Mixture Problem Solver

The Mixture Problem Solver covers the five most common concentration and mixture word problems in one place: blending two solutions to find the combined concentration, finding the missing volume of one solution that brings a mix to a target concentration, diluting a strong solution with pure solvent (typically water), strengthening a weak solution by adding pure solute, and the classic drain-and-replace problem where part of a tank's contents is swapped for another mix. Type in the concentrations and volumes in your preferred units — percent, decimal, or per-mille — and the solver applies a single conservation-of-mass identity, walks through the algebra step by step in LaTeX, and shows an animated three-beaker visualization with color-coded fill levels that respond to the strength of each solution.

How to use this solver

1. Pick the scenario that matches your problem from the dropdown — blend, target, dilute, strengthen, or drain-and-replace.
2. Choose the concentration unit (percent, decimal, or per-mille) and the volume or mass unit (mL, L, gal, cup, fl oz, g, kg, lb). All inputs use the same units.
3. Enter the concentration and volume of solution A. For blend, target, and replace scenarios, also enter the concentration of solution B (and its volume for blend).
4. For all scenarios except plain blend, enter the target concentration you want at the end.
5. Click Solve. The headline value is the missing quantity — final concentration, volume of B to add, water to dilute, pure solute to strengthen, or amount to drain.
6. Watch the beakers fill with color that reflects each solution's concentration. The result beaker shows the final mixture.

The five formulas at a glance

The conservation principle (the key idea)

Every mixture problem reduces to one identity: the mass of solute is conserved. If you mix two streams and nothing chemically reacts, the amount of solute in the final mixture equals the sum of solute in each input.

\[ c_1 V_1 + c_2 V_2 \;=\; c_f (V_1 + V_2) \]

Each scenario in this calculator is just a different unknown in this same equation:

• Blend — solve for \( c_f \) given \( c_1, V_1, c_2, V_2 \).
• Target — solve for \( V_2 \) given \( c_1, V_1, c_2, c_t \).
• Dilute — set \( c_2 = 0 \) (pure water) and solve for \( V_w \).
• Strengthen — set \( c_2 = 1 \) (pure solute) and solve for \( V_s \).
• Replace — keep \( V_1 \) constant; replace volume \( V_r \) of A with B.

Worked example: blending two acid solutions

A chemistry student mixes 300 mL of a 20% acid solution with 200 mL of a 50% acid solution. What is the final concentration?

1. Solute in A: \( 0.20 \times 300 = 60 \) mL of pure acid.
2. Solute in B: \( 0.50 \times 200 = 100 \) mL of pure acid.
3. Total solute: \( 60 + 100 = 160 \) mL.
4. Total volume: \( 300 + 200 = 500 \) mL.
5. Final concentration: \( c_f = \dfrac{160}{500} = 0.32 = 32\% \).

Worked example: diluting alcohol

You have 250 mL of 70% rubbing alcohol but you need 40% strength for a topical preparation. How much water do you add?

1. Mass of solute is conserved: \( 0.70 \times 250 = 0.40 \times (250 + V_w) \).
2. 175 = 100 + 0.40 V_w → \( V_w = \dfrac{75}{0.40} = 187.5 \) mL.
3. Add 187.5 mL of water; final volume is 437.5 mL.

Worked example: drain-and-replace antifreeze

A car radiator holds 8 L of 20% antifreeze. The owner wants 50% antifreeze. They will drain part of the mix and replace it with the same volume of 90% antifreeze. How much do they drain?

1. The mass of solute after draining: \( 0.20 (8 - V_r) \).
2. After refilling: \( 0.20 (8 - V_r) + 0.90 V_r = 0.50 \times 8 \).
3. 1.6 − 0.20 V_r + 0.90 V_r = 4 → 0.70 V_r = 2.4 → \( V_r = 3.43 \) L.
4. Drain about 3.43 L of the existing mix and pour in 3.43 L of 90% antifreeze.

Common pitfalls and how to avoid them

• Target outside the input range — you cannot blend two solutions to a concentration outside their min/max. To go below the lower or above the higher input, you need pure solvent or pure solute.
• Mixing percent and decimal — 50% is 0.50, not 50. The calculator does the conversion for you when you pick the right unit, but on paper, always convert percentages to decimals before arithmetic.
• Mass vs volume — for liquids near room temperature the formulas hold for both, but if densities differ greatly (e.g. mixing alcohol and oil) you should use mass, not volume, to keep the conservation law exact.
• Forgetting to add the new volume — the denominator of \( c_f \) is \( V_1 + V_2 \), not just \( V_1 \). Beginners often divide solute by the original volume only, which gives a wrong answer.
• Drain-and-replace is symmetric — replacing 3 L of 20% with 3 L of 90% gives the same final concentration as starting with the leftover 5 L of 20% and adding 3 L of 90%. The drain step never changes the concentration of what's left, only the volume.

Quick conversion reference

Where mixture problems show up in real life

• Chemistry labs — preparing acid or buffer solutions to a precise molarity, diluting concentrated stock with water (the M₁V₁ = M₂V₂ rule is the dilution scenario in this tool).
• Pharmacy — compounding creams and IV fluids to a target percent strength, often by blending two stock concentrations the pharmacist already has.
• Cooking and brewing — adjusting brine salinity, sugar syrups, or beer alcohol by volume by blending stronger and weaker batches.
• Automotive — antifreeze, washer fluid, and DEF (diesel exhaust fluid) often need to be diluted or strengthened to a target percent.
• Algebra textbooks — drain-and-replace, "how much water" and "two-tank" problems are some of the most popular SAT and competition math word problems.

Frequently asked questions