Question

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

Answer 1

Explanation:

The given reaction equation is as follows.

      $HCN + NaOH \rightarrow NaCN + H_{2}O$

Hence, initial moles of HCN will be as follows.

        Moles = Molarity × Volume

                   = $0.4006 M \times 120 ml$

                   = 48.072 mol

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

             $M_{1}V_{1} = M_{2}V_{2}$

            $0.4006 \times 120 = 0.6812 \times V_{2}$  

              $V_{2}$ = 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 $CN^{-}$ is as follows.

               $\frac{48.072}{190.57}$

               = 0.25 M

Now, the equilibrium equation will be as follows.

             $CN^{-} + H_{2}O \rightleftharpoons HCN + OH^{-}$

Initial:          0.25

Equilbm:  0.25 - x           x             x

Expression to find $K_{a}$ is as follows.

       $K_{a} = \frac{x^{2}}{0.25 - x}$

We know that,  $pK_{a} = -log K_{a}$            

          $K_{a}$ = antilog (-9.21)

                      = $6.16 \times 10^{-10}$

As,     $K_{b} = \frac{K_{w}}{K_{a}}$                

     $\frac{10^{-14}}{6.16 \times 10^{-10}} = \frac{x^{2}}{0.25 - x}$

                 x = $0.2 \times 10^{-2} = [OH^{-}]$

Now, we will calculate the pH as follows.

               pOH = $-log[OH^{-}]$

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.