In the previous unit, we learned about molarity and practiced converting between molar units. In this unit, we'll practice how to create a solution of a specific concentration.

To make a solution with a specific molarity, you need to know two things:

  1. How much solute (in grams) to add, and
  2. How much solution (in liters) you want to make.

This is where molecular weight (also called molar mass) comes in: it tells you how many grams are in one mole of a substance. In addition to molarity, molecular weight (MW) is an important conversion factor in these types of laboratory approaches.

Let's go through an example: Tris, or trisaminomethane, is commonly used as a component in buffer solutions. Tris has a molecular weight of 121 g/mole. Let's say that you want to make 1L of a 1.5M Tris buffer. How many grams of Tris would be needed to create this solution? To solve this problem,start with the volume, use a conversion factor of molarity (moles/liter), and then use a conversion factor of molecular weight (MW = g/mole).

\[\mathbf{\mbox{1 L} * (\frac{\mbox{1.5 moles}}{\mbox{1 L}}) * (\frac{\mbox{121 g}}{\mbox{1 mole}}) = \mbox{181.5 g}}\] In other words, 181.5 g of Tris would be added to 1 L of water to create the 1.5 M solution.

Let's practice this concept:

  • You want to make 1 L of a 5 M sodium bicarbonate (MW = 84 g/mol) buffer solution. How many grams of sodium bicarbonate would be needed to create this solution? g

\[\mathbf{\mbox{1 L} * (\frac{\mbox{5 moles}}{\mbox{1 L}}) * (\frac{\mbox{84 g}}{\mbox{1 mole}}) = \mbox{420 g}}\]

  • You want to make 500 mL of a 6 M HEPES (MW = 238 g/mol) buffer solution. How many grams of HEPES would be needed to create this solution? g

  • You want to make 300 mL of a 1 M sodium diphosphate buffer (MW = 141 g/mol) solution. How many grams of sodium diphosphate would be needed to create this solution? g

  • You want to make 500 mL of a 500 mM Tris (MW = 121 g/mol) buffer solution. How many grams of Tris would be needed to create this solution? g

  • You want to make 100 μL of a 100 mM sodium bicarbonate (MW = 84 g/mol) buffer solution. How many milligrams of sodium bicarbonate would be needed to create this solution? mg

  • You want to make a HEPES stock solution and in discover there is 238 g left in the bottle. What is the maximal amount of 2 M buffer you can make with this amount of HEPES (MW = 238 g/mol)? L

  • You want to make a Tris buffer stock solution and in discover there is only 15 g left in the bottle. What is the maximal amount of 500 mM buffer you can make with this amount of Tris (MW = 121 g/mol)? ml

The concentrations of some solutions is not given in molarity but in weight/volume. A w/v (weight/volume) solution is a way of expressing concentration based on the mass of solute (in grams) per volume of solution (in milliliters). Weight/volume solutions are indicated by percentages:

There are two important things to remember when creating a w/v solution:

  1. The volume is always presented in milliliters (not liters).
  2. The conversion factor for w/v is the number of grams per 100 mL.

Let's practice this concept:

  • You want to make 100 mL of a 10% w/v sodium bicarbonate (MW = 84 g/mol) buffer solution. How many grams of sodium bicarbonate would be needed to create this solution? g

\[\mathbf{\mbox{100 mL} * (\frac{\mbox{10 g}}{\mbox{100 ml}}) = \mbox{10 g}}\]

  • You want to make 500 mL of a 10% Tegosept solution. How many grams of Tegosept would be needed to create this solution? g

  • You want to make 400 mL of a 4% SDS (sodium dodecyl sulfate) solution. How many grams of SDS would be needed to create this solution? g

  • You want to make a 0.8% agarose gel. How many grams of agarose would be needed to create 100 ml of this solution? g

  • You want to make a 1.4% saline solution. How many grams of NaCl would be needed to create 600 ml of this solution? g

You may have noticed in the previous module that most of the solutions we worked with had round numbers like 1 M, 5 M, or 10 M. That’s because these are known as stock solutions. A stock solution is a concentrated solution that’s made ahead of time and kept in the lab. It’s like a “master mix” that you can use to quickly make smaller, less concentrated solutions called working solutions when you need them.

Why work from stock solutions? Working from a stock solution saves time, materials, and increases accuracy. Instead of measuring tiny amounts of a substance to make a weak solution, you can take a small volume of a concentrated stock and dilute it to get the exact concentration you want.

When making dilutions from stock solutions, you can use the formula \[\mathbf{C_1*V_1 = C_2*V_2}\]

The relationship between these variables is represented by the equation \[\mathbf{C_1*V_1 = C_2*V_2}\] When making a working solution, you'll often be told the concentration of the stock solution (ie 1 M) and the concentration and total volume of the working solution (ie 10 mL of a 50 mM solution). Therefore, you are given C1, C2, and V2. To solve for V1, divide C2*V2 by C1. In other words:

\[\mathbf{V_1} = \frac{C_2*V_2}{C_1}\] \[\mathbf{{V_1} = \frac{\mbox{0.05 M * 0.01 L}}{\mbox{1 M}} = \mbox{0.0005 L (or 0.5 mL)}}\] In other words, 0.5 mL of 1 M stock solution would be added to 9.5 mL of solute to create 10 mL of a 50 mM (or 0.05 M) solution.

There are two important things to remember when creating a working solution from a stock solution:

  1. Always treat C1 and V1 as the stock solution and C2 and V2 as your working solution.
  2. Make sure your units are in agreement (ie convert from mM to M or from mL to L, etc).

Let's practice this concept:

  • Your lab manager asks you to make 500 mL of a 100 mM solution. You have a 1 M stock solution and a jug of deionized water. To create this solution, how many many mL of the stock solution would used to create 500 mL of the 100 mM working solution? mL

\[\mathbf{C_1*V_1 = C_2*V_2}\] \[\mathbf{(\mbox{1 M})*V_1 = (\mbox{0.1 M})*(\mbox{500 ml})}\] \[\mathbf{V_1 = \mbox{50 ml}}\]

In other words, 50 mL of a 1 M stock solution would be added to 450 mL of water to create a 100 mM (or 0.1 M) working solution.

  • Your lab manager asks you to make 15 mL of a 100 μM solution. You have a 10 mM stock solution and a jug of deionized water. To create this solution, how many many μL of the stock solution would used to create 15 mL of the 100 μM working solution? μL

  • Your lab manager asks you to make 210 mL of a 40 μM solution. You have a 500 μM stock solution and a jug of deionized water. To create this solution, how many many mL of the stock solution would used to create 210 mL of the 40 μM working solution? mL

This next set of questions asks you to calculate the amount of stock solution that would be used as well as the amount of water. Note that these numbers should add up to the total volume.

  • Your lab manager asks you to make 50 mL of a 25 mM solution. You have a 1 M stock solution and a jug of deionized water. To create 50 mL of a 25 mM solution, add mL of the 1 M solution to mL of water.

  • Your lab manager asks you to make 40 mL of a 0.2 μM solution. You have a 100 μM stock solution and a jug of deionized water. To create 40 mL of a 0.2 μM solution, add μL of the 100 μM solution to mL of water.

  • Your lab manager asks you to make 10 mL of a 25 μM solution. You have a 1 M stock solution and a jug of deionized water. To create 10 mL of a 25 μM solution, add μL of the 1 M solution to mL of water. NOTE: Report value to the nearest tenth and round up (e.g. an answer of 10.38 would be reported as 10.4).

    • This solution is problematic because

A solution to this conundrum is to

Congrats on completing the unit on Making Solutions! Click here to go to the next module: Creating antibody solutions.





Did you notice a typo, an incorrect answer, or another aberration? Send an email here to let us know.