Chimactiv - Interactive numerical educational resources for the analysis of complex media

Calculating a dilution

The quantity of substance introduced via a stock solution is the same as that of the final solution n_i = n_f

Starting from a solution with a known concentration

Example: prepare 100 mL of a 10^-1 mol/L solution of Na⁺ from a 1 mol/L solution

Starting from the above equation, we obtain:

$\large \LARGE n{_{i}} = n{_{f}} \rightarrow C{_{i}} \cdot V{_{i}} = C{_{f}} \cdot V{_{f}}$

Where
C_i = stock solution concentration (in mol/L)
V_i = stock solution volume (in L)
C_f = final solution concentration (in mol/L)
V_f = final solution volume (in L)

Thus, the volume to be withdrawn from the the stock solution of known concentration is:

$\large \large \LARGE V{_{i}} = \frac {C{_{f}} \cdot V{_{f}}}{C{_{i}}}$

$\large \LARGE V{_{i}} = \frac {10{^{-1}} \cdot 0.1}{1} = 10{^{-2}}L = 10mL$

In order to prepare a 100 mL solution of Na⁺ at 10^-1 mol/L from a 1 mol/L solution, we will need to withdraw 10 mL then make it up to 100 mL with the solvent (water or other solution).

Starting from a commercial solution

Example: prepare a 100 mL solution at 10^-1 mol/L of HCl from a commercialised solution.

Most labels of commercial solutions do not give the concentration level but instead the molar mass: M (36.46g/mol), the density: d (1.18) and the percentage by mass: %m (0.35)

Calculating a commercial solution's concentration

$d = \frac {\mu}{\mu{_{0}}}$

µ = volumic mass of the stock solution in kg/m³ (g/L)
µ0 = volumic mass of water = 1 000 kg/m³ (g/L)

$\LARGE C{_{sol}} = \frac {\mu{_{0}} \cdot d \cdot \%m} {M{_{sol}}}$

$\large \%m = \frac {m{_{HCl}}}{m{_{sol}}}$

Where
C_sol = commercial solution concentration
µ₀ = water density = 1000 kg/m³ (g/L)
d = commercial solution density
%m = percentage by mass of commercial solution
M_sol = molar mass of the commercial solution

Calculating the volume to be withdrawn for a dilution

$\large \LARGE V{_{sol}} = \frac {C{_{f}} \cdot V{_{f}} \cdot M{_{sol}}} {\mu{_{0}} \cdot d \cdot \%m}$

Numerical application:
$\large V{_{sol}} = \frac {10{^{-1}} \cdot 0.1 \cdot 36.46}{1000 \cdot 1.18 \cdot 0.35} = 8.83 \cdot 10{^{-4}}L = 883 \mu L$
In order to prepare a 100 mL solution of HCl at 10^-1 mol/L from a (pure) commercial solution, we need to withdraw a sample of 883 µl then make up to 100 mL with the solvent (water or other solution).