# Question #69999

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 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.

$\text{Volume of NH"_3 = "3.5 g H"_2"A" × ("1mol H"_2"A")/("749.7 g H"_2"A") × ("2 mol NH"_3)/("1 mol H"_2"A") × ("17.03 g NH"_3)/("1 mol NH"_3) × ("100 g NH"_3" soln")/("25 g NH"_3) × ("1mL NH"_3" soln")/("0.911 g NH"_3" soln") = "0.70 mL NH"_3" soln}$

Step 2

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

$\text{Volume of NH"_3 = "0.25 g H"_2"A" × ("1mol H"_2"A")/("754.7 g H"_2"A") × ("2 mol NH"_3)/("1 mol H"_2"A") × ("17.03 g NH"_3)/("1 mol NH"_3) × ("100 g NH"_3" soln")/("25 g NH"_3) × ("1mL NH"_3" soln")/("0.911 g NH"_3" soln") = "0.050 mL NH"_3" soln}$

Step 3

Total volume = 0.70 mL + 0.050 mL = 0.75 mL