a. You titrate 25.0 mL of 0.60 M NH3 with 0.60 M HCl.
Calculate the pH of the solution after adding 5.00, 15.0, 22.0, and 30.0 mL of the acid.
Ka = 5.6×10-10
pH(5.00 mL added) ------------------------------
pH(15.0 mL added)------------------------------
pH(22.0 mL added) ----------------------------
pH(30.0 mL added)----------------------------
b. itration of 27.9 mL of a solution of the weak base aniline, C6H5NH2, requires 28.64 mL of 0.160 M HCl to reach the equivalence point.
C6H5NH2(aq) + H3O+(aq) ⇆ C6H5NH3+(aq) + H2O(ℓ)
What is the pH of the solution at the equivalence point?
Ka = 2.5×10-5
pH--------------------------
c. Consider the titration of 25.0 mL of 0.065 M HCN with 0.065 M NaOH.
What is the pH when 105% of the NaOH required to reach the equivalence point has been added?
pH -----------------------------
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