A science educator needs 500 mL of a 20% acid solution. She has 10% and 30% solutions. How much 30% solution is needed?
["How to Create a 20% Acid Solution: A Science Educator’s Guide", "A science educator often faces practical problems that blend chemistry and real-world application. One common task is preparing a precise acid solution, such as creating exactly 500 mL of a 20% acid solution using only available 10% and 30% solutions. This situation isn’t just theoretical—it’s a classic mixture problem that showcases key chemistry principles like concentration, dilution, and solution blending.", "Understanding the Problem", "The educator needs 500 mL of a 20% acid solution but only has access to two stock solutions:\n- 10% acid solution\n- 30% acid solution", "She must determine exactly how many milliliters of the 30% solution to mix with the 10% solution to obtain the desired 20% concentration in the total volume.", "Using the Concept of Mixture Equations", "Let’s define variables:\n- Let ( x ) = volume (in mL) of the 30% acid solution\n- Then ( 500 - x ) = volume of the 10% acid solution", "The total amount of pure acid in the final solution must equal 20% of 500 mL:\n[\n0.20 \ imes 500 = 100 \ ext{ grams of acid (since 1% of 100 mL = 1 gram)}\n]", "The acid contribution comes from both solutions:\n- Acid from 30% solution: ( 0.30x ) grams\n- Acid from 10% solution: ( 0.10(500 - x) ) grams", "Setting up the equation:\n[\n0.30x + 0.10(500 - x) = 100\n]", "Solving the Equation Step-by-Step", "1. Expand the equation:\n[\n0.30x + 50 - 0.10x = 100\n]", "2. Combine like terms:\n[\n(0.30 - 0.10)x + 50 = 100 \implies 0.20x + 50 = 100\n]", "3. Subtract 50 from both sides:\n[\n0.20x = 50\n]", "4. Divide both sides by 0.20:\n[\nx = \frac{50}{0.20} = 250\n]", "Conclusion", "The science educator needs 250 mL of the 30% acid solution. The remaining volume—500 mL minus 250 mL—will be the 10% solution, or 250 mL.", "Final Formula Summary", "[\n\ ext{Volume of 30% solution} = \frac{(C_f - C_d) \ imes V_f}{C_f - C_d + C_i - C_f}\n]", "Where:\n- ( C_f = 20% ) (final concentration)\n- ( C_d = 10% ) (stock 30% solution)\n- ( C_i = 30% ) (stock 30% solution)\n- ( V_f = 500 ) mL (final volume)", "Why This Matters for Science Education", "Understanding how to blend solutions helps students grasp real-world applications of proportional reasoning, stoichiometry, and analytical thinking. It’s a valuable hands-on example of how theoretical concentration calculations directly support laboratory accuracy—a cornerstone skill for future science careers.", "Takeaway\nMixing acid solutions is a manageable problem when approached systematically. By calculating with variables, setting up a balance equation, and solving step-by-step, any educator or student can confidently determine precise volumes—ensuring safe, accurate results in the lab."]
