Chem 121                                  Tutorial on Dilution & Related Chemistry Problems

Fall 2004

INTRODUCTION

Knowing how to solve “dilution problems” is immensely important for students in chemistry, biology and related fields.  Often you will encounter these kinds of problems in future science courses as well as in future place of employment (in biology labs, in pharmacy school, in nursing school, etc.)  For example, you may be asked to prepare a solution of a particular concentration, starting with a more concentrated solution, or you may be asked what the concentration of a solution is if you added a known amount of water to a given solution.

First, let us think about what “dilution” means.  The process of dilution involves starting with a solution of a particular concentration and adding more solvent.  In this tutorial we will focus on aqueous solutions, so the solvent would be water.

It is important to realize in dilution what gets changed and what does NOT get changed.  Obviously when we add more water, the total volume is increased and the concentration is decreased.  Now, think carefully what is NOT changed……                       The answer is the solute is unchanged.  There is no chemical reaction involved, so we still have the same solute.  Also, the amount (# of moles and # of grams) of the solute is unchanged.  This is an important point.

Before we proceed, it is important to remember that a solution labeled

“6M” means 6 moles solute/L solution.  The units of molarity are mol/L.

Now, if we multiply M with V (in units of L), the units tell us we would get moles:

MxV = mol/L x L = mol

You should commit this to memory before proceeding with this tutorial because you will often need to calculate the moles of solute for various problems.

EXAMPLES OF DILUTION PROBLEM

Example 1

Let us consider 5.00 mL of a solution labeled “6.00 M HCl” and we now add enough water to give a total volume of 14.00 mL.  What is the concentration of the new solution?

Strategy:  Think of this as involving two conditions: Condition 1 for the original solution (with M1 and V1) and Condition 2 for the final solution (with M2 and V2).  For Condition 1, the # mol HCl = M1V1 and for Condition 2, the # mol HCl = M2V2.

Remember that the # mol of HCl is unchanged, so M1V1 = M2V2.  This is the equation we will apply to all “dilution problems.”  It also applies to situations where the concentration is increased (by evaporating off some of the water).  It is important to note that this equation should be used ONLY where there is no chemical reaction (no change in the amount of solute).

Consider the question below:

40.0 mL of 6.00 M H2SO4 is neutralized with 50.0 mL of a NaOH solution.  What is the concentration of the NaOH solution?

Before going any further, you should quickly realize this is NOT a “dilution problem” and you should not be using the equation M1V1 = M2V2.  (How you would solve this problem is discussed elsewhere.  I am bringing this up as an example of when you should not be using the dilution equation).

Now, let’s get back to the original question:

Let us consider 5.00 mL of a solution labeled “6.00 M HCl” and we now  add enough water to give a total volume of 14.00 mL.  What is the concentration of the new solution?

Solution:

First decide whether this is a “dilution problem.”

Having determined that it is, translate the word problem into a useful form:

M1 = 6.00 M           M2 = ?

V1 = 5.00 mL          V2 = 14.00 mL

State the equation that applies:

M1V1 = M2V2

Next, solve for the unknown.  In this case, the unknown is M2: Now, plug in the given values, including the units: ANS. The concentration of the new solution is 2.14M HCl.

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Example 2:

How would you prepare 25 mL of 0.010M NaCl starting with a 0.500 M NaCl solution?

Solution:

Is this a “dilution problem?”  Ans. Yes

M1 = 0.010 M         M2 = 0.500 M

V1 = 25 mL             V2 = ?

(Note: It doesn’t matter whether you assign the final condition or the initial condition as Condition 1, as long as you are consistent.  For example, you should be careful not to assign M1 to the final condition and V1 to the initial condition!  Here, in this example, I happen to assign the final condition as Condition 1.)

M1V1 = M2V2 ANS. I would measure out 0.50 mL of the 0.500 M NaCl solution into a 25-mL volumetric flask and dilute it to the mark with deionized water.  Then I would shake it to mix thoroughly.

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PRACTICE PROBLEMS

Below are a few problems for you to practise with.  Take time to write out the setups rather than just punching numbers into your calculator.  After you have the answer you can check it by clicking on the “Ans” link.

1.   Describe how you would prepare 2.00 L of 0.100 M K2CrO4 from 1.75 M K2CrO4 stock solution.

2.   50.00 mL of concentrated nitric acid (16M HNO3) is diluted to 1.000 L.  What is the concentration of the new solution?

3.   A solution is prepared by dissolving 10.8 g ammonium sulfate in enough water to make 100.0 mL stock solution.  A 10.00-mL sample of this stock solution is then placed in a 50.00-mL volumetric flask and diluted to the mark with water.  What is the molarity of the new solution?

4.   30.00 mL of water is added to 20.00 mL of 0.0500 M sodium carbonate.  What is the molarity of sodium ions in the new solution?

5.   What volume of 0.100 M NaCl is needed to react completely with 5.00 mL of 0.200 M Pb(NO3)2 ?

6.   We need a 0.60 M NaCl solution.  Starting with 50.00 mL of 0.300 M NaCl, what must you do?  (Assume there is no other source of NaCl available to you.)

Measure 114 mL of the 1.75 M solution into a 2.00-L volumetric flask and dilute to the mark with water.  Shake it to mix thoroughly.

The concentration of the new solution is 0.80M HNO3.

The molarity of the new solution is 0.164M ammonium sulfate.

(The original solution has a molarity of 0.818M.)

First, you need to be able to figure out that ammonium sulfate is (NH4)2SO4.  Thus, its molar mass is 132.1 g/mol.  Starting with 10.8g (NH4)2SO4, you use the molar mass to convert this to # moles of (NH4)2SO4.  Knowing the volume of the stock solution, you can calculate the molarity of this original solution. Next you can apply the M1V1 = M2V2 equation to give you the ans of 0.164M AmmSulf

The molarity of Na+ is approximately 0.0400M.

(30.00 mL of water added to 20.00 mL of solution gives you approximately 50.00 mL of solution.  Remember that volume is not additive.  By applying the M1V1 = M2V2 equation, the answer gives you the molarity of sodium carbonate.  You need to know the formula of sodium carbonate.  It is Na2CO3 because it tells you that for each unit of Na2CO3 there are two Na+.  Thus the concentration is doubled:

After applying the M1V1 = M2V2 equation, you get M2 = 0.0200 M Na2CO3. Ans. is 0.0400 M Na+

Ans to Question 5:

20.0 mL of 0.100 M NaCl is needed.

This is not a dilution problem!  This means the M1V1 = M2V2 equation should not be applied!

How you solve this problem is not shown here.  This is a “stoichiometry problem.”

0.60 M is twice the concentration of the original solution.  The 50.00 mL of the original solution needs to be concentrated down to half its volume.  So, we need to evaporate the solution down to approximately 25 mL.

M1V1 = M2V2 where M1 = 0.300 M, V1 = 50.00 mL and M2 = 0.60M