Mixture Concentration: Combining Two Solutions

What This Problem Teaches

• Setting up mass balance equations for mixture problems
• Working with concentrations as decimal multipliers
• Recognizing that mixture problems often have only one unknown when total volume is fixed
• Understanding that final concentrations are weighted averages, not simple averages
• Translating word problems into algebraic equations systematically

Visualizing the Problem

The key insight: pure acid from both solutions must equal pure acid in the final mixture

Solution: The Mass Balance Approach

Step 1 — Define the variable

Let x = liters of the 2% solution needed.

Since we need 60 liters total, we need (60 - x) liters of the 12% solution.

Step 2 — Set up the mass balance equation

The amount of pure acid from both solutions must equal the amount of pure acid in the final mixture:

Step 3 — Express each term mathematically

• Pure acid from 2% solution: 0.02 × x = 0.02x liters
• Pure acid from 12% solution: 0.12 × (60 - x) = 0.12(60 - x) liters
• Pure acid in final mixture: 0.05 × 60 = 3 liters

Step 4 — Write and solve the equation

Distribute the 0.12:

Combine like terms:

Subtract 7.2 from both sides:

Divide by -0.10:

Step 5 — Find both solution amounts

• 2% solution needed: 42 liters
• 12% solution needed: 60 - 42 = 18 liters

Solution: Method 2 — The Concentration Difference Approach

Step 1 — Find the distances from target concentration

Our target is 5%. Let's see how far each starting concentration is from this target:

• Distance from 2% to 5%: 5 - 2 = 3 percentage points
• Distance from 12% to 5%: 12 - 5 = 7 percentage points

Step 2 — Apply the inverse ratio rule

In mixture problems, the ratio of volumes is inversely related to the ratio of concentration differences. We need:

This means for every 7 parts of 2% solution, we need 3 parts of 12% solution.

Step 3 — Calculate actual volumes

Total ratio parts: 7 + 3 = 10 parts

Each part represents: 60 ÷ 10 = 6 liters

• 2% solution: 7 × 6 = 42 liters
• 12% solution: 3 × 6 = 18 liters

Verification

Let's verify our answer by checking that the pure acid amounts balance:

Pure acid from each solution:

• From 42 L of 2% solution: 0.02 × 42 = 0.84 L
• From 18 L of 12% solution: 0.12 × 18 = 2.16 L
• Total pure acid: 0.84 + 2.16 = 3.00 L ✓

Final concentration check:

Concentration in final mixture: 3.00 ÷ 60 = 0.05 = 5% ✓

Volume check:

Total volume: 42 + 18 = 60 L ✓

Does This Seem Reasonable?

Our answer says we need 42 liters of the weak solution and only 18 liters of the strong solution. This makes intuitive sense because:

If we had used equal amounts (30L each), the concentration would be:

That's too high, confirming we need more of the weaker solution.

Watch Out For These

The Pattern Behind This

All two-component mixture problems follow the same template:

Where:

• C₁, C₂ = concentrations of the two starting solutions (as decimals)
• V₁, V₂ = volumes of the two starting solutions
• C_final = target concentration (as a decimal)
• V_total = total volume of final mixture

When the total volume is fixed, you only have one unknown: if V₁ = x, then V₂ = V_total - x.

Where This Shows Up in Real Life

• Pharmacy: Pharmacists regularly mix medications of different concentrations to prepare custom dosages for patients.
• Manufacturing: Chemical plants mix solutions to achieve precise concentrations for industrial processes.
• Agriculture: Farmers dilute concentrated fertilizers and pesticides to safe application strengths.
• Food industry: Commercial kitchens mix juices or vinegars of different acid concentrations for consistent flavoring.

What If?