Chp2_SerialDilute

# Chp2_SerialDilute - Let us make 55mL of solution(to allow...

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Serial Dilutions Suppose I wish to make ~ 50mL of 0.1M; 0.01M; 0.001M; 0.0001M; and 0.00001M solution starting with a 0.1M “stock” solution. Use: C i V i = C f V f Method I. ( 29 ( 29 ( 29 ( 29 stock mL x M mL M mL x 5 01 . 0 50 1 . 0 = = to which we add 45 mL of water to make 50 mL final volume. If we continue this process using the original 0.1M stock solution , then we find: mL of 0.1M stock Vol water Makes 50mL of this conc. 5 45 0.01 0.5 49.5 0.001 0.05 49.95 0.0001 0.005 49.995 0.00001 Note, for example, the increase in “sig figs” in the water volume. The problem here is not that it is impossible to measure 0.005 mL of stock, but making such small additions to relatively large volumes of solvent (water vol is 10,000 times as large) leads to large errors. Method II. Here, we shall use each successively more dilute solution as the “stock” for the next most dilute.
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Unformatted text preview: Let us make 55mL of solution (to allow for the fact that, as we shall see, we’ll be using some of it). To make 55mL of 0.01M solution, ( 29 ( 29 ( 29 ( 29 stock mL x M mL M mL x 5 . 5 01 . 55 1 . = = Now, take the 0.01M solution as the new “stock” to make 55mL of 0.001M solution: ( 29 ( 29 ( 29 ( 29 stock M of mL x M mL M mL x 01 . 5 . 5 001 . 55 01 . = = As you see, we make 55mL of each more dilute solution, then use 5.5 mL of that, to make the next more dilute member of the set. There is a means to calculate the amounts needed “exactly” but this simplistic example helps to make the case....
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