posted by Ross .
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
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+)