Question

A student needs to prepare a stock solution for the lab - 500.0 ml of 0.750 m nitric acid. The student is provided with concentrated nitric acid (15.9 m) and also with distilled water. How many ml of distilled water would the student need to add in order to prepare the needed stock solution?

Answer 1

Molarity of concentrated nitric acid = 15.9 M

Volume of the stock solution to be prepared = 500.0 mL

Concentration of the stock that is to be prepared = 0.750 M

Calculating moles from molarity and volume of stock:

$500.0mL*\frac{1L}{1000mL}*0.750\frac{mol}{L}  =0.375 mol HNO_{3}$

Calculating volume of concentrated nitric acid to be taken for the preparation of stock solution:

$0.375mol*\frac{1L}{15.9mol} =0.0236 L$

Converting L to mL:

$0.0236L*\frac{1000mL}{1L}$ = 23.6 mL

Volume of distilled water to be added to 23.6 mL of 15.9 M nitric acid to get the given concentration = 5000.0mL-23.6mL=976.4 mL

Therefore, 976.4 mL distilled water is to be added to 23.6 mL of 15.9 M nitric acid solution to prepare 500.0 mL of 0.750M nitric acid.