Ammonium Iodide
NH4I + H2O --> NH4OH + H+ + I-HI being a strong acid, remains almost wholly ionised as H+ & I- ionNH4OH being a weak base so you have some more H+ left therefore it is acidic solution. (problem: NH4OH is not an exsisting compound, most of it will react with the strong acid H+ and the remainig will be fully ionized into NH4+)Added:(problem solved: not using NH4OH as it is annon-exsisting compound)NH4I will completely dissolve in ammonium and iodide ions:NH4I --> I- + NH4+of which I- is neutral and NH4+ is quite a weak acid (pKa = 9.25). (The molar standard solution of NH4I will have pH approx. 4.6)
Oxidation numbers of the elements in NH4I are N: -3 H: +1 (4 times) I: -1 Overall: NH4+ : +1 and I- : -1
NH4I, ammonium iodide.
It is a strong base.
yes
Nh4i
Ammonium Iodide
NH4I + H2O --> NH4OH + H+ + I-HI being a strong acid, remains almost wholly ionised as H+ & I- ionNH4OH being a weak base so you have some more H+ left therefore it is acidic solution. (problem: NH4OH is not an exsisting compound, most of it will react with the strong acid H+ and the remainig will be fully ionized into NH4+)Added:(problem solved: not using NH4OH as it is annon-exsisting compound)NH4I will completely dissolve in ammonium and iodide ions:NH4I --> I- + NH4+of which I- is neutral and NH4+ is quite a weak acid (pKa = 9.25). (The molar standard solution of NH4I will have pH approx. 4.6)
Oxidation numbers of the elements in NH4I are N: -3 H: +1 (4 times) I: -1 Overall: NH4+ : +1 and I- : -1
NH4I, ammonium iodide.
It is a strong base.
The chemical formula NH4I is for ammonium iodide.
Ammonium Iodide solution.
It is a strong base.
It is a strong base.
KOH is a strong base.