Serial Vs Parallel Dilution Method Concentration

Serial Dilutions. A dilution series is a succession of step dilutions, each with the same dilution factor, where the diluted material of the previous step is used to make the subsequent dilution. This is how standard curves for ELISA can be made. To make a dilution series, use the following formulas: Move Volume = Final Volume / (DF -1). As this first step is the highest concentration of compound, the change in buffer solution could cause material to precipitate out of solution. Subsequent dilution of the first sample in a serial dilution method may transfer the compound at a far lower concentration than anticipated. The generated IC5.

By Real-life chemists in real-life labs don’t make every solution from scratch. Instead, they make concentrated stock solutions and then make dilutions of those stocks as necessary for a given experiment. To make a dilution, you simply add a small quantity of a concentrated stock solution to an amount of pure solvent. The resulting solution contains the amount of solute originally taken from the stock solution but disperses that solute throughout a greater volume. Therefore, the final concentration is lower; the final solution is less concentrated and more dilute. How do you know how much of the stock solution to use and how much of the pure solvent to use?

It depends on the concentration of the stock and on the concentration and volume of the final solution you want. You can answer these kinds of pressing questions by using the dilution equation, which relates concentration (C) and volume ( V ) between initial and final states: C 1 V 1 = C 2 V 2 You can use the dilution equation with any units of concentration, provided you use the same units throughout the calculation. Because molarity is such a common way to express concentration, the dilution equation is sometimes expressed in the following way, where M 1 and M 2 refer to the initial and final molarity, respectively: M 1 V 1 = M 2 V 2 For example, how would you prepare 500. ML of 0.200 M NaOH( aq) from a stock solution of 1.5 M NaOH? Start by using the dilution equation, M 1 V 1 = M 2 V 2 The initial molarity, M 1, comes from the stock solution and is therefore 1.5 M. The final molarity is the one you want in your final solution, which is 0.200 M.

The final volume is the one you want for your final solution, 500. ML, which is equivalent to 0.500 L.

Using these known values, you can calculate the initial volume, V 1: The calculated volume is equivalent to 67 mL. The final volume of the aqueous solution is to be 500 mL, and 67 mL of this volume comes from the stock solution. The remainder, 500 mL – 67 mL = 433 mL, comes from pure solvent (water, in this case). So to prepare the solution, add 67 mL of 1.5 M stock solution to 433 mL water.

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Mix and enjoy! T ry another problem: What is the final concentration in molarity of a solution prepared by diluting 2.50 mL of 3.00 M KCl( aq) up to 0.175 L final volume? You can use the dilution equation, M 1 V 1 = M 2 V 2 In this problem, the initial molarity is 3.00 M, the initial volume is 2.50 mL or 2.50 x 10 –3 L and the final volume is 0.175 L. Use these known values to calculate the final molarity, M 2: So, the final concentration in molarity of the solution is 4.29 x 10 –2 M.

Each calibration standard solution is prepared based on the previous calibration standard. The process involves taking a portion of the previous standard and diluting it with the solvent to obtain the next calibration standard. The errors introduced with each successive dilution drops proportionately with the solution concentration. Preparing a series of calibration standards by this method reduces the amount of required time. Most calibration standards span a large range of concentrations, so the accuracy of the calibration standard prepared increases.