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For the titration of 20.0 mL of 0.1500 mol/L NH3(aq) with 0.1500 mol/L HI(aq)(the titrant), calculate

a) the pH before any HI(aq) is added
b) the pH at the equivalence point

1 answer

At the beginning of the titration (before any HI is added), you have pure 0.1500 M NH3(aq).
NH3 + HOH ==> NH4^+ + OH^-
Kb = (NH4^+)(OH^-)/(NH3(g)
Set up an ICE chart and calculate OH^-, the pOH and pH from that.

At the equivalence point, what do you have.
NH3 + HCl ==> NH4Cl in water.
So set up the hydrolysis of NH4^+, use Ka and solve for the *H^+) and pH.
NH4^+ + HOH ==> NH3 + H3O^+
Ka = Kw/Kb = (NH3)(H3O^+)/(NH4+)