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Making solutions

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  • 1. Making Solutions Biotechnology I
  • 2. Introduction  Making solutions is important in any area of biotechnology  R&D QC Mfg  Accuracy is critical as an incorrect solution can destroy months or years worth of work and could delay a critical drug from being available to the market place
  • 3. Key Terms  Solute: substances that are dissolved  Solvent: substances in which solutes are dissolved (often times this is water or a buffer)  Concentration: amount per volume  Solution: a homogeneous mixture in which one or more substances are dissolved in another.
  • 4. Concentration 5 mL If each star represents 1 mg of NaCl, what is the total amt. of NaCl in the tube? What is the conc. of NaCl in the tube in mg/mL? 4 mg 4 mg / 5mL = 0.8 mg/mL
  • 5. Calculations  There are a few equations you need to remember in order to determine the required amount of solute and solvent required.  Take your time with your calculations  Remember to cancel units  Ask yourself if the answer makes sense  Record all calculations and formulas used
  • 6. Weight per Volume  Simplest way of expressing a concentration  Example: 2mg / mL  2mg is the weight of something in 1 mL of solvent
  • 7. Percents  May be expressed as weight per volume which is grams per 100 mL (w/v%)  Example: 20 grams of KCL in 100 mL of solvent is a 20% solution (w/v)
  • 8. Practice Problem  How would you prepare 500 mL of a 5% (w/v) solution of NaCl?  5% = 5g/100mL  5 = x 100 500mL 100x = 2500 x=25grams of NaCl  Weigh out 25 grams of NaCl and add it to a 500 mL volumetric flask. Add approximately 400 mL and mix. Once dissolved, QS to 500 mL. Cap and store.
  • 9. Percents continued  May be expressed in volume percent which is (v/v): mL of solute 100 mL of solvent  May be expressed in weight per weight which is (w/w): grams solute 100 grams solvent
  • 10. Practice Problem  How would you make 500 grams of a 5% solution of NaCl by weight (w/w)? % strength is 5% w/w, total weight is 500 grams 5 grams x 500 g = 25 grams of NaCl 100 grams 500 grams total weight – 25 grams NaCl = 475 grams of solvent needed Weigh out 25 grams of NaCl and add it to 475 grams of water.
  • 11. Molarity  Most common solution calculation used  Definition: molarity (M) is equal to the number of moles of solute that are dissolved per Liter of solvent.  In order to calculate molarity, you need to know moles.  Definition: moles are the molecular weight or formula weight of a particular chemical.  FW or MW can be found on the chemical bottle or by adding up atomic weights of each atom in that molecule.  For example: what is the MW of H20  2x1 + 16 = 18
  • 12. Practice Calculations  What is the MW for Sulfuric Acid?  H2SO4  2 x 1.00 = 2.00  1 x 32.06= 32.06  4 x 16 = 64.00 98.06  A 1M solution of sulfuric acid contains 98.06 g of sulfuric acid in 1 L of total solution  Mole is an expression of amount  Molarity is an expression of concentration
  • 13. Key unit conversions  Millimolar (mM)  A millimole is 1/1000 of a mole  Micromolar (µM)  A µmole is 1/1,000,000 of a mole
  • 14. Practice Problem  Prepare 100 mL of a 0.1M Tris buffer  Step 1: what is the MW of Tris  FW = 121.1 g/mole  What formula are you going to use  Molarity  moles per liter  121.1 grams per 1L or 1000 mL = moles  How much Tris are you going to weight out?
  • 15. Practice Problem continued  Start with what you know  1M = 121.1 grams/1L  To make 100 mL of 0.1M solution you would add how much Tris buffer?  Formula= FW x molarity x volume (L) = g needed 121.1 x 0.1 x 0.1L = 1.211 grams Tris Weight out 1.211 grams of Tris and add it to 80 mL of water, mix and QS to 100 mL of 0.1M Tris soln.
  • 16. Practice Problem  How much solute is required to make 400 mL of 0.8 M CaCl2  First what is the FW of CaCl2?  40.08 + (35.45x2) = 110.98  110.98 x 0.8 x 0.4L = 35.51 grams CaCl2 required  Weigh out 35.51 grams CaCl2 and add it to approximately 300 mL of water. Mix and then QS to 400 mL using a graduated cylinder.
  • 17. Preparing dilute solutions from concentrated solutions  Formula: C1xV1 = C2xV2  C1: concentration of stock solution  V1: volume of stock solution required  C2: concentration you want your dilute solution to be  V2: how much of the dilute solution is required
  • 18. Practice Problem  How would you prepare 500 mL of a 1M Tris buffer from a 3M stock solution?  What is C1? 3M  What is V1? Unknown  What is C2? 1M  What is V2? 500 mL  3x = 1(500)  X=166.67 mL  Measure 166.67 mL of 3M Tris and add it to 333.33 mL of water to make 500 mL of 1M Tris
  • 19. Serial Dilutions or Stock Solutions Formula  C1 V1 = C2 V2  Example: Prepare 100 mL of 0.15M KCL from 10 M KCL stock solution (0.15M)(100mL) = (10M)(x mL) 15 = 10x X=1.5 mL You would add 1.5 mL of the 10M KCL solution to 98.5 mL of water for a total volume of 100 mL and a final concentration of 0.15M
  • 20. Serial Dilutions  In this type of dilution, an aliquot is taken from a stock tube and placed in the first tube. A volume from the second tube is then transferred to the 3rd tube. 1:2 Add 1.0 mL to 1.0 mL 1:4 1:8

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