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

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

Explanation:

At the equivalence point,

So, first we will calculate the moles of as follows.

                     = 0.0845 mol

Now, volume of present will be calculated as follows.

          Volume =

                        =

                        = 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,

                        = 0.283 M

Chemical equation for this reaction is as follows.

As,              

                     =

                     =

Now,  

                      =

                      =

Now, pH will be calculated as follows.

              pH =

                    =

                    = 5.59

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

it takes on an electronical energy