Sanitizer Dilution with Limited Stock
LAB-3 must prepare 12 L of a 4.5% by-volume sanitizer working solution from an 18% stock solution. The mixing carboy is 15 L and has graduation marks every 0.5 L. The shelf holds 2.5 L of sealed 18% stock; opening any additional stock requires a supervisor's approval that takes one day. A batch of samples arrives in four hours. The written procedure sets the acceptable working concentration at 4.0% to 5.0%; below 4.0% the samples must be re-run, and above 5.0% the sanitizer may damage the sample containers. The technician assumes volumes are additive and that mixing causes no measurable contraction. A colleague suggests simply filling the carboy to 12 L with the 2.5 L on hand and proceeding, arguing that “2.5 L is close to 3 L.” The technician has a 500 mL graduated cylinder and can request exactly one additional 500 mL bottle of 18% stock without a full approval cycle.
1. Calculate the volume of 18% stock needed to make 12 L of 4.5% solution, and determine whether the 2.5 L on hand is sufficient. Show the balance equation.
Read the worked answer
Use the volume balance (C_1V_1=C_2V_2). Active ingredient needed: (0.045\times 12=0.54) L. Stock concentration is 0.18, so (V_1=0.54/0.18=3.0) L. The shelf has only 2.5 L, so the shortfall is 0.5 L. Proceeding with 2.5 L and filling to 12 L would give (0.18\times2.5/12=0.0375), or 3.75%, which is below the 4.0% lower limit. Therefore the technician should not proceed to the 12 L target with the stock on hand; the correct judgement is to request the extra bottle or make a smaller batch.
2. If the technician uses all 2.5 L of stock and fills the carboy to the 12 L mark, what concentration results, and does it fall in the acceptable range? What is the largest volume of acceptable solution that could be made from 2.5 L?
Read the worked answer
Using all 2.5 L: active (=0.18\times2.5=0.45) L; concentration (=0.45/12=0.0375=3.75%). Not acceptable because below 4.0%, so samples would need re-running. Largest acceptable volume is set by the lower limit: (V_{\max}=0.45/0.04=11.25) L at 4.0%. The smallest volume within the upper limit is (0.45/0.05=9.0) L at 5.0%. At the 4.5% target, (V=0.45/0.045=10.0) L. So the stock can make between 9.0 L and 11.25 L of acceptable solution, and exactly 10 L at target, but never the requested 12 L.
3. A supervisor can release one additional 500 mL bottle of 18% stock. If the technician adds it to the 2.5 L and then fills to 12 L, what is the concentration? Is it enough to proceed? If instead the target were 5.0%, would the extra bottle suffice?
Read the worked answer
Adding one 500 mL bottle gives 3.0 L of stock, containing (0.18\times3.0=0.54) L active. Filling to 12 L gives (0.54/12=0.045), or 4.5%, exactly the target. So one bottle is enough for the 12 L batch at 4.5%. For a 5.0% target, active needed is (0.05\times12=0.60) L, requiring (0.60/0.18=3.333) L of stock. With 3.0 L the technician is still 0.333 L short, so the extra bottle alone would not meet a 5.0% target; at least another 340 mL would be needed.