Question

A concentrated phosphoric acid solution is 85% H₃PO₄ by mass and has a density of 1.69 g/mL at 25^oC. How many mL of concentrated solution is required to prepare 250 mL of 0.3 N H₃PO₄ solution when (MW for H₃PO₄: 97.994 g/mol)

Answer 1

given,

density of H3PO4 = 1.69g/ml

at temperature = 25 degree celsius

molecular weight of H3PO4 = 97.994g/mol

and concentrated phosphoric solution is 85% H3PO4 by mass

now, let us assume that there are 100 g of phosphoric solution

then the mass of H3PO4 = 85.5g

then the moles of H3PO4 = mass of H3PO4/molecular weight of H3PO4

= 85.5 g /97.994 g/mol

= 0.872 mol

volume of solution = mass of phosphoric solution / density of H3PO4

= 100 /1.69 = 59.17mL = 0.05917L

molarity 1 = number of moles/ volume of solution = 0.872 / 0.05917 = 14.73 M

As the H3PO4 can give 3 H+ ions, so n-factor is 3.

normality = n factor x molarity

0.3 = 3 x molarity

molarity 2 = 0.1 M

molarity1 x volume1 = voume2 x molarity 2

14.73 x volume of concentrate solution = 250 x 0.1

volume of concentrate solution = 25 / 14.73 = 1.69 mL