes. It is usually the same dilution for tube \#2 through to the last tube, but not always. 3. Multiply the dilution for the first tube by the dilution for the second and successive tubes. You may use the second dilution to the power of (n-1), where n equals the number of the final tube, for the second and successive tubes if the dilution is the same in all of them. Example problem: A serial dilution is performed by placing 9.0 mL of diluent in the first tube and 2.0 mL of diluent in each successive tube. 1.0 mL of serum is added to the first tube and mixed. 1.0 mL of the mixture in the first tube is serially diluted through the remaining tubes. What is the total dilution in tube \#5? The dilution in tube \#1 is 1/10. The dilution in tube \#2 is 1/10 x 1/3 = 1/30 (one mL of the dilution from tube \#1 is added to the 2 mL of diluent in tube \#2 to give a further 1/3 dilution in that tube, 1/10 × 1/3 × 1/3 = 1/90). The dilution in tube \#5 is 1/10 x 1/3 x 1/3 x 1/3 x 1/3 = **1/810 or 1:810(parts:total)** **Or** **1/10 × 1/3 ^(5\ -1)^ = 1/810 or 1:810 (parts:total)** From these examples it can be seen that from a very small amount of serum and relatively small amounts of diluent, a large dilution can be accomplished using the serial dilution technique. This type of serial dilution may be used in testing to detect endpoint. When reporting these semi-quantitative results to the physician, the dilution of the last tube that reacted, or the titer may be reported. The **titer** is the reciprocal of the dilution in the last tube where a reaction occurred. In the previous example if tube \#5 was the last tube where a reaction occurred the dilution would be 1/810 and the titer would be 810. To Determine the Concentration in Any Tube of a Serial Dilution In some uses of serial dilution it may be necessary to determine the concentration of the diluted substance in a particular tube. To do this the same formula as used for single dilution can be used. First, determine the total dilution in that tube, then multiply the original concentration by that dilution. **Example problems:** 1. 50 mg/L × 1/160 = 0.32 mg/L 2. - - - - b. Exercise Three 1\. In a serial dilution: 1. a. What is the serum dilution in tubes \#4, \#6, and \#7? b. If the original serum glucose concentration is 20.0 mmol/L what would the concentration be in tube \#5? c. If the concentration of urea in tube \#3 was 0.050 mmol /L what was the concentration of urea in the patient's undiluted sample? **Exercise One Answers** 1. How many moles of: a. silver nitrate are in 78.20 g? b. NaCl are in 92 mg? 2. How would you prepare: c. 500 mL of 0.050 M sodium sulfate? d. 200 mL of 0.40 M cupric sulfate pentahydrate? 3. Calculate the osmolarity of the following solutions: a. 2.5 M H~2~SO~4~ **2.5 moles/L x 3 osmoles/mole = 7.5 osmoles/L** b. 1.5 M HCl **1.5 moles/L x 2 osmoles/mole = 3.0 osmoles/L** c. 0.30 M NaOH **0.30 moles/L x 2 osmoles/mole = 0.60 osmoles/L** 4. How would you prepare: a. 500 mL of 0.290 osM NaCl? b. 250 mL of 0.06 osM sodium carbonate 5. How would you prepare 500 mL of 5% (W/V) KCl? 6. How would you prepare 250 mL of 0.85% (W/V) NaCl? 7. A solution of ammonium chloride is 0.50 M. a. What is the concentration expressed as %(W/V)? b. What is the concentration expressed as osmolarity? 8. How would you prepare: a. 500 mL of 28 %(V/V) aqueous solution of methanol from 95 % methanol? b. 250 mL of 85 % (V/V) ethanol from absolute ethanol? 9. How would you prepare a 0.50 molal solution of CaCl~2~ containing 0.500 kg solvent? 10. How would you prepare approximately 1 kg: a. a 0.500 osmolal solution of NaCl? b. a 2.00 osmolal solution of NaCl? 11\. Calculate the pH of a 0.06 mol/L solution of HCl. 12\. What is the pH of solution when one gram of NaOH is added to 500 mL of water? (mw NaOH= 40 g/mol) Moles NaOH = 1 g / 40 g/mol = 0.025 mol in 0.5 L = 0.05 mol/L NaOH **Exercise Two Answers** 1. A 250 mmol/L solution of urea is diluted 1/10. a. b. c. 2. a\. An 'intermediate' standard is prepared by diluting 10 mL of a 1 mmol/L stock standard up to 100 mL in a volumetric flask. What is the concentration of this standard? 1. 3. Working standards for KCl containing 2 mmol/L, 5 mmol/L, and 10 mmol/L were prepared by diluting a stock standard 1/50, 1/20, and 1/10. Using any *one* of these working standards, determine how much KCl had to be used to prepare 1000 mL of the stock standard? **Exercise Three Answers** 1. In a serial dilution: a. What is the serum dilution in tubes \#4, \#6, and \#7? Dilution in tube \#4 = 1/20 x ½ x ½ x ½ = 1/160 b. If the original serum glucose concentration is 20.0 mmol/L what would the concentration be in tube \#5? c. If the concentration of urea in tube \#3 is 0.050 mmol /L what is the concentration of urea in the patient's undiluted sample?