Solution Mixing Calculator
Calculate final concentration when mixing solutions.
Solution Mixing Calculator
C₁V₁ + C₂V₂ = C₃V₃
Parameters
Solution Mixing
This calculator solves the mixing equation C₁V₁ + C₂V₂ = C₃V₃, where C is concentration and V is volume. Since V₃ = V₁ + V₂, you can solve for the two unknown volumes if you know the three concentrations and the final volume.
Understanding Solution Mixing
Calculating the Concentration of Mixtures.
What is Solution Mixing?
Solution Mixing is the process of combining two or more solutions to create a new, final solution. In chemistry, this most often involves calculating the final concentration of a solute after mixing solutions of different concentrations.
The fundamental principle behind all mixing calculations is the conservation of moles. The total amount of solute (in moles) in the final mixture is simply the sum of the amounts of solute contributed by each of the initial solutions.
Similarly, the final volume of the mixture is assumed to be the sum of the individual volumes of the solutions being mixed (unless stated otherwise).
Example: If you mix a light-blue sugar solution with a dark-blue sugar solution, the final mixture will have a color and concentration that is an average of the two, weighted by their volumes.
The Mixing Formula
The process of calculating the final concentration is governed by a straightforward formula derived from the principle of mole conservation.
Since Moles = Molarity × Volume (M * V), the total moles are M₁V₁ + M₂V₂. The total volume is V₁ + V₂.
The final molarity (M_f) is the total moles divided by the total volume:
M_f = (M₁V₁ + M₂V₂) / (V₁ + V₂)
Example:This equation can be extended for mixing any number of solutions: M_f = (M₁V₁ + M₂V₂ + M₃V₃ + ...) / (V₁ + V₂ + V₃ + ...).
Components of the Equation
Each part of the mixing formula represents a specific quantity:
M₁ and V₁: The Molarity and Volume of the first solution.
M₂ and V₂: The Molarity and Volume of the second solution.
M_f: The final Molarity of the resulting mixture.
It is crucial that the units for volume (e.g., mL or L) are consistent for all terms in the calculation.
Example:This formula essentially calculates a weighted average of the initial concentrations, where the weighting is based on the volume of each solution.
Real-World Application: Laboratories and Healthcare
Solution mixing is one of the most common procedures in any laboratory or medical setting.
Laboratory Research: Scientists constantly need to prepare solutions of a specific concentration for experiments. They often do this by mixing a concentrated stock solution with a diluent (like water) or another solution.
Pharmaceutical Compounding: Pharmacists use these principles to prepare custom medications, such as IV drips or liquid medicines, by mixing different stock drug solutions to achieve the exact concentration prescribed by a doctor.
Industrial Processes: In manufacturing, large vats of chemical solutions are often adjusted by mixing in other solutions to maintain optimal concentrations for a chemical process.
Example:A hospital pharmacist preparing an intravenous saline solution for a patient must perform a precise mixing calculation to ensure the final concentration is correct and safe for the patient.
Key Summary
- **Solution Mixing** calculations are based on the **conservation of moles**.
- The final molarity of a mixture is the total moles of solute divided by the total volume.
- The key formula is **M_f = (M₁V₁ + M₂V₂) / (V₁ + V₂)**.
- This is a fundamental skill for preparing solutions in laboratory, medical, and industrial settings.
Practice Problems
Problem: If you mix 300 mL of a 2.0 M HCl solution with 200 mL of a 4.0 M HCl solution, what is the final molarity of the mixture?
Use the mixing formula: M_f = (M₁V₁ + M₂V₂) / (V₁ + V₂).
Solution: M_f = ((2.0 M * 300 mL) + (4.0 M * 200 mL)) / (300 mL + 200 mL) = (600 + 800) / 500 = 1400 / 500 = 2.8 M.
Problem: You need to prepare 1.0 L of a 1.5 M NaCl solution. You have a large supply of 1.0 M NaCl and 2.5 M NaCl. What volume of each solution should you mix?
This is a system of two equations. Let V₁ be the volume of the 1.0M solution and V₂ be the volume of the 2.5M solution. Eq 1: V₁ + V₂ = 1.0 L. Eq 2: (1.0*V₁ + 2.5*V₂) / 1.0 = 1.5 M. Solve the system.
Solution: From Eq 1, V₁ = 1.0 - V₂. Substitute into Eq 2: 1.0*(1.0 - V₂) + 2.5*V₂ = 1.5. => 1.0 - V₂ + 2.5*V₂ = 1.5. => 1.5*V₂ = 0.5. => V₂ = 0.333 L. Then V₁ = 1.0 - 0.333 = 0.667 L. You should mix 667 mL of the 1.0 M solution and 333 mL of the 2.5 M solution.
Frequently Asked Questions
What is the difference between mixing and dilution?
Dilution is a specific type of mixing where one of the 'solutions' is a pure solvent (like water), which has a concentration of zero. You can see this in the mixing formula: if M₂ = 0, the formula simplifies to M_f = M₁V₁ / (V₁ + V₂), which is the dilution equation.
What happens if you mix solutions of two different chemicals that react with each other?
If the solutes react, it is no longer a simple mixing problem. It becomes a stoichiometry problem where you must first calculate the moles of each reactant, determine the limiting reactant, and then calculate the final concentration of all species remaining after the reaction is complete.
Are volumes always additive?
In general chemistry, volumes are assumed to be perfectly additive. In high-precision physical chemistry, this is not always true. Sometimes, due to intermolecular interactions, the final volume can be slightly different from the sum of the individual volumes. However, for most practical purposes, the assumption of additivity is very accurate.
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