How to prepare solutions and dilutions

## Calculating a dilution

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

### 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:

$\large \LARGE n{_{i}} = n{_{f}} \rightarrow C{_{i}} \times V{_{i}} = C{_{f}} \times V{_{f}}$
Where
Ci = stock solution concentration (in mol/L)
Vi = stock solution volume (in L)
Cf = final solution concentration (in mol/L)
Vf = final solution volume (in L)

Thus, the volume to be withdrawn from the the stock solution of known concentration is:

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

$\LARGE V{_{i}} = \frac {10{^{-1}} \times 0.1}{1} = 10{^{-2}}L = \mathbf{{\color{DarkGreen}10\ mL }}$

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

$\large d = \frac {\mu}{\mu{_{0}}}$
µ = volumic mass of the stock solution in kg/m3 (g/L)
µ0 = volumic mass of water = 1 000 kg/m3 (g/L)
$\LARGE C{_{sol}} = \frac {\mu{_{0}} \times d \times \%m} {M{_{sol}}}$
$\large \%m = \frac {m{_{HCl}}}{m{_{sol}}}$
Where
Csol = commercial solution concentration
µ0 = water density = 1000 kg/m3 (g/L)
d = commercial solution density
%m = percentage by mass of commercial solution
Msol = molar mass of the commercial solution

#### Calculating the volume to be withdrawn for a dilution

$\large \LARGE V{_{sol}} = \frac {C{_{f}} \times V{_{f}} \times M{_{sol}}} {\mu{_{0}} \times d \times \%m}$
Where
Csol = commercial solution concentration
µ0 = water density = 1000 kg/m3 (g/L)
d = commercial solution density
%m = percentage by mass of commercial solution
Msol = molar mass of the solution

Numerical application:
$\large V{_{sol}} = \frac {10{^{-1}} \times 0.1 \times 36.46}{1000 \times 1.18 \times 0.35} = 8.83 \times 10{^{-4}}L = \mathbf{{\color{DarkGreen} 883 \mu L}}$
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).