Overview

Dilution occurs when a substance containing none of the component being tracked is added to an existing mixture. The most familiar example is adding water to a salt, acid, alcohol, or chemical solution.

The total quantity of the mixture increases, but the amount of the tracked component stays the same.

For example, if a solution contains \(4\) liters of salt before water is added, it still contains \(4\) liters of salt afterward. The salt has simply been spread through a larger total volume.

This makes dilution a direct application of the Mixture Principle:

\[ A=CT. \]

The central idea is

\[ \boxed{\text{component amount before dilution} = \text{component amount after dilution}}. \]

A Zero-Concentration Substance

Suppose the component being tracked is salt.

Pure water contains no salt, so its salt concentration is

\[ 0\%=0. \]

If \(x\) liters of water are added, the amount of salt contributed by that water is

\[ 0(x)=0. \]

This is the dilution counterpart to adding a pure substance. A pure component has concentration \(1\), while a substance containing none of the component has concentration \(0\).

Both fit naturally into the same Mixture Principle.

The General Dilution Model

Suppose a mixture initially has concentration \(C\) and total quantity \(T\).

The amount of the tracked component is

\[ CT. \]

Now suppose \(x\) units of a \(0\%\) substance are added.

The added substance contributes

\[ 0(x)=0 \]

units of the tracked component.

The new total quantity becomes

\[ T+x. \]

If the final concentration is \(C_f\), then conservation gives

\[ CT+0(x)=C_f(T+x). \]

So the standard dilution equation is

\[ \boxed{CT=C_f(T+x)}. \]

Example 1: Diluting a Salt Solution

A tank contains \(10\) liters of a \(40\%\) salt solution. How much water must be added to reduce the concentration to \(25\%\)?

Let \(x\) be the number of liters of water added.

The original amount of salt is

\[ 0.40(10)=4. \]

The water contributes no salt.

After adding \(x\) liters of water, the total volume is

\[ 10+x. \]

At a final concentration of \(25\%\), the salt amount is

\[ 0.25(10+x). \]

Conservation gives

\[ 4=0.25(10+x). \]

Expand:

\[ 4=2.5+0.25x. \]

Subtract \(2.5\):

\[ 1.5=0.25x. \]

Therefore,

\[ x=6. \]

So

\[ \boxed{6\text{ liters of water}} \]

must be added.

Checking the Result

The solution originally contained

\[ 4 \]

liters of salt.

After \(6\) liters of water are added, the total volume becomes

\[ 10+6=16. \]

The concentration is therefore

\[ \frac{4}{16}=0.25. \]

Thus,

\[ \boxed{25\%}. \]

The amount of salt did not change. Only the total volume changed.

Organizing Dilution in a Table

Material Concentration Total Quantity Component Amount
Original mixture \(0.40\) \(10\) \(4\)
Water \(0\) \(x\) \(0\)
Final mixture \(0.25\) \(10+x\) \(0.25(10+x)\)

Reading across the component-amount column gives

\[ 4+0=0.25(10+x), \]

or simply

\[ 4=0.25(10+x). \]

Why Dilution Lowers Concentration

Concentration compares the amount of one component with the entire mixture:

\[ C=\frac{A}{T}. \]

During dilution, the numerator \(A\) remains fixed while the denominator \(T\) becomes larger.

For example,

\[ \frac{4}{10}=0.40, \]

but after adding \(6\) liters of water,

\[ \frac{4}{16}=0.25. \]

The component itself has not disappeared. Its share of the larger whole has decreased.

This is the essential mathematical meaning of dilution.

Example 2: Diluting an Alcohol Solution

A container holds \(18\) liters of a \(50\%\) alcohol solution. How much water must be added to reduce the concentration to \(30\%\)?

Let \(x\) be the amount of water added.

The original amount of alcohol is

\[ 0.50(18)=9. \]

The final quantity is

\[ 18+x. \]

Conservation gives

\[ 9=0.30(18+x). \]

Expand:

\[ 9=5.4+0.30x. \]

Subtract \(5.4\):

\[ 3.6=0.30x. \]

Therefore,

\[ x=12. \]

So

\[ \boxed{12\text{ liters of water}} \]

must be added.

Solving the General Dilution Equation

Begin with

\[ CT=C_f(T+x). \]

Expand:

\[ CT=C_fT+C_fx. \]

Subtract \(C_fT\):

\[ CT-C_fT=C_fx. \]

Factor \(T\):

\[ T(C-C_f)=C_fx. \]

Divide by \(C_f\):

\[ \boxed{ x=\frac{T(C-C_f)}{C_f} }. \]

This formula gives the amount of a \(0\%\) substance needed to dilute a mixture from concentration \(C\) to a lower concentration \(C_f\).

What the Dilution Formula Reveals

Consider

\[ x=\frac{T(C-C_f)}{C_f}. \]

The formula shows several relationships.

If the original quantity \(T\) doubles while the concentrations remain the same, the required amount of diluent also doubles.

If the desired concentration is only slightly lower than the original concentration, relatively little diluent is needed.

If the target concentration becomes much smaller, a much larger amount must be added.

As

\[ C_f\rightarrow0, \]

the denominator approaches zero, causing the required amount of diluent to grow without bound.

Can Dilution Reach Exactly 0%?

Suppose an original mixture contains a positive amount of the tracked component. Can adding a finite amount of water make its concentration exactly \(0\%\)?

No.

The original component remains present. Adding more water makes its proportion smaller and smaller, but does not remove it.

For example, if the mixture contains \(4\) liters of salt, then after adding \(x\) liters of water its concentration is

\[ C(x)=\frac{4}{T+x}. \]

As \(x\) becomes very large,

\[ C(x)\rightarrow0, \]

but for every finite value of \(x\),

\[ C(x)>0. \]

Dilution can approach \(0\%\), but it cannot reach exactly \(0\%\) unless the component itself is removed.

Example 3: Approaching a Very Low Concentration

A \(10\)-liter solution is \(20\%\) component. How much water must be added to reduce it to \(5\%\)?

The original component amount is

\[ 0.20(10)=2. \]

Let \(x\) be the amount of water added.

Then

\[ 2=0.05(10+x). \]

Expand:

\[ 2=0.5+0.05x. \]

Subtract \(0.5\):

\[ 1.5=0.05x. \]

Therefore,

\[ x=30. \]

So

\[ \boxed{30\text{ liters of water}} \]

must be added.

Notice that reducing the concentration from \(20\%\) to \(5\%\) requires adding three times the original volume of water.

An Alternate Way to Think About Dilution

Instead of immediately writing a mixture equation, sometimes it is useful to determine what the final total quantity must be.

Suppose the original mixture contains \(4\) liters of component and the desired concentration is \(25\%\).

Since

\[ C=\frac{A}{T}, \]

the final total must satisfy

\[ 0.25=\frac{4}{T_f}. \]

Therefore,

\[ T_f=\frac{4}{0.25}=16. \]

If the original mixture contained \(10\) liters, the amount of water added must be

\[ 16-10=6. \]

This gives the same result as the conservation equation.

Example 4: Finding the Final Concentration

A \(12\)-liter solution is \(30\%\) acid. If \(6\) liters of water are added, what is the final concentration?

The original acid amount is

\[ 0.30(12)=3.6. \]

The final total quantity is

\[ 12+6=18. \]

Therefore,

\[ C_f=\frac{3.6}{18}. \] \[ C_f=0.20. \]

The resulting solution is

\[ \boxed{20\%}. \]

Example 5: Finding the Original Concentration

A \(10\)-liter solution has an unknown concentration. After \(5\) liters of water are added, the final \(15\)-liter mixture is \(20\%\). Find the original concentration.

Let \(C\) be the original concentration.

The original component amount is

\[ 10C. \]

Water adds none of the component.

The final component amount is

\[ 0.20(15)=3. \]

Therefore,

\[ 10C=3. \]

So

\[ C=0.30. \]

The original concentration was

\[ \boxed{30\%}. \]

Dilution and Proportional Reasoning

Dilution can also be understood through inverse proportional reasoning.

If the component amount remains fixed,

\[ A=CT. \]

Therefore,

\[ CT=\text{constant}. \]

Solving for concentration gives

\[ C=\frac{A}{T}. \]

So as the total quantity increases, the concentration decreases inversely.

For example, doubling the total quantity while keeping the component amount unchanged cuts the concentration in half.

A \(40\%\) solution diluted to twice its original total volume becomes

\[ \boxed{20\%}. \]

Dilution as a Function

Suppose a mixture initially contains \(A\) units of component in \(T\) units of total mixture.

If \(x\) units of water are added, then the concentration becomes a function of \(x\):

\[ \boxed{ C(x)=\frac{A}{T+x} }. \]

For example, a \(10\)-liter mixture that is \(40\%\) component contains

\[ A=0.40(10)=4. \]

Its concentration after adding \(x\) liters of water is

\[ C(x)=\frac{4}{10+x}. \]

As \(x\) increases, \(C(x)\) decreases.

This gives dilution problems a natural connection to rational functions and inverse relationships.

Pure Addition and Dilution Are Opposite Cases

The previous lesson examined adding a pure component with concentration

\[ 1. \]

Dilution examines adding a substance with concentration

\[ 0. \]

The two situations form useful opposites.

Process Added Concentration Effect
Adding pure component \(1\) Raises concentration
Adding diluent \(0\) Lowers concentration

Both are ordinary mixture problems. The only difference is the concentration of the material being added.

Some Dilution Targets Are Impossible

Suppose a solution is currently \(30\%\) component.

Can adding water make it \(50\%\)?

No.

Water contains none of the tracked component, so adding it cannot increase the concentration.

For dilution by a \(0\%\) substance,

\[ \boxed{0 for any positive finite amount added to a mixture that initially contains some component.

If the target is higher than the starting concentration, a stronger mixture or pure component must be added instead.

Method for Dilution Problems

  1. Identify the component being tracked. Determine what substance or property the concentration describes.
  2. Find the original component amount. Use \[ A=CT. \]
  3. Recognize the diluent as \(0\%\). The added substance contributes no amount of the tracked component.
  4. Represent the amount added. Let \(x\) be the unknown amount of diluent when necessary.
  5. Find the final total. If \(x\) units are added to \(T\), then \[ T_f=T+x. \]
  6. Apply conservation. Use \[ CT=C_f(T+x). \]
  7. Solve the equation. Use ordinary algebraic techniques.
  8. Check the result. The final concentration should be lower than the original concentration but greater than \(0\%\) for a finite dilution.

Common Errors

Practice Problems

  1. A \(12\)-liter solution is \(50\%\) salt. How much water must be added to reduce the concentration to \(30\%\)?
  2. A \(20\)-liter solution is \(35\%\) alcohol. How much water must be added to reduce the concentration to \(25\%\)?
  3. A \(10\)-liter solution is \(40\%\) acid. If \(5\) liters of water are added, what is the final concentration?
  4. A \(16\)-liter solution is \(25\%\) component. How much water must be added to reduce the concentration to \(10\%\)?
  5. A \(15\)-liter mixture has an unknown concentration. After \(5\) liters of water are added, the final mixture is \(30\%\). Find the original concentration.
  6. A \(24\)-liter solution is \(60\%\) component. How much water must be added to reduce it to \(40\%\)?
  7. Can water be added to a \(25\%\) solution to produce a \(35\%\) solution? Explain.
  8. A \(30\)-liter solution is \(20\%\) component. How much water must be added to reduce the concentration to \(5\%\)?

Practice Answers

  1. \[ 0.50(12)=0.30(12+x) \] \[ 6=3.6+0.30x \] \[ 2.4=0.30x \] \[ \boxed{x=8\text{ liters}}. \]
  2. \[ 0.35(20)=0.25(20+x) \] \[ 7=5+0.25x \] \[ 2=0.25x \] \[ \boxed{x=8\text{ liters}}. \]
  3. Original component: \[ 0.40(10)=4. \] Final quantity: \[ 10+5=15. \] Therefore, \[ C_f=\frac{4}{15}\approx0.2667. \] So \[ \boxed{26.67\%}. \]
  4. \[ 0.25(16)=0.10(16+x) \] \[ 4=1.6+0.10x \] \[ 2.4=0.10x \] \[ \boxed{x=24\text{ liters}}. \]
  5. Let \(C\) be the original concentration: \[ 15C=0.30(20). \] \[ 15C=6. \] \[ C=0.40. \] Therefore, \[ \boxed{40\%}. \]
  6. \[ 0.60(24)=0.40(24+x) \] \[ 14.4=9.6+0.40x \] \[ 4.8=0.40x \] \[ \boxed{x=12\text{ liters}}. \]
  7. \[ \boxed{\text{No}}. \] Water has \(0\%\) concentration and can only lower the concentration of a \(25\%\) solution.
  8. \[ 0.20(30)=0.05(30+x) \] \[ 6=1.5+0.05x \] \[ 4.5=0.05x \] \[ \boxed{x=90\text{ liters}}. \]

Summary

Dilution occurs when a substance containing none of the tracked component is added to a mixture.

The added concentration is

\[ \boxed{0\%=0}. \]

Because no component is added, the component amount remains constant:

\[ \boxed{\text{component before}=\text{component after}}. \]

For an original mixture with concentration \(C\), quantity \(T\), and \(x\) units of diluent added,

\[ \boxed{CT=C_f(T+x)}. \]

Solving for the amount of diluent gives

\[ \boxed{ x=\frac{T(C-C_f)}{C_f} }. \]

The most important conceptual distinction is that dilution changes the concentration by increasing the total quantity while leaving the component amount unchanged.

Continue the Exploration

Dilution adds a \(0\%\) substance without removing any of the original mixture. The next lesson introduces a more complex process: removing part of a mixture and replacing it with another substance. Because the material removed contains some of the tracked component, the component amount itself changes.