Nitric acid is usually purchased in a concentrated form that is 70.3% HNO3 by mass and has a density of 1.41 g/mL. How much concentrated solution would you take to prepare 1.15 L of 0.115M HNO3 by mixing with water?

to prepare a 0.115 M HNO3 solution, 11.85 mL of the concentrate solution is needed and water is added until the desired volume of 1.15 L is reached

Explanation:

%w/w = 70.3% = g HNO3/g sln * 100

⇒ 0.703 = g HNO3/g sln

let g sln = 1000g:

⇒ g HNO3 = 0.703 * 1000g = 703g HNO3

⇒ mol HNO3 = 703g * (mol/63.01g HNO3) = 11.157 mol HNO3

for 1 L of solution:

⇒ C HNO3 = 11.157 mol / L

dilution formula:

V1*C1 = V2*C2

∴ C1 = 11.157 mol/L

∴ C2 = 0.115 mol/L

∴ V2 = 1.15 L

⇒ V1 = V2*C2 / C1

⇒ V1 = ((1.15 L) * (0.115 mol/L)) / 11.157 mol/L

⇒ V1 = 0.01185 L (11.85 mL)