Comparing Mitosis and Meiosis

Solution Preparation and Dilutions

Part A. Making Solutions using Mass/Volume ConcentrationsMany solutions are prepared with a certain mass of solute in a certain volume of solvent. Any metric mass in any metric volume is possible, but the most common units of mass/volume concentrations are as follows:

g/mLgrams per milliliter

g/Lgrams per liter

mg/mLmilligrams per milliliter

g/mLmicrograms per milliliter

g/Lmicrograms per microliter

ng/Lnanograms per liter

ng/Lnanograms per microliter

To determine how to prepare a certain volume of a solution at a certain mass/volume concentration, use the equation below. Convert units as necessary to make sure units that are used can be cancelled out.

Mass/Volume Concentration Equation

concentration desired x total volume desired = mass of solute in the total volume desired

(ex. g/mL) (ex. mL) (ex. g)Ex. A technician needs 50 mL of 15 mg/mL pepsin solution for an experiment. Using the Mass/Volume Equation, the calculation would be as follows:

15 mg/mL x 50 mL = 750 mg = 0.75 g pepsin

The technician would add 0.75 g pepsin to a container and fill up to the 50 mL mark with solvent (usually deionized water). Most balances weigh in grams, so the conversion from mg to g was necessary. Determine the calculations for the solutions in the Practice Problems section for Part A. Making Solutions Using Mass/Volume Concentrations at the end of this handout.

Part B. Making SolutionsUsing Molarity Concentrations

The concentration of many solutions is reported as moles/liter (mol/L or M; the M is spoken molar) or some function of those units. This concentration measurement is called molarity. Molarity is sometimes a challenging concept to understand. However, with your recently acquired solution preparation skills, you will see that making molar solutions requires only one extra calculation.

To understand how to make a solution of a given molarity, you must know what a mole is. A mole of a compound is equal to 6.02 x 1023 molecules, but that is not really a very useful number. So, in biotechnology, it is easier to use this definition: The unit 1 mole is the mass, in grams, equal to the molecular weight (MW), also called formula weight (FW), of the substance. The FW can be determined by using a Periodic Table or by adding the atomic weights of the atoms in the molecule. An easy way, though, is to just read the label of a chemical reagent bottle, which lists the MW or FW. The molecular weight of NaCl is 58.5 atomic mass units (amu) since the Na atom weighs 23 amu, and a Cl atom weighs 35.5 amu.

Molarity concentrations are reported as the number of moles per liter (mol/L or M). If the concentration is very low, then the concentration could be reported in millimoles/liter (mmol/L or mM). If you wanted a 1-M NaCl solution, you would measure out 1 mole of NaCl (58.5 g) and dissolve it in water to a total volume of 1 L. This gives you 1 mole of NaCl per liter of solution, 1 M NaCl.

A liter of solution is a large volume for most research and development purposes. In research and development labs, mL or L quantities are usually used. To determine how to mix up a smaller volume of a solution of some molarity, follow the example below.

Multiply the volume desired (L) by the concentration (molarity) desired (mol/L), as you did in the mass volume calculations. Then, multiply the result by the compounds molecular weight (g/mol) to account for measuring in moles, as in the following equation:Molarity Concentration Equation

volume xmolarity xmolecular weight =grams of solute to be dissolved in

wanteddesired of the solutesolvent to the final desired volume

(L)(mol/L) (g/mol)

Convert smaller or larger units to these as necessary. The L units cancel out and the mol units cancel out, leaving the mass in grams of the solutes needed to make the solution.

Ex. A technician needs 50 mL of 0.5 M NaCl solution for an experiment. Using the Molarity Concentration Equation, the calculation would be as follows:

0.05 L x 0.5 mol/L x 58.5 g/mol = 1.46 g NaClThe technician would add 1.46 g NaCl to a container and fill up to the 50 mL mark with solvent.

Determine the calculations for the solutions in the Practice Problems section for Part B. Making Solutions Using Molarity Concentrations at the end of this handout.Part C. Making Dilutions of Concentrated Solutions

Making dilutions of concentrated solutions is a common practice in a biotechnology lab. A concentrated solution is generally called a stock solution, and the diluted solution is called the working solution. Preparing a concentrated stock solution saves a lot of time and is easier to store than large volumes of diluted working solutions. Making a working solution simply requires diluting some volume of stock solution to the concentration needed. When a number of dilutions must be made, and each is proportionally the same dilution as the one before, it is called a serial dilution. Doing a serial dilution makes sense for many experiments when many samples of varying concentrations are needed. A serial dilution is also useful for preparing very dilute solutions that are hard to make from scratch, because the solute masses can be too small to measure on a balance. Each succeeding sample is made with the

same ratio of sample and diluent as the one before

1 M NaCl 0.5 M NaCl

0.25 M NaCl0.125 M NaCl

Each of the above dilutions is one part previous sample and one part solvent. This is called a 1:2 dilution, or one part sample to two total parts.

To figure out how to prepare a working solution from a stock solution, we use the process bulleted below: Restate the problem:

What do you have, i.e., what is the concentration of the stock solution and how much of it do you have?

What is needed, i.e., what is the concentration of the working solution and how much of it do you need?

Convert all concentrations and volumes to the same units.

Calculate the Dilution Factor: Concentration of stock/Concentration of working solution

Calculate the volume of stock needed: Volume needed/Dilution factor

Calculate the volume of solvent (usually water) needed: Volume of working solution needed Volume of stock needed

For example, a technician needs to prepare 150 mL of 0.1 M TRIS (the working solution), from 100 ml of a stock soltuion of 1 M TRIS. What you have: 100 ml of 1.0 M Tris What you need: 150 ml of 0.1M Tris

All volumes and concentrations are already in the same units

The dilution factor = 1M Tris/0.1M Tris = 10

The volume of stock needed = 100/10 = 10ml

The volume of solvent needed = 100 ml 10 ml = 90 ml of water.

Determine the calculations for the solutions in the Practice Problems section for Part C. Making Dilutions of Concentrated Solutions at the end of this handout.

Practice ProblemsPart A. Making Solutions Using Mass/Volume Concentrations

1. Describe how you would prepare 25 mL of a NaCl solution at a concentration of 2.5 g/mL.

2. Describe how you would prepare 2 L of a 0.5 g/mL dextrose solution.

Part B. Making Solutions Using Molarity Concentrations

1. Describe how you would prepare 125 mL of a 10 M NaOH solution.

2. Describe how you would prepare 75 mL of a 0.1 M NaCl solution.

Part C. Making Dilutions of Concentrated Solutions

1. Describe how you would prepare 950 mL of a 1M CuSO45H2O solution from a 25M CuSO45H2O stock.2. Describe how you would prepare 50 mL of a 5 mM NaCl solution from a 1 M NaCl stock. 500 mL

0.25M NaCl

500 mL

0.5M NaCl

500 mL

1M NaCl

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Solution Preparation 6