I'm trying to prepare a basic alcohol solution (mixture of ammonium hydroxide & isopropanol) but having difficulty as it is not conventional dilution using water. Any assistance is dearly appreciated.
56.6 wt% NH4OH approximately equals 28 wt% NH3.
We can retrieve densities of ammonia (aq. sol.) from (e.g.) Table 2-32, in: R.H. Perry, D.W. Green, J.O. Maloney (Eds.), "Perry's Chemical Engineers’ Handbook", 7th ed., McGraw-Hill, 1997.
For ammonia (aq. sol.) of label concentration 28 wt% as NH3 (approximate) at 20 ºC; ρ = 0.898 g/cm3. The molecular mass of NH3 is 17.03 g/mol. Hence, the molarity in terms of NH3 would be: 0.28(g NH3 / g aq. sol.)·0.898(g aq. sol. / cm3)·(1000 cm3/dm3)/(17.03 g NH3/mol NH3) = 14.8 M as NH3 (approximate).
The concentration of the obtained ammonia sol. should, in principle, be taken as approximate, namely for quantitative analytical purposes. It is convenient to dilute that solution by some appropriate dilution factor (f) and then titrate the obtained solution with standardized strong mineral acid aq. solution, to accurately determine its molarity. Alternatively, a known excess of standardized strong mineral acid aq. sol. can be added, which excess is then back-titrated with standardized strong base aq. sol., to find the stoichiometric amount of acid previously consumed by the ammonia solution.
The pH of pure ammonia (NH4OH) aq. sol. can be predicted as I have shown elsewhere at this forum:
https://www.researchgate.net/post/How_can_i_calculate_PH_of_an_ammunia_solution_without_using_a_PH_meter
- Badges
- Science method