Dilution of Ingested Glucose

In the last section on Low Glycemic Index Foods and Blood Glucose Concentration we saw two examples where it is convenient to determine the concentration of a solution made by diluting a solution of known concentration.

First, we calculated the plasma glucose concentration that would result from ingesting 24.71 cm³ of a sample of 1.30 M Karo syrup if it were diluted to our blood volume of about 4.7 L (Example $\PageIndex{2}$ below). Next, we noted the importance of creating standard solutions of exactly known glucose concentration, in order to calibrate a glucometer. Often several solutions of different concentrations are necessary, so a rapid method of preparing them is desired. Dilution of a single stock solution of known concentration provides a rapid method. Aliquots (carefully measured volumes) of the stock solution can then be diluted to any desired volume. See Example 2 Below.

In other cases it may be inconvenient to weigh accurately a small enough mass of sample to prepare a small volume of a dilute solution. Each of these situations requires that a solution be diluted to obtain the desired concentration.

Example $\PageIndex{1}$: Concentration of Diluted Solution

A buret is used to measure 24.71 ml of 1.30 M glucose, which is ingested and diluted to 4.7 l in the circulatory system. What is the concentration of the diluted solution?

Solution

To calculate concentration, we first obtain the amount of glucose in the 24.71 ml (24.71 cm³) of solution:

$$n_{glucose} = \text{24.71 mL} \times \dfrac{\text{10}^{-3}\text{L}}{\text{1mL}} \times \dfrac{\text{1.30 mol}}{\text{1 L}} = \text{0.0321 mol}$$

Dividing by the new volume gives the concentration

$$c_{glucose} = \dfrac{n_{glucose}}{\text{V}} = \dfrac{\text{0.0321 mol}}{\text{4.7 L}} = \text{0.00684} \dfrac{mol}{L}$$

Thus the new solution is 0.00684 M. This is not high enough to be classified as hyperglycemic (>0.007 M or 7 mM).

Alternatively,

$$n_{glucose} = \text{24.71 cm}^3 \times \dfrac{\text{1.30 mmol}}{\text{1 cm}^3} = \text{32.1 mmol}$$

Dividing by the new volume gives the concentration

$$c_{glucose} = \dfrac{n_{glucose}}{\text{V}} = \dfrac{\text{32.1 mmol}}{\text{4700 cm}^3} = \text{0.00684} \dfrac{mmol}{cm^3}$$

Example $\PageIndex{2}$: Volume Needed

What volume of 0.1000 M glucose solution is necessary to make 500 mL of a 0.006000 M standard solution?

Solution

Using the volume and concentration of the desired solution, we can calculate the amount of glucose required. Then the concentration of the original solution (0.1000 M) can be used to convert that amount of glucose to the necessary volume. Schematically

$$\text{V}_{new} \xrightarrow {c_{new}} n_{glucose} \xrightarrow {c_{old}} \text{V}_{old}$$

$$\text{V}_{old} = \text{500.00 cm}^3 \times \dfrac{\text{0.00600 mmol}}{\text{1 cm}^3} \times \dfrac{\text{1 cm}^3}{\text{0.100 mmol}} = \text{30.00 cm}^3$$

Thus we should dilute a 30-ml aliquot of the stock solution to 500.00 ml. This could be done by measuring 30.00 ml from a buret into a 500.00-ml volumetric flask and adding water up to the mark.

Note that the calculation above can be simplified, since the concentration and volume of a concentrated solution (C_conc and V_conc) were used to calculate the amount of solute, and this amount was entirely transferred to the dilute solution:

$$\text{C}_{conc} \times \text{V}_{conc} = \text{n}_{conc} = \text{n}_{dil} = \text{C}_{dil} \times \text{V}_{dil}\nonumber$$

So

$$\text{C}_{conc} \times \text{V}_{conc} = \text{C}_{dil} \times \text{V}_{dil}\nonumber$$

So for Example $\PageIndex{2}$,

$$\text{(0.006000 M) } \times \text{V}_{conc} = \text{(500.00 mL)} \times \text{ (0.1000 M)}$$

$\text{V}_{conc} = \text{30.00 mL}$ which will be diluted to 500.00 mL as before. 

Note

The calculated volume will have the same dimensions as the input volume, and dimensional analysis tells us that in this case we do not need to convert to liters, since L cancels when we divide M (mol/L) by M (mol/L).

From ChemPRIME: 3.11: Diluting and Mixing Solutions

Contributors and Attributions

• Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.

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