Best examples of finding final concentration after mixing solutions

If you’re working with chemistry, medicine, or even water treatment, you will run into problems that require finding the concentration of a mixture. That’s where good, clear **examples of finding final concentration after mixing solutions** become your best friend. Instead of memorizing formulas in a vacuum, it’s far more helpful to see how they play out in real scenarios: mixing IV bags, combining cleaning products, or blending lab buffers. In this guide, you’ll walk through several **real examples of finding final concentration after mixing solutions** step by step. We’ll move from simple classroom-style questions to situations that look a lot more like what professionals actually do in labs, hospitals, and industry. Along the way, you’ll see the same core ideas repeat: conservation of moles, volume additivity (and when that’s an approximation), and how units can quietly sabotage you if you’re not paying attention. By the end, you’ll have a practical toolkit, not just a formula on a page.
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Starting with simple examples of finding final concentration after mixing solutions

Let’s begin with the most common classroom scenario: two solutions of the same solute, different concentrations, mixed together. These are the cleanest examples of finding final concentration after mixing solutions because the chemistry is straightforward and the volume change is usually small enough that we treat volumes as additive.

The core idea is conservation of moles:

\[ n_{total} = n_1 + n_2 = C_1 V_1 + C_2 V_2 \] \[ C_{final} = \dfrac{n_{total}}{V_{total}} = \dfrac{C_1 V_1 + C_2 V_2}{V_1 + V_2} \]

Here’s a basic example of that in action.

Example 1: Mixing two salt solutions in a beaker

Say you mix 100.0 mL of 0.50 M NaCl with 200.0 mL of 0.20 M NaCl. This is the kind of example of final concentration that shows up on exams constantly.

  • Convert to liters:
    • \(V_1 = 0.1000\,\text{L}, C_1 = 0.50\,\text{M}\)
    • \(V_2 = 0.2000\,\text{L}, C_2 = 0.20\,\text{M}\)
  • Moles from each portion:
    • \(n_1 = C_1 V_1 = 0.50 \times 0.1000 = 0.050\,\text{mol}\)
    • \(n_2 = C_2 V_2 = 0.20 \times 0.2000 = 0.040\,\text{mol}\)
  • Total moles: \(n_{total} = 0.050 + 0.040 = 0.090\,\text{mol}\)
  • Total volume: \(V_{total} = 0.1000 + 0.2000 = 0.3000\,\text{L}\)
  • Final concentration:

\[ C_{final} = \dfrac{0.090}{0.3000} = 0.30\,\text{M} \]

This is one of the best examples for building intuition: the final concentration (0.30 M) lies between the two starting concentrations (0.50 M and 0.20 M), closer to the one that had more volume.


Medical and lab examples of finding final concentration after mixing solutions

Real-world chemistry doesn’t live in a vacuum. Nurses, pharmacists, and lab techs routinely apply these same ideas. These applied situations are some of the best examples of finding final concentration after mixing solutions because they force you to manage units, safety limits, and real constraints.

Example 2: Adjusting an IV drip concentration

Suppose a patient is receiving 250 mL of a 0.9% (w/v) NaCl solution (normal saline), and a physician wants to increase sodium chloride concentration by adding 50 mL of a 3% (w/v) NaCl solution. You want the final concentration.

For mass/volume percent, \(%\,\text{w/v} = \frac{\text{grams solute}}{100\,\text{mL solution}} \times 100\). So:

  • 0.9% NaCl: 0.9 g per 100 mL
  • 3% NaCl: 3.0 g per 100 mL

First, find grams of NaCl in each portion:

  • From 250 mL of 0.9%:
    • \(0.9\,\text{g} / 100\,\text{mL} \times 250\,\text{mL} = 2.25\,\text{g}\)
  • From 50 mL of 3%:
    • \(3.0\,\text{g} / 100\,\text{mL} \times 50\,\text{mL} = 1.50\,\text{g}\)

Total NaCl: \(2.25 + 1.50 = 3.75\,\text{g}\)

Total volume: \(250 + 50 = 300\,\text{mL}\)

Final % (w/v):

\[ \%\,\text{w/v} = \dfrac{3.75\,\text{g}}{300\,\text{mL}} \times 100 = 1.25\% \]

So the new IV bag is 1.25% NaCl. This is a practical example of final concentration calculation that has direct clinical relevance. Organizations such as the CDC and NIH publish guidance that assumes this kind of quantitative accuracy in solution prep for research and infection control.

Example 3: Preparing a working buffer from two stock solutions

In molecular biology labs, you often mix a concentrated stock buffer with water or another buffer to hit a target working concentration. Here’s a realistic scenario.

You have:

  • 50.0 mL of a 0.20 M Tris buffer at pH 8.0
  • 150.0 mL of a 0.05 M Tris buffer at pH 8.0

You mix them to get 200.0 mL of a combined buffer. Because the solute and pH are the same, no reaction occurs; you just average the concentrations based on moles.

Moles of Tris:

  • From 0.20 M: \(0.20 \times 0.0500 = 0.0100\,\text{mol}\)
  • From 0.05 M: \(0.05 \times 0.1500 = 0.00750\,\text{mol}\)

Total moles: \(0.0100 + 0.00750 = 0.01750\,\text{mol}\)

Total volume: \(0.0500 + 0.1500 = 0.2000\,\text{L}\)

Final concentration:

\[ C_{final} = \dfrac{0.01750}{0.2000} = 0.0875\,\text{M} \]

This kind of calculation shows up constantly in protocols from research institutions like Harvard or NIH labs, even if the math is tucked away in the methods section.


Mixed units: real examples of finding final concentration after mixing solutions

In the real world, units rarely cooperate. One of the more realistic examples of finding final concentration after mixing solutions is when one solution is given in molarity and the other in mass percent.

Example 4: Mixing a molar solution with a % (w/v) solution

You mix:

  • 100 mL of 0.10 M glucose (C\(_6\)H\(_{12}\)O\(_6\))
  • 50 mL of a 5.0% (w/v) glucose solution

You want the final concentration in molarity.

Step 1: Moles from the 0.10 M solution.

  • \(V_1 = 0.100\,\text{L}, C_1 = 0.10\,\text{M}\)
  • \(n_1 = 0.10 \times 0.100 = 0.0100\,\text{mol}\)

Step 2: Convert the 5.0% (w/v) to moles.

  • 5.0% (w/v) means 5.0 g per 100 mL
  • In 50 mL, mass of glucose:
    • \(5.0\,\text{g}/100\,\text{mL} \times 50\,\text{mL} = 2.50\,\text{g}\)

Molar mass of glucose ≈ 180.16 g/mol, so:

  • \(n_2 = \dfrac{2.50\,\text{g}}{180.16\,\text{g/mol}} \approx 0.0139\,\text{mol}\)

Step 3: Total moles and volume.

  • \(n_{total} = 0.0100 + 0.0139 = 0.0239\,\text{mol}\)
  • \(V_{total} = 100\,\text{mL} + 50\,\text{mL} = 150\,\text{mL} = 0.150\,\text{L}\)

Step 4: Final concentration.

\[ C_{final} = \dfrac{0.0239}{0.150} \approx 0.159\,\text{M} \]

This is a good example of how you sometimes need to translate everything into moles before you can even think about a final concentration.


Industrial and environmental examples of finding final concentration after mixing solutions

Industry and environmental science provide some of the best real examples of finding final concentration after mixing solutions because the numbers get big and the stakes are high.

Example 5: Wastewater dilution in a treatment plant

A treatment facility receives two wastewater streams containing the same contaminant, say nitrate (NO\(_3^-\)). You’re told:

  • Stream A: 10,000 L/day at 40 mg/L nitrate
  • Stream B: 5,000 L/day at 10 mg/L nitrate

They combine in a common tank. Assuming volumes add, what is the final nitrate concentration?

First, treat mg/L as mass per liter directly.

Total mass of nitrate per day:

  • Stream A: \(40\,\text{mg/L} \times 10{,}000\,\text{L} = 400{,}000\,\text{mg}\)
  • Stream B: \(10\,\text{mg/L} \times 5{,}000\,\text{L} = 50{,}000\,\text{mg}\)

Total mass: \(450{,}000\,\text{mg}\)

Total volume: \(10{,}000 + 5{,}000 = 15{,}000\,\text{L}\)

Final concentration:

\[ C_{final} = \dfrac{450{,}000\,\text{mg}}{15{,}000\,\text{L}} = 30\,\text{mg/L} \]

Regulators such as the U.S. EPA and public health agencies like the CDC care about exactly this sort of calculation when setting or checking water quality limits.

Example 6: Blending commercial cleaning solutions

Suppose a janitorial staff member mixes:

  • 2.0 L of a disinfectant at 4.0% (v/v) active ingredient
  • 6.0 L of water (0% active ingredient)

You want the final % (v/v) of active ingredient.

Volume of active ingredient in the disinfectant:

  • 4.0% (v/v) means 4.0 mL active per 100 mL solution
  • In 2.0 L (2,000 mL):
    • \(4.0\,\text{mL} / 100\,\text{mL} \times 2{,}000\,\text{mL} = 80\,\text{mL}\)

Total mixture volume:

  • \(2.0\,\text{L} + 6.0\,\text{L} = 8.0\,\text{L} = 8{,}000\,\text{mL}\)

Final % (v/v):

\[ \%\,\text{v/v} = \dfrac{80\,\text{mL}}{8{,}000\,\text{mL}} \times 100 = 1.0\% \]

This is another everyday example of final concentration after mixing a concentrated solution with water to reach a safer working strength. Health resources like Mayo Clinic and WebMD frequently emphasize correct dilution of disinfectants and antiseptics, which rests on this exact math.


Temperature and density: more advanced examples of finding final concentration after mixing solutions

Up to now, we’ve quietly assumed that volumes simply add and that temperature doesn’t matter. In many lab and classroom settings, that’s fine. But if you want the best, most realistic examples of finding final concentration after mixing solutions, you need to at least acknowledge that density and temperature can shift things.

Example 7: Mixing ethanol solutions where volume contraction matters

Mixing water and ethanol is a classic case where volumes are not perfectly additive. If you need very accurate alcohol content (for example, in pharmaceutical or analytical work), you’d rely on density tables instead of simple volume addition.

Imagine:

  • 100 mL of 95% (v/v) ethanol
  • 100 mL of water

Naively, you might say total volume is 200 mL and final ethanol concentration is 47.5% (v/v). In reality, the mixture’s volume is slightly less than 200 mL due to volume contraction when ethanol and water interact.

The rigorous approach:

  • Convert each component to mass using density data (for example, from NIST or a standard CRC Handbook)
  • Sum masses
  • Use mixture density to get the actual final volume
  • Then compute final % (v/v) or % (w/w)

This is a good example of how high-precision work moves beyond the simple formula and into density-based calculations.

Example 8: Temperature-dependent solubility in a lab

Consider a lab preparing saturated solutions of potassium nitrate (KNO\(_3\)) at two temperatures, then mixing them. At 20 °F versus 68 °F (roughly −6.7 °C vs 20 °C), solubility and density differ, which means the same mass of solute occupies different volumes. If you mix:

  • A cold, more concentrated KNO\(_3\) solution
  • A warmer, less concentrated one

You can still apply the same conservation of moles idea, but you must be careful with reported concentrations (they’re usually tied to a specific temperature). Modern data sheets and online resources from universities (for example, chemistry departments at major U.S. universities) routinely flag this, especially in 2024–2025 lab manuals where climate control and reproducibility are getting more attention.

While this is a more conceptual example of final concentration after mixing solutions, it mirrors real experimental design: you can’t ignore temperature forever.


General strategy pulled from all these examples

Across all these examples of finding final concentration after mixing solutions, the same pattern keeps showing up:

  • Translate everything into moles, grams, or volume of pure solute
  • Add those amounts to get a total quantity of solute
  • Add volumes (or, for high-precision work, use density to get actual volumes)
  • Divide total solute by total volume in consistent units

Whether you’re mixing NaCl in a beaker, glucose in an IV bag, or nitrate in wastewater, this is the backbone. The variety of real examples of finding final concentration after mixing solutions is mostly about context and unit juggling, not different chemistry.


FAQ: Common questions and examples of final concentration calculations

How do I handle examples of final concentration when the solutes are different?

If the solutions contain different solutes that do not react, you cannot talk about a single final concentration in the same units unless you track each solute separately. You might say, for instance, “The final solution is 0.10 M NaCl and 0.05 M KCl,” but there is no single combined molarity that replaces both. If the solutes react (for example, acid and base), you must first run a stoichiometry calculation, then compute the final concentration of products and any leftover reactants.

Can you give an example of finding final concentration after mixing an acid and water?

Take 25.0 mL of 6.0 M HCl added to enough water to reach a final volume of 250.0 mL. This is technically a dilution rather than mixing two independent solutions, but the math is almost identical. Moles of HCl stay the same:

  • \(n = 6.0 \times 0.0250 = 0.150\,\text{mol}\)
  • Final volume: 0.2500 L
  • Final concentration: \(C_{final} = 0.150 / 0.2500 = 0.60\,\text{M}\)

This fits neatly alongside the other examples of finding final concentration after mixing solutions, just with one of the starting concentrations effectively being zero (pure water).

In real examples, when should I worry about non-additive volumes?

You start worrying when:

  • You’re working with high alcohol content, concentrated acids, or other strongly interacting mixtures
  • You need high precision (analytical chemistry, pharmaceutical formulation, regulatory testing)

In those cases, you look up densities and use mass-based calculations. For routine classroom and many lab situations, assuming volumes add is good enough, and most examples of finding final concentration after mixing solutions quietly make that assumption.

Are there online tools that can check my examples of final concentration calculations?

Many university chemistry departments and educational sites host molarity and dilution calculators. They’re helpful for checking your work, but they’re only as smart as the numbers you feed them. You still need to understand what you’re mixing, which units you’re using, and whether your situation matches the simple formulas used in those tools.


Across all these scenarios, the pattern doesn’t change: track the amount of solute, track the total volume, and you can handle almost any examples of finding final concentration after mixing solutions that show up in class, on exams, or in your actual job.

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