A chemical reaction requires a 5% saline solution. If you have 300 mL of a 2% saline solution, how much pure salt must be added to achieve the required concentration?
["How to Create a 5% Saline Solution: A Step-by-Step Guide", "When working in laboratories, medicine, or industrial chemistry, precise saline solutions are essential — especially when a specific concentration like 5% is required. You may wonder: If I start with 300 mL of a 2% saline solution, how much pure salt must I add to reach exactly 5%? This article walks you through solving this chemistry problem step by step while exploring the significance of accurate saline preparation.", "---", "### Understanding Saline Solutions", "A saline solution’s concentration is defined as the mass of salt (in grams) per volume of solution, typically expressed as a percentage. Pure salt (sodium chloride, NaCl) is dissolved in water to form this solution. For this scenario:", "- Starting volume: 300 mL\n- Starting concentration: 2%\n- Target concentration: 5%", "The key fact: adding pure salt increases both the salt mass and the total solution volume — but carefully, to maintain the correct ratio.", "---", "### Step 1: Calculate the Salt Mass in the Initial Solution", "Since concentration = (mass of salt / total mass of solution) × 100%, but we can work directly with mass assuming a constant solution density (close enough for aqueous solutions), we approximate:", "At 2% concentration:", "[\n\ ext{Mass of salt} = 2% \ ext{ of total initial solution mass}\n]", "Assuming 1 mL of solution ≈ 1 gram (standard approximation):", "[\n\ ext{Initial salt mass} = 2% \ imes 300\ \ ext{mL} = 0.02 \ imes 300 = 6\ \ ext{grams}\n]", "[\n\ ext{Initial solution mass} \approx 300\ \ ext{g}\n]", "---", "### Step 2: Let x = grams of pure salt to add", "We add x grams of pure salt to the 300 g solution.", "After addition:", "[\n\ ext{New salt mass} = 6 + x\ \ ext{grams}\n]\n[\n\ ext{New total mass} = 300 + x\ \ ext{grams}\n]", "We want the final concentration to be 5%, so:", "[\n\frac{6 + x}{300 + x} = 0.05\n]", "---", "### Step 3: Solve the equation", "[\n6 + x = 0.05(300 + x)\n]\n[\n6 + x = 15 + 0.05x\n]\n[\nx - 0.05x = 15 - 6\n]\n[\n0.95x = 9\n]\n[\nx = \frac{9}{0.95} \approx 9.47\ \ ext{grams}\n]", "---", "### Final Answer", "To increase 300 mL of a 2% saline solution to a 5% saline solution, you must add approximately 9.47 grams of pure sodium chloride.", "---", "### Why Precision Matters in Saline Solutions", "Even small errors in salt concentration can significantly impact experimental results or medical safety. For instance, in saline IV solutions or lab reagents, precise salinity ensures compatibility with biological systems and accurate reaction conditions. The calculation above demonstrates how something as simple as dissolving a measured amount of salt demonstrates both foundational chemistry principles and real-world applicability.", "---", "### Resources for Further Reading:", "- Understanding Common Saline Concentrations\n- Measuring and Preparing Standard Solutions\n- The Importance of Accuracy in Chemical Reactions", "---", "Keywords: saline solution, sodium chloride, salt concentration, 2% saline, 5% saline, chemistry calculation, lab preparation, add salt solution, solution prep, chemical reactions, precise mixing, molarity solution.", "---", "Perfect consistency, accurate concentration — whether in science or daily lab work, every molecule counts!"]
