Hi all, I need to prepare 28% ammonium hydroxide solution from 25% extra pure ammonia. Can anyone tell me how to prepare it.
I would suggest this working with concentrated ammonia may be hazardous, so definately a fume hood in a chemistry lab and perhaps other personal protection. I would suggest you might want to purchase it from a laboratory supply. Ammonia has weight of 17.04 g/mole and ammonium hydroxide has 35.04 g/mole. But I am personally confused about a solution after thinking about this, I think you need someone who can tell you exactly how it should be done and done safely. I am only sending this poor response to make sure you are aware and be cautious.
@Hansen Thanks for ur comments. I bought 28% NH3 from shop. Thanks once again
I think you made a great choice. Glad you are safe, from mixing. Still be careful in your use of it.
I think you made a great choice. Glad you are safe, from mixing. Still be careful in your use of it.
You can increase strength of 25 % ammonia to 28 % If you cool the prepared solution below 10 oC and then slowly adsorb ammonia gas. then check its strength by using British Pharmacopoeia method" Assy of strong ammonia solution"
Ammonia (aq. sol.) 28 wt% (as NH3) can be obtained by either adding water to a 30 wt% (as NH3) sol., or by mixing both 25wt% and 30wt% ammonia solutions.
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 25 wt% as NH3 (approximate) at 20 ºC; ρ = 0.907 g/cm3 (interpolated). The molecular mass of NH3 is 17.03 g/mol. Hence, the molarity in terms of NH3 would be: 0.25(g NH3 / g aq. sol.)·0.907(g aq. sol. / cm3)·(1000 cm3/dm3)/(17.03 g NH3/mol NH3) = 13.3 M as NH3 (approximate).
For ammonia (aq. sol.) of label concentration 30 wt% as NH3 (approximate) at 20 ºC; ρ = 0.892 g/cm3. Hence, the molarity in terms of NH3 would be: 0.30(g NH3 / g aq. sol.)·0.892(g aq. sol. / cm3)·(1000 cm3/dm3)/(17.03 g NH3/mol NH3) = 15.7 M (approximate).
For ammonia (aq. sol.) of label concentration 28 wt% as NH3 (approximate) at 20 ºC; ρ = 0.898 g/cm3. 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.
We can then accurately calculate the the weight (mass) % concentration for the obtained solution (~28 wt%) based on its standardized molarity, by reversing the above exemplified units conversion.
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