30. Answers to Practice Problems in Chap. 3, Dilutions and Concentrations

A diagram of this scheme is shown here.

The calculations are performed as follows:

A diagram of this scheme is shown below (note that the first dilution is “dilute to,” while the last two dilutions are “dilute with”).

The calculations are performed as follows:

A diagram of this scheme is shown below (note that all of these are “dilute with,” the standards are prepared in parallel, and 100 μL = 0.1 mL).

The calculations are performed as shown below. For the stock solution:

For the standard solutions:

First, 160 g/L = 160 mg/mL. Therefore, large dilutions are needed to get from 160 mg/mL to 0–0.5 mg/mL. From this point, there are several approaches that could be used. One approach would be to dilute the stock solution down to the most concentrated standard (0.5 mg/mL) and then do further dilutions from that concentration. To do this, we calculate the DF:

From this DF, we can calculate the ratio of volumes needed. Recall that:

Therefore, we need to find a ratio of volumes equal to 1/320 to do this dilution. You do not have a 320 mL volumetric flask to do a simple 1 mL into 320 mL total dilution. However, you could use a 250 mL volumetric flask for the final volume and calculate the starting volume of the stock solution needed:

Therefore, the 0.5 mL standard is made by diluting 0.781 mL (using a 1 mL adjustable pipettor) to 250 mL final volume. You could use a variety of different dilutions to get the same final concentrations, as long as you don’t use less than 0.2 mL as your starting volume (e.g., you could also dilute 0.313 mL to 100 mL and get the same concentration). From the 0.5 mg/mL standard, the other standards (0–0.3 mg/mL) can easily be made in 2 mL volumes by combining variable volumes of the 0.5 mg/mL standard with water to achieve a total volume of 2 mL. These volumes are calculated as follows: and so forth for each solution. The volume of water is the amount needed to bring the starting volume up to 2 mL: and so forth. The dilutions from the diluted 0.5 mg/mL stock solution are shown in Table 30.1:

A diagram of this scheme is shown here:

We know that the anthocyanin concentration of the juice is probably somewhere between 750 and 3000 μg/mL (0.75–3 mg/mL). However, it could literally be anywhere within this range. Samples with 0.75 and 3 mg/mL anthocyanins would require very different dilutions to get near the center of the standard curve (~0.25 mg/mL)—how do we handle this? The solution is to design a dilution scheme with various dilutions of the same sample, which will likely yield at least 1 dilution in the acceptable range that can be used for quantification. To do this, we assume that the anthocyanin concentration in the juice could be on the extremes (~0.75 and ~3 mg/mL) and right in the middle of those extremes (~1.125 mg/mL). Then, we can calculate 3 different dilutions (assuming the 3 different anthocyanin levels and diluting to the middle of the curve) so that we will have at least 1 useable sample regardless of the actual sample concentration. Let’s assume that we want a final sample volume of 10 mL (we could pick any volume that corresponds to a volumetric flask we have available, but it wouldn’t make sense to make 50–1000 mL of diluted sample if we only need 2 mL for the analysis):

This problem is set up as shown below. Note that all of the minerals from 2.8 mL milk are diluted to a total volume of 50 mL, so these steps can be simplified somewhat in the calculation as shown in the figure.

The problem can then be set up and solved as a multistep dilution as follows:

First, you need to get the sample and standards into the same units. Convert the sample caffeine concentration to mM:

Then, determine the starting volume needed to dilute to 250 mL total volume and obtain a concentration at the center of the standard curve (50 μM caffeine):