How to prepare solutions and dilutions

## Calculating a mass to be weighed

### For a mass concentration

Example: prepare a 100 mL solution of 1 mg/L of Na+ with NaCl salt

1 mg NaCl ≠ 1 mg Na+

MNaCl = 58.44 g/mol ; MNa = 22.99 g/mol ; MCl = 35.45 g/mol $\large \Rightarrow \% M{_{Na}} = \frac{M{_{Na}}}{M{_{NaCl}}} = 39.3\%$

It can be seen here that in 1 mg of NaCl, there is in fact only 0.39 mg of Na+. This ratio must therefore be taken into account when calculating the mass of salt to be weighed in order to obtain a concentration of 1mg/L of Na+.

#### Calculating the mass of salt to be weighed

$\LARGE m{_{salt}} = \frac{M{_{salt}}}{M{_{element}}} \times C{_{f}} \times V{_{f}}$
Where
msalt = mass of salt to be weighed (in g)
Msalt = molar mass of salt (in g/mol)
Melement = molar mass of the element to consider (in g/mol)
Cf = final mass concentration (in g/L)
Vf = final volume (in L)
##### Mass of NaCl to be weighed

$\large m{_{NaCl}} = \frac{58.44}{22.99} \times 10{^{-3}} \times 0.1 = 2.54 \times 10{^{-4}}g = \mathbf{{\color{DarkGreen} 0.254mg}}$

Thus in order to obtain a 100 mL solution of 1mg/L of Na+, we will need to weigh 0.254 mg of salt. In reality, weighing 0.254 mg of salt is practically impossible because the most precise scales have a precision of 0.1 mg. We will need to prepare a more concentrated solution and then dilute it in order to obtain the desired concentration level of Na+.

### For a molar concentration

Example: prepare a 100 mL solution of 1 mol/L Na+ with Na2SO4 salt

Pay attention to the ratio between the number of moles of salt and of ion.
Here 1 mole of Na2SO4 contains 2 moles of Na+
$\LARGE m{_{salt}} = \frac{n{_{salt}}}{n{_{element}}} \times C{_{f}} \times M{_{salt}} \times V{_{f}}$
Where
msalt = mass of salt to be weighed (in g)
Msalt = molar mass of salt (in g/mol)
nsalt/nelement = molar ratio between the element and the salt
Cf = final molar concentration (in mol/L)
Vf = final volume (in L)

Thus, the mass of salt to be weighed in order to prepare a 1 mol/L solution of Na+ is :

$\large m{_{NaCl}} = \frac{1}{2} \times 1 \times 142.04 \times 0.1 = \mathbf{{\color{DarkGreen} 7.1g}}$