Question #69999

1 Answer
Feb 3, 2015

The calculation makes several assumptions, but I think you need 0.75 mL of 25 % ammonia solution.

Step 1

HPMCP is hydroxypropylmethylcellulose phthalate.

The repeating unit in HPMCP is

pub2.hi2000.com

This gives a formula of C₃₅H₄₁O₁₈ and a molar mass of 749.7 g.

HPMCP is a diprotic acid. Let's write it as H₂A.

Then

H₂A + 2NH₃ → 2NH₄⁺ + 2A⁻

The density of 25 % aqueous NH₃ is 0.911 g/cm³.

Volume of NH3=3.5 g H2A×1mol H2A749.7 g H2A×2 mol NH31 mol H2A×17.03 g NH31 mol NH3×100 g NH3 soln25 g NH3×1mL NH3 soln0.911 g NH3 soln=0.70 mL NH3 soln

Step 2

HPMCAS is hydroxypropylmethylcellulose acetate succinate. The repeating unit in HPMCAS is

www.harke.com

This gives a formula of C₃₂H₅₀O₂₀ and a molar mass of 754.7 g.

HPMCAS is also a diprotic acid. Let's write it as H₂A.

Then

H₂A + 2NH₃ → 2NH₄⁺ + 2A⁻

Volume of NH3=0.25 g H2A×1mol H2A754.7 g H2A×2 mol NH31 mol H2A×17.03 g NH31 mol NH3×100 g NH3 soln25 g NH3×1mL NH3 soln0.911 g NH3 soln=0.050 mL NH3 soln

Step 3

Total volume = 0.70 mL + 0.050 mL = 0.75 mL