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 M

Remember that the # mol of HCl is
unchanged, so M_{1}V_{1} = M_{2}V_{2}. 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 H _{2}SO_{4} 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 M

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:*

M_{1} = 6.00 M M_{2} = ?

V_{1} = 5.00 mL V_{2} = 14.00 mL

*State
the equation that applies:*

M_{1}V_{1} = M_{2}V_{2}

*Next,
solve for the unknown. In this case, the
unknown is M _{2}:*

_{}

*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*

M_{1} = 0.010 M M_{2} = 0.500 M

V_{1} = 25 mL V_{2} = ?

*(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 M_{1} to the final condition and V_{1} to the initial
condition! Here, in this example, I
happen to assign the final condition as Condition 1.)*

M_{1}V_{1} = M_{2}V_{2}

_{}

**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 K_{2}CrO_{4} from 1.75
M K_{2}CrO_{4} stock solution.

2. 50.00 mL
of concentrated nitric acid (16M HNO_{3}) 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(NO_{3})_{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.

Click here to go on to Question 2.

The concentration
of the new solution is 0.80M HNO_{3}.

Click here to go on to Question 3.

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 (NH_{4})_{2}SO_{4}. Thus, its molar mass is 132.1 g/mol. Starting with 10.8g (NH_{4})_{2}SO_{4},
you use the molar mass to convert this to # moles of (NH_{4})_{2}SO_{4}. Knowing the volume of the stock solution, you
can calculate the molarity of this original solution.

_{}

Next you can
apply the M_{1}V_{1} = M_{2}V_{2} equation to
give you the ans of 0.164M AmmSulf

Click here to go on to Question 4.

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 M

After
applying the M_{1}V_{1} = M_{2}V_{2} equation,
you get M_{2} = 0.0200 M Na_{2}CO_{3}.

_{}

**Ans. is 0.0400 M Na ^{+}**

Click here to go on to Question 5.

20.0 mL of 0.100 M NaCl is needed.

__This
is not a dilution problem! This means
the M _{1}V_{1} = M_{2}V_{2} equation should not
be applied!__

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

Click here to go on to Question 6.

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.

M_{1}V_{1} =
M_{2}V_{2} where M_{1} = 0.300 M, V_{1} = 50.00
mL and M_{2} = 0.60M

Click here to go back to the Questions.