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}} \cdot V{_{i}} = C{_{f}} \cdot 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Thus, the volume to be withdrawn from the the stock solution of known concentration is: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}} \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
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Calculating the volume to be withdrawn for a dilution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}} \cdot V{_{f}} \cdot M{_{sol}}} {\mu{_{0}} \cdot d \cdot \%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}} \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).

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