Solution Chemistry

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 • 25mL 1mL • ? = 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 itsgram 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 isw/v.eg. 20g of KCl in 100mL of total solution is a 20%,w/v , solution. • volume percentis used when both the amount of solute and the total solution are liquids. The abbreviation used isv/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” • C1V1=C2V2 , 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. • C1V1 = C2V2 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 < -- STOCK -- > 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. Methods of water purification Resistivity Bacterial counts Pyrogens Organic carbon pH Monitoring Water Quality Resistivity Bacterial counts Pyrogens Organic carbon pH Water quality control processes

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