Read the paper, Select a chemical from the list in the table and conduct an experiment with results that should be placed in graphs just as the students did in the paper. Explain your results and grap

Bio 3201 Name______________________

Solution Preparation and Dilutions

Solutions are required for all functions in a laboratory setting. These reagents can be prepared based on four sets of calculations: mass/volume concentrations, % mass/volume concentrations, molarity concentrations, and dilutions of concentrated solutions.

In this laboratory you will:

1. learn proper metric conversions,

2. learn the calculations required for each of the solution types above,

3. determine the calculations for solutions required during the semester, and

4. prepare solutions

Materials

Lab handout

Calculator

material to make solutions

Background

Solution: salt water

Solute: substance that is being dissolved

(salt)

Solvent: substance in which solute is dissolved (water)

Concentration-amount of solute in a volume of solution

Expressed as a: mg/ml, percent, molar, or X solution

Part A. Making Metric Conversions

When preparing solutions, metric units must match for the proper calculations to occur. Use Table 1 to learn how metric units are properly converted.

Table 1. Commonly used metric units in the biotechnology field.

3 g---3000mg---3,000,000 μg

3 L---3000ml---3,000,000 μl

Part A. Making Metric Conversions

Table 1. Conversion Table

Part B. Making Solutions of Differing Mass/Volume Concentrations

Solutions are prepared with a certain mass of solute in a certain volume of solvent or mass/volume concentrations are as follows:

g/L

g/mL

mg/mL

μg/mL

μg/μL

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

mg/mL x mL = mg or g

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

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

1mL ? mg

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.

Part B. Making Solutions of Differing Mass/Volume Concentrations

1. 1. Describe how you would prepare 25 mL of a NaCl solution at a concentration of 0.1 g/mL. After your calculations, MAKE THIS SOLUTION!

Part C. Making Solutions of Differing % Mass/Volume Concentrations

Make a solution at any of % mass/volume. Most commonly, solutions are made with concentrations reported in one of these three measurements:

Measurement Example

mass/volume 4 mg/mL salmon sperm DNA solution

% (in mass/volume or volume/volume) 2% sucrose solution

molarity (moles/liter) 0.5 M TRIS solution

A 1% solution contains 1 g of solute in a total volume of 100 ml

EXAMPLE: a 5% solution has 5g of sugar for 100ml of solution or 5g/100ml

Use the % Mass/Volume Equation shown below to calculate the mass of each solute needed for a solution at some volume. Make sure units that are used can be cancelled out and convert units as necessary.

% Mass/Volume Concentration Equation

Step 1. Convert the % to a decimal

% (percent value) = decimal value of the g/mL (grams needed divided by 100 mL)

Step 2. Determine the amount needed

decimal (g/mL) of the % concentration x total volume desired (mL) = grams of solute

Ex. A technician needs 5 mL of 10% NaOH solution for an experiment. Using the % Mass/Volume Equation, the calculation would be as follows:

Step 1. 10% = 10 g/100 mL = 0.1 g/mL

Step 2. 0.1 g/mL x 5 mL = 0.5 g NaOH

The technician would add 0.5 g NaOH to a container and fill up to the 5 mL mark with solvent.

Part C. Making Solutions of Differing % Mass/Volume Concentrations

1. 1. Describe how you would prepare 200 mL of a 3% NaCl solution. After your calculations, MAKE THIS SOLUTION!

Part D. Making Solutions of Differing 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. A mole of a compound is equal to 6.02 x 1023 molecules. But 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 easier way is to just read the label of a chemical reagent bottle for the listed “MW” or “FW.” For example, the molecular weight of NaCl is 58.5 atomic mass units (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) or millimoles/liter (mmol/L or mM) for smaller concentrations. 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.

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 compound’s molecular weight (g/mol) to account for measuring in moles, as in the following;

Molarity Concentration Equation

volume x molarity x molecular weight = grams of solute to be dissolved in

wanted desired of the solute solvent to the final desired volume

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

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 NaCl

The technician would add 1.46 g NaCl to a container and fill up to the 50 mL mark with solvent.

REMEMBER: 0.5M = 0.5mol/L, 50ml=0.05L

Part D. Making Solutions of Differing Molarity Concentrations

1. Describe how you would prepare 75 mL of a 0.1 M NaCl solution. After your calculations, MAKE THIS SOLUTION & KEEP FOR PART E!!

Part E. 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.” Making a working solution simply required diluting some volume of stock solution to the concentration needed.

The working concentration of a solution is represented as 1X. A concentrated solution could be represented as 10X if it has 10 times the amount of solute per unit volume compared with the working solution. For example, a buffer may be used at a concentration of 0.01 M TRIS. This is the working concentration of the TRIS solution (1X). But because of shipping costs, a small amount of the enzyme buffer is shipped as a 10X solution with a concentration of 0.1 M TRIS. When the technician is ready to use the buffer, it is diluted with deionized water down to 1X (0.01 M TRIS).

If the technician needs 150 mL of the 0.1 M TRIS, then a dilution of the concentrated 1 M TRIS can be calculated. To figure out how to dilute something from a concentrated solution, we use a simple ratio equation as shown in the following equation:

Diluting Concentrated Solutions Equation

C1V1 = C2V2

C1 = the concentration of the concentrated stock solution (the starting solution)

V1 = the volume to use of the stock solution in the diluted sample

C2 = the desired concentration of the diluted sample

V2 = the desired volume of the diluted sample

The C1V1 = C2V2 equation may be used with any concentration units (i.e., mass/volume, %, or molar). Using the equation with the scenario above in which the technician needs 150 mL of the 0.1 M TRIS,

C1 = 0.1 M TRIS (the starting solution)

V1 = amount of the concentrate to use for the dilution

C2 = 0.01 M TRIS (the working solution)

V2 = 150 mL

(0.1 M)(V1) = (0.01 M)(150 mL)

V1 = (0.01 M)(150 mL)/(0.1 M)

V1 = 15 mL

Therefore, to make 150 mL of 0.01 M TRIS from the concentrated stock, measure out 15 mL of the concentrated 0.1 M TRIS stock and add 135 mL of deionized water to is and mix well.

1. Describe how you would prepare 50 mL of a 5 mM NaCl solution from a 0.1 M NaCl stock. After your calculations, MAKE THIS SOLUTION!

Solution Preparation 11