Question

A chemist titrates 110.0 mL of a 0.7684 M methylamine (CH3NH2) sotion with 0.4469 M HNO3 solution at 25 °C. Calculate the pH at equivalence. The p K, of methylamine is 3.36 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 HNO3 solution added

Answer 1

Answer: The pH at equivalence point for the given solution is 5.59.

Explanation:

At the equivalence point,

            $n_{HNO_{3}} = n_{CH_{3}NH_{2}}$

So, first we will calculate the moles of $CH_{3}NH_{2}$ as follows.

      $n_{CH_{3}NH_{2}} = 0.764 M \times \frac{110 ml}{1000 ml/L}$      

                     = 0.0845 mol

Now, volume of $HNO_{3}$ present will be calculated as follows.

          Volume = $\frac{\text{no. of moles}}{\text{Molarity}}$

                        = $\frac{0.0845}{0.4469 M}$

                        = 0.1891 L

Therefore, the total volume will be the sum of the given volumes as follows.

                    110 ml + 189.1 ml

                  = 299.13 ml

or,               = 0.2991 L

Now, $[CH_{3}NH_{3}^{+}] = \frac{0.0845 mol}{0.2991 L}$

                        = 0.283 M

Chemical equation for this reaction is as follows.

     $CH_{3}NH_{3}^{+} + H_{2}O \rightleftharpoons CH_{3}NH_{2} + H_{3}O^{+}$

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

                     = $\frac{10^{-14}}{10^{-3.36}}$

                     = $2.29 \times 10^{-11}$

Now,   $[HNO_{3}] = \sqrt{k_{a}[CH_{3}NH_{3}^{+}]}$

                      = $\sqrt{2.29 \times 10^{-11} \times 0.283}$

                      = $2.546 \times 10^{-6}$

Now, pH will be calculated as follows.

              pH = $-log [H_{3}O^{+}]$

                    = $-log (2.546 \times 10^{-6})$

                    = 5.59

Thus, we can conclude that pH at equivalence point for the given solution is 5.59.