Diluting Solutions to Prepare Workable Solutions

Introduction

Chemicals are shipped around the world in their most concentrated forms (solids, concentrated acids, etc.)

If this is not the case, we'd be shipping lots of water along with the chemicals; and this results less cost effective.  Therefore, we need to be able to make solutions of any concentration from a more concentrated source.

Key Idea

• Moles of solute is constant (i.e. the only difference is that there's more water in the less concentrated solution)

Formula

moles solute before = moles solute after

M1L1 = M2L2

M = molarity

L = volume

1 = before

2 = after

Sample Question

Concentrated NaCl is 10 mole/L.  How to make up 2.0 L of 3.0 mole/L?

Step one: decide what which variable we need to find out

M1 = 10 mol/L

M2 = 3.0 mol/L

L1 = ?

L2 = 2.0 L

We need to find L1.

Step two: write down the formula and plug the already known variables in the formula

M1L1 = M2L2

10 mol/L * L1 = 3.0 mole/L * 2.0 L

Step three: solve for the unknown variable

L1 = 3.0 * 2.0 / 10

L1 = 0.60 L

Step four: find out the amount of water needed by subtraction

2.0 L - 0.60 L = 1.4 L of water

Therefore, we need 0.60 L of NaCl and 1.4 L of water to make up 2.0 L of the solution.

Things to keep in mind...

• Watch out for significant figures when solving the questions.
• Note that: solute = chemical in smaller quantity, solvent = chemical in larger quantity

Links for your reference

http://dl.clackamas.cc.or.us/ch105-05/dilution.htm (this is a site with detailed examples and explanations of how to solve dilution problems)

http://www.800mainstreet.com/9/0009-008-dilution.html (a site with examples of dilution problems in various forms)

http://www.graphpad.com/quickcalcs/molarityform.cfm (molarity calculator.  Use it to check your answer)