Dilution of Ingested Glucose - ChemPRIME

# Dilution of Ingested Glucose

back to Diluting and Mixing Solutions

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 (See Example 1 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 concenrations 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 1 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:

$n_{\text{glucose}}=\text{24}\text{.71 ml} ~\times~ \frac{\text{10}^{-3}\text{L}}{\text{1 ml}} ~\times ~\frac{\text{1}\text{.30 mol}}{\text{L}} ~=~\text{0}\text{.0321 mol}$

Dividing by the new volume gives the concentration

$c_{\text{glucose}}=\frac{n_{\text{glucose}}}{V}=\frac{\text{0}\text{.0321 mmol}}{\text{4.7 L}}=\text{0.00684 mol/L}$

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,
$n_{\text{glucose}}=\text{24}\text{.71 cm}^{\text{3}}\text{ }\times \text{ }\frac{\text{1}\text{.30 mmol}}{\text{1 cm}^{\text{3}}}=\text{32}\text{.1 mmol}$

Dividing by the new volume gives the concentration

$c_{\text{glucose}}=\frac{n_{\text{glucose}}}{V}=\frac{\text{32}\text{.1 mmol}}{\text{4700 cm}^{\text{3}}}=\text{0.00684 mmol cm}^{\text{-3}}$

EXAMPLE 2

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

\begin{align} & V_{\text{new}}\xrightarrow{c_{\text{new}}}n_{\text{glucose}}\xrightarrow{c_{\text{old}}}V_{\text{old}} \\ & \\ & V_{\text{old}}=\text{500}\text{.00 cm}^{\text{3}}\text{ }\times \text{ }\frac{\text{0}\text{.006000 mmol}}{\text{1 cm}^{\text{3}}}\text{ }\times \text{ }\frac{\text{1 cm}^{\text{3}}}{\text{0}\text{.1000 mmol}}\text{ }=\text{30}\text{.00 cm}^{\text{3}} \\ \end{align}

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 (Cconc and Vconc) were used to calculate the amount of solute, and this amount was entirely transferred to the dilute solution:

Cconc  x  Vconc   =   nconc   =   ndil  =  Cdil  x  Vdil

So
Cconc  x  Vconc  =  Cdil  x  Vdil

So for Example 2,

(0.006000 M)   x   (Vconc)   =   (500.00 ml)   x   (0.1000 M)

Vconc= 30.00 mL, which will be diluted to 500.00 mL as before.

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