The reaction we are analysing:
NH4+(aq) + H2O(l)-->H30+(aq)+NH3(aq)
is mainly the equilibrium between NH3(aq) and acidic NH4+
note that
NH4+ is acidic since it can donate H+ ions as follow:
NH4+ <-> NH3 + H+
therefore
if we add NaOH, a strong base
expect:
NaOH(aq) --> Na+ + OH-
OH- will react wit NH4+ as follows:
NH4+ + OH -- NH3 + OH
therefore, we have 2 important things to consider:
NH4+ is decreasing in concentration
NH3(aq) is increasing in concentration
these two actions will shift strongly toward MORE NH4+ and H2O production, i.e. the left side
since we must counter balance the loss in NH4+ and gain in NH3
NaOH is added to the following reaction: NH4+(aq) + H2O(l)-->H30+(aq)+NH3(aq) Using Le Châtelier's Principle, state in...
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Calculate the concentration of OH- at equilibrium in the following reaction: NH3 (aq) + H2O (l) -> NH4+ (aq) + OH- (aq) (Kb = 1.8 x10^(-5)