Question

It is necessary to make 225 mL of 0.222 M solution of nitric acid. Looking on the shelf, you see only 16 M nitric acid. How much concentrated nitric acid is required to make the desired solution?

Answer 1

Explanation:

The required concentration of $HNO_3$ M1 =0.222 M.

The required volume of $HNO_3$ is V1 =225 mL.

The standard solution of $HNO_3$ is M2 =16 M.

The volume of standard solution required can be calculated as shown below:

Since the number of moles of solute does not change on dilution.

The number of moles $n=molarity * volume$

$M_1.V_1=M_2.V_2$

$V2=\frac{M_1.V_1}{M_2} \\=0.222M x 225 mL / 16 M\\=3.12 mL$

Hence, 3.12 mL of 16 m nitric acid is required to prepare 0.222 M and 225 mL of nitric acid.