Question

# a. You titrate 25.0 mL of 0.60 M NH3 with 0.60 M HCl. Calculate the pH...

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 -----------------------------

Answer #1

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