Question

A chemist must prepare 700.0mL of nitric acid solution with a pH of 1.50 at 25°C. He will do this in three steps: Fill a 700.0mL volumetric flask about halfway with distilled water. Measure out a small volume of concentrated (6.0M) stock nitric acid solution and add it to the flask. Fill the flask to the mark with distilled water. Calculate the volume of concentrated nitric acid that the chemist must measure out in the second step. Round your answer to 2 significant digits.

Answer 1

`V_(conc)=3.7mL`

Step-by-step explanation:

Hello,

In this case, a problem is about dilution, which is a process wherein from a concentrated acid, a less concentrated solution is obtained via adding an extra volume. In such a way, since the required acid has a pH of 1.50, it means that it has a hydrogen ions concentration of:

`[H]^+=10^(-pH)=10^(-1.50)=0.0316M`

Thus, since nitric acid is a strong acid, the concentration of hydrogen ions, equals the concentration is the acid due to complete dissociation, hence:

`[H]^+=[HNO_3]=0.0316M`

Thereby, it is concentration of the diluted acid. Now, as during a dilution process the moles of the acid are kept constant we obtain:

`n_(concentrated)=n_(diluted)`

That in terms of molarities and volume result:

`V_(dil)M_(dil)=V_(conc)M_(conc)`

Thus, solving for the used volume of concentrated acid, we obtain:

`V_(conc)=(V_(conc)M_(dil))/(M_(conc)) =(700mL*0.0316M)/(6.0M) \\\\V_(conc)=3.7mL`

Best regards.