The ph of aqueous ammonia is 11. Find molarity of solution kb of nh4oh is 10^-5
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PH = 11.50
pOH = 2.50
[OH-] = 0.00316 M
A bottle labeled "ammonium hydroxide" contains very little in the way of ammonium or hydroxide ions and no molecules of NH4OH. In fact, the solution in the bottle is mostly aqueous ammonia.
NH3(aq) + HOH(l) <==> NH4+ + OH- ..... Kb = 1.8x10^-5
?M ........................... .0.00316M...0.00316M
Kb = [NH4+] [OH-] / [NH3]
[NH3] = Kb / ([NH4+] [OH-])
[NH3] = 1.8x10^-5 / (0.00316)²
[NH3] = 1.8 M
this is your answer
pOH = 2.50
[OH-] = 0.00316 M
A bottle labeled "ammonium hydroxide" contains very little in the way of ammonium or hydroxide ions and no molecules of NH4OH. In fact, the solution in the bottle is mostly aqueous ammonia.
NH3(aq) + HOH(l) <==> NH4+ + OH- ..... Kb = 1.8x10^-5
?M ........................... .0.00316M...0.00316M
Kb = [NH4+] [OH-] / [NH3]
[NH3] = Kb / ([NH4+] [OH-])
[NH3] = 1.8x10^-5 / (0.00316)²
[NH3] = 1.8 M
this is your answer
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