A chemist titrates 120.0 mL of a 0.4006 M hydrocyanic acid (HCN) solution with 0.6812 MNaOH solution at 25 °C. Calculate the pH at equivalence. The pKa of hydrocyanic acid is 9.21. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of NaOH solution added

Explanation:

The given reaction equation is as follows.

Hence, initial moles of HCN will be as follows.

        Moles = Molarity × Volume

                   =

                   = 48.072 mol

Now, we will calculate the volume of NaOH as follows.

              = 70.57 ml

At the equivalence point, moles of both HCN and NaOH will be equal. And, total volume will be as follows.

               120 ml + 70.57 ml = 190.57 ml

Initial concentration of is as follows.

               = 0.25 M

Now, the equilibrium equation will be as follows.

Initial:          0.25

Equilbm:  0.25 - x           x             x

Expression to find is as follows.

We know that,              

          = antilog (-9.21)

                      =

As,                    

                 x =

Now, we will calculate the pH as follows.

               pOH =

as pOH = 2.698

And,      pH + pOH = 14

               pH = 14 - 2.698

                     = 11.30

Thus, we can conclude that pH at equivalence is 11.30.        

h