1. Solution Chemistry

2. Lab solutions ………“like a recipe!”  Components required (or materials)  Quantities required (correct volumes and concentrations)  Preparation instructions (or a procedure or protocol)

3. Types of Concentrations  Weight per volume (amount/volume).  Molarity (M): (# moles solute/liter).  Percent concentration. - weight/volume % - volume/volume % (volume percent) - weight/weight % (weight percent) Parts. - parts per million (ppm) - parts per billion (ppb)  Molality (m) - # of molecular weights solute (g)/kilograms  Normality (N) - # of “equivalent weights” solute/liter of solution

4. Weight per volume  The numerator is the amount of solute and the denominator is the total volume of solvent. This is the simplest way to express concentration. For example 2mg/mL of KCl is 2mg in a total volume of 1milliter.  Example:  How much alpha-amylase is required to make 25mL of solution at a concentration of 5mg/mL?  Solution:  ? = 5mg alpha-amylase  25mL1mL  ? = 125mg = amount of alpha-amylase required.

5. Molarity (M)  The numerator is the number of moles of a solute and the denominator is one liter of solution.  Thus a 1 molar solution of any compound contains 1 mole of that compound dissolved in 1 liter. Key terms for review:  The weight of a mole of a given substance = atomic weight (from the periodic table) in grams or its gram atomic weight.  Thus, 1 mole of NaCl = 22.99 + 35.45 = 58.44g dissolved in 1 liter. In this example the gram molecular weight (MW) sometimes called the gram formula weight (FW) of sodium chloride is said to be 58.44g.

6. Molarity cont.  Solute required = (grams/1 mole)(Molarity)(Volume) Example:  How would you prepare 125mL of a 10mM solution of calcium chloride? Solution:  Step 1: Convert 125mL to liters, 125/1000 = 0.125L  Step 2: Convert molarity concentration into moles/L. i.e.,10mM which is equal to 0.010M.  Step 3: Calculate the formula weight (FW) of calcium chloride i.e., 40.08 + 35.45 = 75.53g.  Step 4: substitute values into above formula  Solute required = (75.53g/1mole)(0.010 moles/1liter)(0.125L) = 0.0944g.

7. Percent concentrations  The numerator is the amount of solute and the denominator is 100 units of total solution.  weight per volume % is the weight of solute (in grams) per 100 mL of total solution. The abbreviation used is w/v. eg. 20g of KCl in 100mL of total solution is a 20%,w/v, solution.  volume percent is used when both the amount of solute and the total solution are liquids. The abbreviation used is v/v. eg. 5mL of methanol in 100mL of total solution is a 20% by volume solution.  weight percent (w/w) is the weight of solute in the total weight of solution. This type of expression if often used for viscous materials whose volumes are difficult to measure..

8. Parts  Parts solutions tell you how many parts of each component are mixed together. Each component must have the same units!  Parts per million (ppm): number of parts of solute per 1 million parts of total solution.  Parts per billion (ppb): number of parts of solute per 1 billion parts of total solution. These units are primarily used to express a very small amount of something in a large volume of solvent eg. pollutants in a lake.

9. Molality (m)  These are the concentration units used commonly in biological and chemistry laboratories.  Molality is "the number of molecular weights of solute (in grams) per kilogram of solvent".  This concentration (molality) is often confused with molarity.  - preparing a 1 molal (1 m) solution requires adding 1 mole of solute to 1 kilogram of water.  - preparing a 1 molar (1 M) solution requires adding the water to the solute until the final volume is 1 liter.

10. Normality (N)  Normality is used primarily by chemists.  Reactive equivalents (or equivalent weights) are the ions that participate in the reaction. Thus if we consider acids and bases; - Acids - 1 equivalent weight = number of grams required to produce 1 mole of H + ions. - Bases - 1 equivalent weight = number of grams required to produce 1 mole of OH - ions.  A 1 normal solution of a compound contains 1 equivalent weight of the compound dissolved in a total volume of 1 liter.

11. Dilutions  “ Working solution” and “stock solutions”  C 1 V 1 =C 2 V 2, where, 1 =initial or stock and 2 = final Remember determining how to make a diluted solution from a stock solution is a two step process!

12.  Example:  How would you prepare 250mL of a 0.5M phosphate buffer from a 2 M stock solution of this buffer?  Step 1: Substitute in the known concentrations and volumes to the equation.  C 1 V 1 = C 2 V 2 2 M x (?) = 0.5 M x (250mL) ? = 125 M (mL) 2 M ? = 62.5 mL  Step 2 : Take 62.5 mL of the concentrated stock solution and bring it up to a final volume of 250mL.

13. Dilution Series  Often used in biology or microbiology  “….group of solutions that have same components but at different concentrations”. Each solution is made independently beginning with the stock solution. 1/50 dilution 1/10 dilution (1mL + 49mL) (1mL + 9mL)

14. Serial Dilution  Dilutions are NOT independent of each other.  Used when a dilution cannot be done accurately in one step. Original  a  b  c = 1/1,000,000 dilution Broth 1/100 1/100 1/100

15. Labeling  Identification - chemical name, CAS #, Source, Lot#.  Qualification - grade or purity, expiration date.  Precautions - light sensitivity, temperature and humidity restrictions.

16. Water…the “stuff” of life! What is pure water?  Dissolved inorganics  Dissolved organics  Suspended particles  Dissolved gases  Microorganisms  Pyrogens/endotoxins

17. Water quality control processes Methods of water purification  Resistivity  Bacterial counts  Pyrogens  Organic carbon  pH Monitoring Water Quality  Resistivity  Bacterial counts  Pyrogens  Organic carbon  pH

18. Storage of Biological Solutions  Bactericidal agents  Decreased temperature - liquid nitrogen - rapid/snap freezing - freeze/thaw cycles - addition of glycerol  Lyophilization  Nonionic detergent