Related Products:

Society for Amateur Scientists

Determining the Amount of Reagent Needed for Solution Preparation

Before you prepare a solution you must calculate the correct amount of reagent that is to be diluted to make the solution. Several examples of how to make such calculations are given below:

 Approximate Dilution of a Concentrated Reagent
What volume concentrated nitric acid is needed for the preparation of 100 mL of 6.0 M HNO₃? The calculation is as follows:

(moles of reagent in concentrated solution) = (moles of reagent in diluted solution)

(15.8 moles/liter conc. HNO₃)(X liters) = (6.0 moles/liter dilute HNO₃)(0.100 liter)

X = 0.038 L = 38 mL concentrated HNO₃

Use a graduated cylinder to measure 38 mL of concentrated nitric acid and add this amount to approximately 60 mL of distilled water in a plastic bottle. Mix thoroughly, let the solution cool, then cap and label the container.

Finding the molarity

  If the molarity of the reagent is not known, you must find it from the given value of specific gravity and percent composition. For all practical purposes, the specific gravity of a solution is numerically equal to its density. For example, what is the molarity of concentrated nitric acid given that it is 70% HNO₃ by mass and has a specific gravity of 1.42?

First, calculate the mass of one liter of solution:

1.42 g conc. soln.      1000 mL conc. soln.     1.42 x 10³ g soln.
--------------- x ----------------- = ---------------
1.0 mL conc. soln.      1.0 L conc. soln.        1.0 L conc. soln.
 

The solution is 70% by mass, so 100 g of solution contains 70 g of pure HNO₃. The mass of HNO₃ in one liter of concentrated solution is:

1.42 x 10³ g conc. soln.       70 g HNO₃        9.94 x 10² g HNO₃
------------------- x ---------------= ------------------
    1.0 L conc. soln.         100 g conc. soln.        1.0 L conc. soln.

To find the number of moles in one liter of solution (the molarity):

9.94 x 10² g HNO₃      1.0 mole HNO₃       15.8 moles HNO₃
----------------- x -------------- = ---------------
  1.0 L conc. soln.         63.0 g HNO₃         1.0 L solution

Quantitative Dilution of a Standard Reagent

(moles of reagent in concentrated solution) = (moles of reagent in dilute solution)

(2.00 M concentrated HCl)(X liters) = (0.100 M dilute HCl)(0.100 L)
 

      (0.100 mole/L HCl)(0.100L)
X = ---------------------- = 5.00x10^-3 L = 5.00 mL HCl
         2.00 moles/liter HCl
 

The solution is then prepared by pipetting 5.00 mL of 2.00 M HCl solution and diluting it in a 100-mL volumetric flask as described previously.

Preparation of a Solution from a Solid Suppose a you wish to prepare one liter of a 20.0% (w/v) sodium chloride solution. Find the amount of NaCl needed.

  20.0 g NaCl
------------- x 1000 mL soln. = 2.00x10² g NaCl
100 mL solution
 

Dissolve 2.00x10² g of NaCl in a 1-Liter volumetric flask as described previously.
 

As another example, what mass of AgNO₃ is needed to prepare 250 mL of a 1.00 M solution?

1.00 mole AgNO₃                         107 g AgNO₃
-------------- x 0.250 L solution x ------------ = 26.8 g AgNO₃
 1.0 L solution                              mole AgNO₃

Dissolve 26.8 g of silver nitrate in a 250-mL volumetric flask as described previously.

Reprinted from: