Dilutions

Dilutions

Introduction

Preparing solutions of a specific concentration is one of the most common tasks performed in a chemistry laboratory. Rather than weighing out solute every time a solution is needed, chemists often prepare or purchase a concentrated stock solution and then dilute it to the desired working concentration.

The mathematics of dilution is based on a simple but extremely important principle:

Adding solvent changes the volume of a solution, but it does not change the number of moles of solute present.

Because the amount of solute remains constant during dilution, a relationship can be established between the initial concentration and volume and the final concentration and volume.

This relationship leads directly to the dilution equation:

M1V1=M2V2M_1V_1=M_2V_2

where:

Table showing the relationship between initial molarity, volume of stock solution, final molarity after dilution, and final total volume after dilution for various variables.

Key Point:

The dilution equation works because the number of moles of solute remains constant during dilution. Only the volume changes.

Figure 1 . Dilution of a Stock Solution

Illustration of diluting a concentrated stock solution by transferring 10.00 mL into a volumetric flask and adding distilled water to achieve a final concentration of 0.100 M.

The Dilution Equation

The dilution equation can be derived directly from the definition of molarity.

Recall:

M=mol soluteL solutionM=\frac{\text{mol solute}}{\text{L solution}}

Before dilution:

mol solute=M1V1\text{mol solute}=M_1V_1

After dilution:

mol solute=M2V2\text{mol solute}=M_2V_2

Since the number of moles remains unchanged:

M1V1=M2V2M_1V_1=M_2V_2

The equation can be rearranged to solve for any unknown variable.

V1=M2V2M1V_1=\frac{M_2V_2}{M_1}

M2=M1V1V2M_2=\frac{M_1V_1}{V_2}

V2=M1V1M2V_2=\frac{M_1V_1}{M_2}

M1=M2V2V1M_1=\frac{M_2V_2}{V_1}

Volumes may be expressed in liters or milliliters as long as both volumes use the same unit.

Key Point:

Both volume terms must be expressed in the same units. Both can be liters or both can be milliliters.

Understanding What Changes During Dilution

During a dilution:

• The number of moles of solute remains constant.

• The volume increases.

• The concentration decreases.

For example, suppose a solution contains:

0.500 mol solute0.500\ \text{mol solute}

in

0.250 L0.250\ \text{L}

The concentration is:

M=0.5000.250=2.00 MM=\frac{0.500}{0.250}=2.00\ \text{M}

If additional water is added until the volume becomes:

1.00 L1.00\ \text{L}

the moles remain:

0.500 mol0.500\ \text{mol}

The new concentration becomes:

M=0.5001.00=0.500 MM=\frac{0.500}{1.00}=0.500\ \text{M}

The concentration decreased because the same amount of solute is spread throughout a larger volume.

Key Point:

Dilution does not remove solute. It simply spreads the same amount of solute through a larger volume of solution.

Figure 2. Concentrated vs. Dilute

Illustrating the dilution principle, the image compares a concentrated solution with a smaller volume and higher concentration to a diluted solution with a larger volume and lower concentration, while maintaining the same number of solute particles.

Worked Example 1: Simple Dilution

How many milliliters of a 6.00 M HCl stock solution are needed to prepare 250 mL of 0.500 M HCl?

Step 1: Identify the Variables

Table showing the concentrations of various solutions, including a missing value, to assist in understanding their relative measurements.

Step 2: Rearrange the Equation

V1=M2V2M1V_1=\frac{M_2V_2}{M_1}

Step 3: Substitute Values

V1=(0.500 M)(250 mL)6.00 MV_1=\frac{(0.500\ \text{M})(250\ \text{mL})}{6.00\ \text{M}}

Step 4: Calculate

V1=20.8 mLV_1=20.8\ \text{mL}

Answer

20.8 mL20.8\ \text{mL}

of 6.00 M HCl is required.

Laboratory Procedure

1. Measure 20.8 mL of 6.00 M HCl.

2. Transfer the acid to a 250 mL volumetric flask.

3. Add distilled water until the solution reaches the calibration mark.

4. Mix thoroughly.

Key Point:

The final volume is 250 mL. You do not add 250 mL of water. You add enough water so that the total solution volume becomes 250 mL.

Figure 3. Making a Solution

Step-by-step instructions for preparing a 250.0 mL diluted sulfuric acid solution from a 4.00 M stock solution using a volumetric flask, ensuring precise concentration adjustment.

Laboratory Safety During Dilutions

Dilutions involving strong acids require special care.

When water is added to concentrated acid, substantial heat may be released.

For safety:

Always add acid to water, never water to concentrated acid.

Adding water directly to concentrated acid can cause rapid boiling and dangerous splattering.

Key Point:

Always add acid to water because the larger volume of water safely absorbs the heat released during mixing.

Figure 4. Do As You Otta, Add Acid to Water

Learn the correct and safe method for diluting concentrated acid by adding acid to water, avoiding the dangerous reaction caused by adding water to acid. Key safety points include wearing protective gear and working in a well-ventilated area.

Worked Example 2: Finding the Final Concentration

A student transfers 15.0 mL of 2.00 M NaCl solution into a flask and dilutes the solution to a final volume of 60.0 mL.

Determine the final concentration.

Step 1: Identify Variables

Table displaying variable values and corresponding measurements for a specific experiment, including unknown quantities.

Step 2: Rearrange the Equation

M2=M1V1V2M_2=\frac{M_1V_1}{V_2}

Step 3: Substitute Values

M2=(2.00 M)(15.0 mL)60.0 mLM_2=\frac{(2.00\ \text{M})(15.0\ \text{mL})}{60.0\ \text{mL}}

Step 4: Calculate

M2=0.500 MM_2=0.500\ \text{M}

Answer

0.500 M0.500\ \text{M}

The volume increased by a factor of four:

60.015.0=4\frac{60.0}{15.0}=4

The concentration therefore decreased by the same factor:

2.004=0.500\frac{2.00}{4}=0.500

Key Point:

Volume and concentration change in opposite directions. If volume increases by a factor of four, concentration decreases by a factor of four.

Dilution Factor

Many dilution calculations can be understood using a dilution factor.

Dilution Factor=V2V1\text{Dilution Factor}=\frac{V_2}{V_1}

or equivalently:

Dilution Factor=M1M2\text{Dilution Factor}=\frac{M_1}{M_2}

For Example 2:

60.015.0=4\frac{60.0}{15.0}=4

The dilution factor is 4.

This means the final concentration is one-fourth of the original concentration.

Key Point:

A dilution factor tells you how many times more spread out the solute becomes after dilution.

Figure 5. The Dilution Factor

The diagram illustrates how dilution affects solution concentration, showing that as volume increases, concentration decreases while the moles of solute remain constant. Key equations and graphs highlight the relationship between initial and final volumes and concentrations.

Serial Dilutions

Sometimes an extremely dilute solution is needed.

Instead of attempting one very large dilution, chemists often perform a series of smaller dilutions.

This process is called a serial dilution.

Serial dilutions are widely used in:

• Microbiology

• Analytical chemistry

• Environmental chemistry

• Pharmacology

• Medical diagnostics

At each stage:

M1V1=M2V2M_1V_1=M_2V_2

is applied independently.

The product of one dilution becomes the stock solution for the next dilution.

Key Point:

Each dilution step is treated as a completely separate dilution problem.

Figure 6. Serial Dilution

Illustrating the concept of progressive dilution, this chart shows how the concentration of a solution decreases by half with each equal-volume step, highlighting changes in color intensity and solute concentration.

Worked Example 3: Serial Dilution

Prepare the following sequence:

3.00 M→2.50 M→2.00 M→1.50 M3.00\ \text{M}\rightarrow2.50\ \text{M}\rightarrow2.00\ \text{M}\rightarrow1.50\ \text{M}

Each solution should have a final volume of 10.0 mL.

Step 1: 3.00 M → 2.50 M

V1=(2.50 M)(10.0 mL)3.00 MV_1=\frac{(2.50\ \text{M})(10.0\ \text{mL})}{3.00\ \text{M}}

V1=8.33 mLV_1=8.33\ \text{mL}

Transfer 8.33 mL of 3.00 M stock and dilute to 10.0 mL.

Result:

10.0 mL of 2.50 M10.0\ \text{mL of 2.50 M}

Step 2: 2.50 M → 2.00 M

The stock is now the 2.50 M solution produced in Step 1.

V1=(2.00 M)(10.0 mL)2.50 MV_1=\frac{(2.00\ \text{M})(10.0\ \text{mL})}{2.50\ \text{M}}

V1=8.00 mLV_1=8.00\ \text{mL}

Result:

10.0 mL of 2.00 M10.0\ \text{mL of 2.00 M}

Step 3: 2.00 M → 1.50 M

The stock is now the 2.00 M solution from Step 2.

V1=(1.50 M)(10.0 mL)2.00 MV_1=\frac{(1.50\ \text{M})(10.0\ \text{mL})}{2.00\ \text{M}}

V1=7.50 mLV_1=7.50\ \text{mL}

Result:

10.0 mL of 1.50 M10.0\ \text{mL of 1.50 M}

Key Point:

The stock solution changes after every step in a serial dilution.

Common Errors

Table showing common errors in volume calculations and dilutions, highlighting misconceptions and clarifying proper practices for accurate results.

Key Point:

The most common dilution mistake is confusing the volume of stock solution used with the final volume of the diluted solution.

Summary

Dilution calculations are based on the conservation of solute moles. When solvent is added, the amount of solute remains unchanged while the solution volume increases, causing the concentration to decrease. This relationship is expressed by the dilution equation M1V1=M2V2M_1V_1=M_2V_2.

Dilution calculations can be used to determine stock volumes, final concentrations, final volumes, or stock concentrations. Serial dilutions apply the same equation repeatedly, using the product of one dilution as the stock for the next. These calculations are among the most frequently used quantitative techniques in chemistry laboratories.

Key Points

• Dilution conserves moles of solute.

• The dilution equation is:

M1V1=M2V2M_1V_1=M_2V_2

• Concentration decreases as volume increases.

• Both volume terms must use the same units.

• Dilution always proceeds from higher concentration to lower concentration.

• Final volume is not the same as solvent volume added.

• Dilution factor describes how much a solution has been diluted.

• Serial dilutions use the previous solution as the stock for the next step.

• Strong acids should always be added to water, not vice versa.

• Many laboratory solution preparations rely on dilution calculations.