Overview
Concentration problems are the most basic form of mixture problem. Two or more mixtures contain different proportions of the same component, and the goal is to determine what happens when those mixtures are combined.
The important quantity is not the concentration by itself. What must be tracked is the actual amount of the component contained in each mixture.
The Mixture Principle gives
\[ A=CT, \]where \(A\) is component amount, \(C\) is concentration, and \(T\) is total quantity.
When mixtures are combined without loss, their component amounts add:
\[ \boxed{C_1T_1+C_2T_2=C_fT_f}. \]Most concentration problems are variations of this single conservation relationship.
The Basic Concentration Model
Suppose one mixture has concentration \(C_1\) and quantity \(T_1\), while another has concentration \(C_2\) and quantity \(T_2\).
The first mixture contains
\[ C_1T_1 \]units of the component being tracked, while the second contains
\[ C_2T_2. \]If the two mixtures are combined, the final quantity is usually
\[ T_f=T_1+T_2. \]Therefore,
\[ \boxed{C_1T_1+C_2T_2=C_f(T_1+T_2)}. \]Depending on the problem, the unknown may be a concentration, one of the starting quantities, or the amount needed to reach a desired final concentration.
Example 1: Finding the Final Concentration
Suppose \(6\) liters of a \(20\%\) solution are mixed with \(4\) liters of a \(50\%\) solution. What is the concentration of the final mixture?
The first solution contributes
\[ 0.20(6)=1.2 \]liters of the measured component.
The second contributes
\[ 0.50(4)=2. \]Therefore, the total component amount is
\[ 1.2+2=3.2. \]The total mixture contains
\[ 6+4=10 \]liters.
The final concentration is therefore
\[ C_f=\frac{3.2}{10}. \] \[ C_f=0.32. \]So the final mixture is
\[ \boxed{32\%}. \]A Reasonableness Check
The answer \(32\%\) should make intuitive sense.
The two original concentrations were
\[ 20\% \]and
\[ 50\%. \]The final concentration must lie between them:
\[ 20\%<32\%<50\%. \]There were also more liters of the weaker \(20\%\) mixture than of the stronger \(50\%\) mixture, so the final concentration should be closer to \(20\%\) than to \(50\%\).
That is exactly what happens.
Concentration as a Weighted Average
The final concentration can also be written directly as
\[ \boxed{ C_f= \frac{C_1T_1+C_2T_2}{T_1+T_2} }. \]This is a weighted average.
The concentrations are weighted according to the quantities of the mixtures to which they belong.
In the previous example,
\[ C_f= \frac{0.20(6)+0.50(4)}{10} \] \[ = \frac{3.2}{10} \] \[ =0.32. \]A simple arithmetic average,
\[ \frac{20\%+50\%}{2}=35\%, \]would be wrong because the two mixtures do not have equal quantities.
When Can Concentrations Be Averaged Directly?
There is one important special case.
If equal quantities of two mixtures are combined, then their concentrations contribute equal weight.
Suppose \(5\) liters of a \(20\%\) solution are mixed with \(5\) liters of a \(50\%\) solution.
The final concentration is
\[ C_f= \frac{0.20(5)+0.50(5)}{10}. \]Factor out \(5\):
\[ C_f= \frac{5(0.20+0.50)}{10}. \] \[ C_f= \frac{0.70}{2}=0.35. \]So in this special case,
\[ \boxed{C_f=35\%}. \]Directly averaging the concentrations works because the quantities are equal.
Example 2: Finding an Unknown Quantity
How many liters of a \(60\%\) solution should be mixed with \(10\) liters of a \(30\%\) solution to produce a \(40\%\) solution?
Let \(x\) be the number of liters of the \(60\%\) solution.
The \(30\%\) solution contributes
\[ 0.30(10)=3 \]liters of the component.
The \(60\%\) solution contributes
\[ 0.60x. \]The final mixture contains
\[ 10+x \]liters at a concentration of \(40\%\), so its component amount is
\[ 0.40(10+x). \]Apply conservation:
\[ 3+0.60x=0.40(10+x). \]Expand:
\[ 3+0.60x=4+0.40x. \]Subtract \(0.40x\):
\[ 3+0.20x=4. \]Subtract \(3\):
\[ 0.20x=1. \]Therefore,
\[ x=5. \]So
\[ \boxed{5\text{ liters}} \]of the \(60\%\) solution are required.
Organizing the Same Problem in a Table
The previous problem can be organized using the concentration, quantity, and component amount of each mixture.
| Mixture | Concentration | Total Quantity | Component Amount |
|---|---|---|---|
| 30% solution | \(0.30\) | \(10\) | \(3\) |
| 60% solution | \(0.60\) | \(x\) | \(0.60x\) |
| Final solution | \(0.40\) | \(10+x\) | \(0.40(10+x)\) |
Reading across the final column immediately produces
\[ 3+0.60x=0.40(10+x). \]The table is particularly useful when a problem contains several percentages and quantities that might otherwise become difficult to track.
Example 3: Finding an Unknown Starting Concentration
A \(12\)-liter solution of unknown concentration is mixed with \(8\) liters of a \(50\%\) solution. The resulting \(20\)-liter mixture is \(38\%\). What was the concentration of the first solution?
Let \(x\) represent the concentration of the first solution as a decimal.
Its component amount is
\[ 12x. \]The second solution contributes
\[ 0.50(8)=4. \]The final mixture contains
\[ 0.38(20)=7.6 \]units of the component.
Conservation gives
\[ 12x+4=7.6. \]Subtract \(4\):
\[ 12x=3.6. \]Divide by \(12\):
\[ x=0.30. \]Therefore, the original solution was
\[ \boxed{30\%}. \]Example 4: Finding Both Quantities from a Fixed Total
A laboratory needs \(20\) liters of a \(35\%\) solution. It has a \(20\%\) solution and a \(50\%\) solution available. How much of each should be used?
Let \(x\) be the amount of the \(20\%\) solution.
Since the final mixture must contain \(20\) liters, the amount of the \(50\%\) solution is
\[ 20-x. \]The first solution contributes
\[ 0.20x. \]The second contributes
\[ 0.50(20-x). \]The final mixture must contain
\[ 0.35(20)=7 \]liters of the measured component.
Therefore,
\[ 0.20x+0.50(20-x)=7. \]Expand:
\[ 0.20x+10-0.50x=7. \]Combine like terms:
\[ 10-0.30x=7. \]Subtract \(10\):
\[ -0.30x=-3. \]Thus,
\[ x=10. \]The amount of the \(20\%\) solution is \(10\) liters, and the amount of the \(50\%\) solution is
\[ 20-10=10. \]Therefore, the laboratory should use
\[ \boxed{10\text{ liters of each solution}}. \]Why the Previous Answer Makes Sense
The desired concentration of \(35\%\) lies exactly halfway between
\[ 20\% \]and
\[ 50\%. \]Because the target is exactly halfway between the two concentrations, equal quantities of the two solutions produce the desired mixture.
This does not replace the algebra, but it gives an excellent check on the result.
The Target Concentration Determines the Balance
Suppose instead that the desired concentration were \(25\%\). Since \(25\%\) is much closer to \(20\%\) than to \(50\%\), we should expect the mixture to contain much more of the \(20\%\) solution.
Likewise, a target of \(45\%\) would require much more of the \(50\%\) solution.
This gives an important qualitative rule:
\[ \boxed{ \text{The final mixture contains more of the source whose concentration is closer to the target.} } \]The exact amounts still depend on the differences between the concentrations, but this observation is useful for checking whether an algebraic answer is plausible.
Example 5: A Target Closer to One Concentration
How many liters of a \(20\%\) solution and a \(50\%\) solution should be combined to make \(30\) liters of a \(25\%\) solution?
Let \(x\) be the amount of the \(20\%\) solution.
Then
\[ 30-x \]is the amount of the \(50\%\) solution.
Apply conservation:
\[ 0.20x+0.50(30-x)=0.25(30). \]Expand:
\[ 0.20x+15-0.50x=7.5. \]Combine like terms:
\[ 15-0.30x=7.5. \]Subtract \(15\):
\[ -0.30x=-7.5. \]Therefore,
\[ x=25. \]So the amount of the \(20\%\) solution is
\[ 25\text{ liters}, \]and the amount of the \(50\%\) solution is
\[ 30-25=5\text{ liters}. \]Thus,
\[ \boxed{25\text{ liters at }20\%\text{ and }5\text{ liters at }50\%}. \]As expected, much more of the weaker solution is needed because the target concentration of \(25\%\) lies much closer to \(20\%\).
A Useful Concentration-Difference Pattern
The previous example reveals an interesting relationship.
The target \(25\%\) is
\[ 25\%-20\%=5\% \]above the weaker concentration, while it is
\[ 50\%-25\%=25\% \]below the stronger concentration.
The quantities appear in the opposite ratio:
\[ 25:5=5:1. \]The concentration differences are
\[ 25:5=5:1. \]In general, when two concentrations are blended to reach a target concentration between them, the quantities satisfy
\[ \boxed{ \frac{T_1}{T_2} = \frac{C_2-C_f}{C_f-C_1} }. \]This relationship follows directly from the ordinary mixture equation. It is not a separate principle, but it can provide a useful shortcut and a deeper view of why weighted mixtures behave as they do.
Why the Difference Relationship Works
Begin with
\[ C_1T_1+C_2T_2=C_f(T_1+T_2). \]Expand the right side:
\[ C_1T_1+C_2T_2=C_fT_1+C_fT_2. \]Move the terms involving \(T_1\) to one side and the terms involving \(T_2\) to the other:
\[ C_fT_1-C_1T_1=C_2T_2-C_fT_2. \]Factor:
\[ (C_f-C_1)T_1=(C_2-C_f)T_2. \]Divide:
\[ \frac{T_1}{T_2} = \frac{C_2-C_f}{C_f-C_1}. \]The amount of one mixture therefore depends on the concentration difference on the opposite side of the target.
Three or More Concentrations
The same conservation principle works when more than two mixtures are combined.
For three mixtures,
\[ \boxed{ C_1T_1+C_2T_2+C_3T_3=C_fT_f }. \]If their quantities simply add,
\[ T_f=T_1+T_2+T_3. \]So
\[ C_1T_1+C_2T_2+C_3T_3 = C_f(T_1+T_2+T_3). \]Nothing fundamentally new has occurred. Each source contributes a component amount, and all contributions must be accounted for in the final mixture.
Example 6: Three Solutions
Suppose \(4\) liters of a \(10\%\) solution, \(6\) liters of a \(30\%\) solution, and \(10\) liters of a \(50\%\) solution are combined. Find the final concentration.
The component amounts are
\[ 0.10(4)=0.4, \] \[ 0.30(6)=1.8, \]and
\[ 0.50(10)=5. \]The total component amount is
\[ 0.4+1.8+5=7.2. \]The total quantity is
\[ 4+6+10=20. \]Therefore,
\[ C_f=\frac{7.2}{20}. \] \[ C_f=0.36. \]The final concentration is
\[ \boxed{36\%}. \]Converting Percentages Correctly
One of the simplest mistakes can completely change the answer: forgetting that a percentage means “per hundred.”
For example,
\[ 45\%=0.45, \]not
\[ 45. \]A \(45\%\) concentration in \(8\) liters therefore contains
\[ 0.45(8)=3.6 \]liters of component.
It does not contain
\[ 45(8)=360 \]liters.
Keeping concentrations in decimal form during calculations usually makes the algebra much cleaner.
Keep the Units Consistent
The quantities being added must use compatible units.
If one solution is measured in liters and another in milliliters, convert one of them before constructing the mixture equation.
For example,
\[ 500\text{ mL}=0.5\text{ L}. \]The concentration itself has no physical unit when written as a decimal fraction, but multiplying concentration by quantity produces the same unit as the total quantity.
Thus,
\[ (0.30)(10\text{ L})=3\text{ L}. \]Tracking units is another useful way to check that the model makes sense.
Physical Restrictions on the Answer
Mixture problems describe real quantities, so not every algebraic answer is physically meaningful.
A volume, mass, or quantity cannot ordinarily be negative:
\[ T\geq 0. \]A concentration expressed as a decimal normally satisfies
\[ 0\leq C\leq 1. \]Equivalently, as a percentage,
\[ 0\%\leq C\leq100\%. \]A calculation that produces a concentration of \(130\%\) or a quantity of \(-5\) liters signals that the equation, assumptions, or interpretation should be checked.
Some Target Concentrations Are Impossible
Suppose the only available solutions are \(20\%\) and \(50\%\).
Can they be mixed to produce a \(70\%\) solution?
No.
No ordinary combination of the two can have a concentration greater than the stronger starting concentration.
Likewise, they cannot produce a \(10\%\) solution without adding some substance whose concentration is below \(20\%\).
When two positive quantities are combined, the target concentration must lie between the starting concentrations:
\[ \boxed{ C_{\min}\leq C_f\leq C_{\max} }. \]Recognizing an impossible target before doing algebra is an important modeling skill.
Method for Concentration Problems
- Identify the component being tracked. Choose one substance or property and follow it consistently.
- Convert percentages to decimals. For example, \(25\%=0.25\).
- Identify the quantity of each mixture. Use a variable for an unknown amount.
- Write each component amount. Use \[ A=CT. \]
- Determine the final total quantity. If mixtures are simply combined, add their quantities.
- Apply conservation. Write \[ \text{component contributions} = \text{final component amount}. \]
- Solve the equation. Use the appropriate algebraic operations.
- Check the result. Verify the units, concentration range, and physical reasonableness.
Common Errors
- Averaging concentrations without considering quantities. Mixture concentrations are generally weighted averages.
- Using percentages as whole numbers. Use \(0.30\), not \(30\), for \(30\%\).
- Adding concentrations instead of component amounts. The quantities that combine are \(C_1T_1\), \(C_2T_2\), and so forth.
- Forgetting that the total quantity changes. If \(x\) liters are added to \(10\) liters, the final amount is \(10+x\).
- Using the target concentration as a component amount. The final component amount is \(C_fT_f\), not merely \(C_f\).
- Accepting an impossible target. A mixture of two concentrations cannot ordinarily exceed the stronger concentration or fall below the weaker one.
- Ignoring the meaning of a negative solution. A negative quantity usually indicates that the proposed mixture cannot satisfy the conditions as stated.
Practice Problems
- \(8\) liters of a \(25\%\) solution are mixed with \(12\) liters of a \(45\%\) solution. Find the concentration of the final mixture.
- \(5\) liters of a \(10\%\) solution are mixed with \(15\) liters of a \(30\%\) solution. Find the final concentration.
- How many liters of a \(50\%\) solution should be mixed with \(8\) liters of a \(20\%\) solution to produce a \(30\%\) solution?
- A \(10\)-liter solution of unknown concentration is mixed with \(10\) liters of a \(60\%\) solution. The final mixture is \(40\%\). Find the unknown concentration.
- How many liters of a \(20\%\) solution and a \(60\%\) solution are needed to make \(40\) liters of a \(35\%\) solution?
- A mixture contains \(6\) liters of a \(15\%\) solution, \(4\) liters of a \(40\%\) solution, and \(10\) liters of a \(50\%\) solution. Find the final concentration.
- Can a \(25\%\) solution and a \(40\%\) solution be combined to produce a \(50\%\) solution? Explain.
- A laboratory has \(30\%\) and \(70\%\) solutions. It needs \(24\) liters of a \(45\%\) solution. How many liters of each should be used?
Practice Answers
- \[ \frac{0.25(8)+0.45(12)}{20} = \frac{7.4}{20} = \boxed{37\%}. \]
- \[ \frac{0.10(5)+0.30(15)}{20} = \frac{5}{20} = \boxed{25\%}. \]
- \[ 0.20(8)+0.50x=0.30(8+x) \] gives \[ \boxed{x=4\text{ liters}}. \]
- \[ 10x+0.60(10)=0.40(20) \] gives \[ 10x+6=8, \] so \[ x=0.20. \] Therefore, \[ \boxed{20\%}. \]
- Let \(x\) be the amount of the \(20\%\) solution: \[ 0.20x+0.60(40-x)=0.35(40). \] Solving gives \[ x=25. \] Therefore, \[ \boxed{25\text{ liters at }20\%\text{ and }15\text{ liters at }60\%}. \]
- \[ C_f= \frac{0.15(6)+0.40(4)+0.50(10)}{20} \] \[ = \frac{7.5}{20} = \boxed{37.5\%}. \]
- \[ \boxed{\text{No}}. \] A mixture of \(25\%\) and \(40\%\) solutions must have a concentration between \(25\%\) and \(40\%\).
- Let \(x\) be the amount of the \(30\%\) solution: \[ 0.30x+0.70(24-x)=0.45(24). \] Solving gives \[ x=15. \] Therefore, \[ \boxed{15\text{ liters at }30\%\text{ and }9\text{ liters at }70\%}. \]
Summary
Concentration problems are built around the Mixture Principle
\[ A=CT \]and the conservation equation
\[ \boxed{ C_1T_1+C_2T_2=C_fT_f }. \]When two mixtures are simply combined,
\[ T_f=T_1+T_2, \]so
\[ \boxed{ C_1T_1+C_2T_2=C_f(T_1+T_2) }. \]The final concentration can also be viewed as a weighted average:
\[ \boxed{ C_f= \frac{C_1T_1+C_2T_2}{T_1+T_2} }. \]The central skill is not memorizing several formulas. It is learning to identify how much of the tracked component each source contributes and then conserving that quantity through the mixing process.
Continue the Exploration
The next lesson examines a special but important case: adding a pure substance to an existing mixture. Because a pure substance has a concentration of \(100\%\), the ordinary Mixture Principle applies without requiring a new formula.
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Continue to: Adding a Pure Substance
Learn how \(100\%\) concentration fits naturally into mixture equations and how adding a pure component changes both the component amount and the total mixture.