I have a problem in calculating the PH.
Imagine that i have a 25% ammonia solution and I want to dilute it by adding water to reach a specific PH ( for example PH=9). do you know how can i calculate the amount of water without using PH meter?
Hi Elnaz,
You can use pH calculator in the following link.
"pH Calculator | Calculate pH of a Solution | Sensorex" https://sensorex.com/ph-calculator/
Please look at the following below links which may help you in your analysis:
https://floridadep.gov/sites/default/files/5-Unionized-Ammonia-SOP_1.pdf
https://naldc.nal.usda.gov/download/56459/PDF
https://www.mt.com/mt_ext_files/Editorial/Generic/1/Guides_to_Electrochemical_Analysis_0x000248ff00025c9a00093c4a_files/guideph.pdf
Thanks
Duried Alwazeer Isam Eldin Hussein Elgailani Thank you so much for helping me.
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).
The pH of not too concentrate pure ammonia (NH4OH) aq. sol. can be predicted as follows:
For NH4OH dissociation: NH4OH ⇌ NH4+ + OH-, Kb = [NH4+]·[OH-]/[NH4OH] = 1.80·10-5 M. Hence: [NH4OH] = [NH4+]·[OH-]/Kb.
Ammonium molar balance writes: CNH4OH = [NH4+] + [NH4OH] + [NH3] ≈ [NH4+] + [NH4OH] =[NH4+]·{1 + [OH-]/Kb}, for enough dil. sol. ([NH3] < CNH4OH/100) and where CNH4OH is the nominal molar concentration of dissolved NH4OH (formality, to be precise). Hence: [NH4+] ≈ CNH4OH/{1 + [OH-]/Kb}.
Charge balance writes: [NH4+] + [H3O+] = [OH-]; or: CNH4OH/{1 + [OH-]/Kb} + [H3O+] = [OH-].
For [H3O+] < [OH-]/100 (pH > 8):
CNH4OH/{1 + [OH-]/Kb} ≈ [OH-]
this solves as:
[OH-] ≈ - ½·Kb + √{CNH4OH·Kb + (1/4)·(Kb)2}
or:
[OH-] ≈ ½√Kb·{-√Kb + √{4CNH4OH + Kb}}
for not too dilute sol., 4CNH4OH > 100Kb:
[OH-] ≈ √Kb·{-½·√Kb + √CNH4OH}
for not too dilute sol., √CNH4OH > 100·½·√Kb; CNH4OH > 104·(1/4)·Kb = 4.5·10-2 M:
[OH-] ≈ √(Kb·CNH4OH)
where pOH = -log10{[OH-]/M}; pH = 14 - pOH
Example: for CNH4OH = 1.00 M; pOH ≈- ½·{log10(1.80) - 5} =- ½·{0.255 - 5} = ½·{4.745 }= 2.37; pH ≈ 14 - 2.37 = 11.63.